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CLP-1 Differential Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Appendix D Hints for Exercises

1 Limits
1.1 Drawing Tangents and a First Limit
1.1.2 Exercises

1.1.2.2.

Hint.
The tangent line to a curve at point \(P\) passes through \(P\text{.}\)

1.1.2.3.

Hint.
Try drawing tangent lines to the following curves, at the given points \(P\text{:}\)

1.2 Another Limit and Computing Velocity
1.2.2 Exercises

1.2.2.3.

Hint.
Where did you start, and where did you end?

1.2.2.4.

Hint.
Is the object falling faster and faster, slower and slower, or at a constant rate?

1.2.2.5.

Hint.
Slope is change in vertical component over change in horizontal component.

1.2.2.6.

Hint.
Sign of velocity gives direction of motion: the velocity is positive at time \(t\) if \(s(t)\) is increasing at time \(t\text{.}\)

1.2.2.7.

Hint.
Velocity is distance over time.

1.2.2.8.

Hint.
Use that \(\frac{\sqrt{a}-b}{c} = \frac{\sqrt{a}-b}{c}\cdot \left(\frac{\sqrt{a}+b}{\sqrt{a}+b}\right) = \frac{a-b^2}{c(\sqrt{a}+b)}\text{.}\)

1.3 The Limit of a Function
1.3.2 Exercises

1.3.2.2.

Hint.
Consider the difference between a limit and a one-sided limit.

1.3.2.3.

Hint.
Pay careful attention to which limits are one-sided and which are not.

1.3.2.5.

Hint.
The function doesn’t have to be continuous.

1.3.2.6.

Hint.
See Question 1.3.2.5

1.3.2.7.

Hint.
See Question 1.3.2.5

1.3.2.8.

Hint.
What is the relationship between the limit and the two one-sided limits?

1.3.2.9.

Hint.
What is the relationship between the limit and the two one-sided limits?

1.3.2.14.

Hint.
What are the one-sided limits?

1.3.2.16.

Hint.
Think about what it means that \(x\) does not appear in the function \(f(x)=\dfrac{1}{10}\text{.}\)

1.3.2.17.

Hint.
We only care about what happens really, really close to \(x=3\text{.}\)

1.4 Calculating Limits with Limit Laws
1.4.2 Exercises

1.4.2.2.

Hint.
Try to make two functions with factors that will cancel.

1.4.2.3.

Hint.
Try to make \(g(x)\) cancel out.

1.4.2.6.

Hint.
Find the limit of the numerator and denominator separately.

1.4.2.7.

Hint.
Break it up into smaller pieces, evaluate the limits of the pieces.

1.4.2.8.

Hint.
First find the limit of the “inside” function, \(\dfrac{4x-2}{x+2}\text{.}\)

1.4.2.9. (✳).

Hint.
Is \(\cos(-3)\) zero?

1.4.2.10. (✳).

Hint.
Expand, then simplify.

1.4.2.14. (✳).

Hint.
Try the simplest method first.

1.4.2.15. (✳).

Hint.
Factor the denominator.

1.4.2.16. (✳).

Hint.
Factor the numerator and the denominator.

1.4.2.17. (✳).

Hint.
Factor the numerator.

1.4.2.18. (✳).

Hint.
Simplify first by factoring the numerator.

1.4.2.19.

Hint.
The function is a polynomial.

1.4.2.20. (✳).

Hint.
Multiply both the numerator and the denominator by the conjugate of the numerator, \(\sqrt{x^2+8}+3\text{.}\)

1.4.2.21. (✳).

Hint.
Multiply both the numerator and the denominator by the conjugate of the numerator, \(\sqrt{x+7}+\sqrt{11-x}\text{.}\)

1.4.2.22. (✳).

Hint.
Multiply both the numerator and the denominator by the conjugate of the numerator, \(\sqrt{x+2}+\sqrt{4-x}\text{.}\)

1.4.2.23. (✳).

Hint.
Multiply both the numerator and the denominator by the conjugate of the numerator, \(\sqrt{x-2}+\sqrt{4-x}\text{.}\)

1.4.2.24. (✳).

Hint.
Multiply both the numerator and the denominator by the conjugate of the denominator, \(2+\sqrt{5-t}\text{.}\)

1.4.2.25.

Hint.
Consider the factors \(x^2\) and \(\cos\left(\frac{3}{x}\right)\) separately. Review the squeeze theorem.

1.4.2.26.

Hint.
Look for a reason to ignore the trig. Review the squeeze theorem.

1.4.2.27. (✳).

Hint.
As in the previous questions, we want to use the Squeeze Theorem. If \(x \lt 0\text{,}\) then \(-x\) is positive, so \(x \lt -x\text{.}\) Use this fact when you bound your expressions.

1.4.2.28.

Hint.
Factor the numerator.

1.4.2.29.

Hint.
Factor the denominator; pay attention to signs.

1.4.2.30.

Hint.
First find the limit of the “inside” function.

1.4.2.31.

Hint.
Factor; pay attention to signs.

1.4.2.32.

Hint.
Look for perfect squares

1.4.2.33.

Hint.
Think about what effect changing \(d\) has on the function \(x^5-32x+15\text{.}\)

1.4.2.34.

Hint.
There’s an easy way.

1.4.2.35. (✳).

Hint.
What can you do to safely ignore the sine function?

1.4.2.36. (✳).

Hint.
Factor

1.4.2.37.

Hint.
If you’re looking at the hints for this one, it’s probably easier than you think.

1.4.2.38.

Hint.
You’ll want to simplify this, since \(t=\frac{1}{2}\) is not in the domain of the function. One way to start your simplification is to add the fractions in the numerator by finding a common denominator.

1.4.2.39.

Hint.
If you’re not sure how \(\dfrac{|x|}{x}\) behaves, try plugging in a few values of \(x\text{,}\) like \(x=\pm 1\) and \(x=\pm 2\text{.}\)

1.4.2.40.

Hint.
Look to Question 1.4.2.39 to see how a function of the form \(\dfrac{|X|}{X}\) behaves.

1.4.2.41.

Hint.
Is anything weird happening to this function at \(x=0\text{?}\)

1.4.2.42.

Hint.
Use the limit laws.

1.4.2.43. (✳).

Hint.
The denominator goes to zero; what must the numerator go to?

1.4.2.45.

Hint.
Try plotting points. If you can’t divide by \(f(x)\text{,}\) take a limit.

1.4.2.46.

Hint.
There is a close relationship between \(f\) and \(g\text{.}\) Fill in the following table:
\(x\) \(f(x)\) \(g(x)\) \(\dfrac{f(x)}{g(x)}\)
\(-3\)      
\(-2\)      
\(-1\)      
\(-0\)      
\(1\)      
\(2\)      
\(3\)      

1.4.2.47.

Hint.
Velocity of white ball when \(t=1\) is \(\displaystyle\lim_{h \rightarrow 0}\dfrac{s(1+h)-s(1)}{h}\text{.}\)

1.4.2.49.

Hint.
When you’re evaluating \(\displaystyle\lim_{x \rightarrow 0^-} f(x)\text{,}\) you’re only considering values of \(x\) that are less than \(0\text{.}\)

1.4.2.50.

Hint.
When you’re considering \(\displaystyle\lim_{x \rightarrow -4^-} f(x)\text{,}\) you’re only considering values of \(x\) that are less than \(-4\text{.}\) When you’re considering \(\displaystyle\lim_{x \rightarrow -4^+} f(x)\text{,}\) think about the domain of the rational function in the top line.

1.5 Limits at Infinity
1.5.2 Exercises

1.5.2.1.

Hint.
It might not look like a traditional polynomial.

1.5.2.2.

Hint.
The degree of the polynomial matters.

1.5.2.3.

Hint.
What does a negative exponent do?

1.5.2.4.

Hint.
You can think about the behaviour of this function by remembering how you first learned to describe exponentiation.

1.5.2.5.

Hint.
The exponent will be a negative number.

1.5.2.6.

Hint.
What single number is the function approaching?

1.5.2.7.

Hint.
The highest-order term dominates when \(x\) is large.

1.5.2.8.

Hint.
Factor the highest power of \(x\) out of both the numerator and the denominator. You can factor through square roots (carefully).

1.5.2.9. (✳).

Hint.
Multiply and divide by the conjugate, \(\sqrt{x^2+5x}+\sqrt{x^2-x}\text{.}\)

1.5.2.10. (✳).

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator.
Remember that \(\sqrt{\ }\) is defined to be the positive square root. Consequently, if \(x \lt 0\text{,}\) then \(\sqrt{x^2}\text{,}\) which is positive, is not the same as \(x\text{,}\) which is negative.

1.5.2.11. (✳).

Hint.
Factor out the highest power of the denominator.

1.5.2.12. (✳).

Hint.
The conjugate of \((\sqrt{x^2+x}-x)\) is \((\sqrt{x^2+x}+x)\text{.}\)
Multiply by \(1=\dfrac{\sqrt{x^2+x}+x} {\sqrt{x^2+x}+x}\) to coax your function into a fraction.

1.5.2.13. (✳).

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator.

1.5.2.14. (✳).

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator.

1.5.2.15. (✳).

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator.

1.5.2.16.

Hint.
Divide both the numerator and the denominator by \(x\) (which is the largest power of \(x\) in the denominator). In the numerator, move the resulting factor of \(1/x\) inside the two roots. Be careful about the signs when you do so. Even and odd roots behave differently-- see Question 1.5.2.10.

1.5.2.17. (✳).

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator.

1.5.2.18.

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator. It is not always true that \(\sqrt{x^2}=x\text{.}\)

1.5.2.19.

Hint.
Simplify.

1.5.2.20. (✳).

Hint.
What is a simpler version of \(|x|\) when you know \(x \lt 0\text{?}\)

1.5.2.22. (✳).

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator. When is \(\sqrt{x}=x\text{,}\) and when is \(\sqrt{x}=-x\text{?}\)

1.5.2.23.

Hint.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator. Pay careful attention to signs.

1.5.2.24. (✳).

Hint.
Multiply and divide the expression by its conjugate, \(\big(\sqrt{n^2+5n}+n\big)\text{.}\)

1.5.2.25.

Hint.
Consider what happens to the function as \(a\) becomes very, very small. You shouldn’t need to do much calculation.

1.5.2.26.

Hint.
Since \(x=3\) is not in the domain of the function, we need to be a little creative. Try simplifying the function.

1.5.2.27.

Hint.
This is a bit of a trick question. Consider what happens to a rational function as \(x\rightarrow\pm \infty\) in each of these three cases:
  • the degree of the numerator is smaller than the degree of the denominator,
  • the degree of the numerator is the same as the degree of the denominator, and
  • the degree of the numerator is larger than the degree of the denominator.

1.5.2.28.

Hint.
We tend to conflate “infinity” with “some really large number.”

1.6 Continuity
1.6.4 Exercises

1.6.4.1.

Hint.
Try a repeating pattern.

1.6.4.2.

Hint.
\(f\) is my height.

1.6.4.3.

Hint.
The intermediate value theorem only works for a certain kind of function.

1.6.4.7.

Hint.
Compare what is given to you to the definition of continuity.

1.6.4.8.

Hint.
Compare what is given to you to the definition of continuity.

1.6.4.9.

Hint.
What if the function is discontinuous?

1.6.4.10.

Hint.
What is \(h(0)\text{?}\)

1.6.4.11.

Hint.
Use the definition of continuity.

1.6.4.12.

Hint.
If this is your password, you might want to change it.

1.6.4.13. (✳).

Hint.
Find the domain: when is the denominator zero?

1.6.4.14. (✳).

Hint.
When is the denominator zero? When is the argument of the square root negative?

1.6.4.15. (✳).

Hint.
When is the denominator zero? When is the argument of the square root negative?

1.6.4.16. (✳).

Hint.
There are infinitely many points where it is not continuous.

1.6.4.17. (✳).

Hint.
\(x=c\) is the important point.

1.6.4.18. (✳).

Hint.
The important place is \(x=0\text{.}\)

1.6.4.19. (✳).

Hint.
The important point is \(x=c\text{.}\)

1.6.4.20. (✳).

Hint.
The important point is \(x=2c\text{.}\)

1.6.4.21.

Hint.
Consider the function \(f(x)=\sin x - x +1\text{.}\)

1.6.4.22. (✳).

Hint.
Consider the function \(f(x)=3^x-x^2\text{,}\) and how it relates to the problem and the IVT.

1.6.4.23. (✳).

Hint.
Consider the function \(2\tan x - x - 1\) and its roots.

1.6.4.24. (✳).

Hint.
Consider the function \(f(x) = \sqrt{\cos(\pi x)} - \sin(2\pi x) -1/2\text{,}\) and be careful about where it is continuous.

1.6.4.25. (✳).

Hint.
Consider the function \(f(x)=1/\cos^2(\pi x)-x-\frac{3}{2}\text{,}\) paying attention to where it is continuous.

1.6.4.26.

Hint.
We want \(f(x)\) to be 0; 0 is between a positive number and a negative number. Try evaluating \(f(x)\) for some integer values of \(x\text{.}\)

1.6.4.27.

Hint.
\(\sqrt[3]{7}\) is the value where \(x^3=7\text{.}\)

1.6.4.28.

Hint.
You need to consider separately the cases where \(f(a) \lt g(a)\) and \(f(a)=g(a)\text{.}\) Let \(h(x)=f(x)-g(x)\text{.}\) What is \(h(c)\text{?}\)

2 Derivatives
2.1 Revisiting Tangent Lines
2.1.2 Exercises

2.1.2.3.

Hint.
Your calculations for slope of the secant lines will all have the same denominators; to save yourself some time, you can focus on the numerators.

2.1.2.4.

Hint.
You can do this by calculating several secant lines. You can also do this by getting out a ruler and trying to draw the tangent line very carefully.

2.1.2.5.

Hint.
There are many possible values for \(Q\) and \(R\text{.}\)

2.1.2.6.

Hint.
A line with slope \(0\) is horizontal.

2.2 Definition of the Derivative
2.2.4 Exercises

2.2.4.1.

Hint.
What are the properties of \(f'\) when \(f\) is a line?

2.2.4.2.

Hint.
Be very careful not to confuse \(f\) and \(f'\text{.}\)

2.2.4.3.

Hint.
Be very careful not to confuse \(f\) and \(f'\text{.}\)

2.2.4.5.

Hint.
The slope has to look “the same” from the left and the right.

2.2.4.6.

Hint.
Use the definition of the derivative, and what you know about limits.

2.2.4.7.

Hint.
Consider continuity.

2.2.4.8.

Hint.
Look at the definition of the derivative. Your answer will be a fraction.

2.2.4.9.

Hint.
You need a point (given), and a slope (derivative).

2.2.4.10.

Hint.
You’ll need to add some fractions.

2.2.4.11. (✳).

Hint.
You don’t have to take the limit from the left and right separately--things will cancel nicely.

2.2.4.12. (✳).

Hint.
You might have to add fractions.

2.2.4.14.

Hint.
Your limit should be easy.

2.2.4.15. (✳).

Hint.
Add fractions.

2.2.4.16. (✳).

Hint.
For \(f\) to be differentiable at \(x=2\text{,}\) two things must be true: it must be continuous at \(x=2\text{,}\) and the derivative from the right must equal the derivative from the left.

2.2.4.17. (✳).

Hint.
After you plug in \(f(x)\) to the definition of a derivative, you’ll want to multiply and divide by the conjugate \(\sqrt{1+x+h}+\sqrt{1+x}\text{.}\)

2.2.4.18.

Hint.
From Section 1.2, compare the definition of velocity to the definition of a derivative. When you’re finding the derivative, you’ll need to cancel a lot on the numerator, which you can do by expanding the polynomials.

2.2.4.19. (✳).

Hint.
You’ll need to look at limits from the left and right. The fact that \(f(0)=0\) is useful for your computation. Recall that if \(x \lt 0\) then \(\sqrt{x^2}=|x|=-x\text{.}\)

2.2.4.20. (✳).

Hint.
You’ll need to look at limits from the left and right. The fact that \(f(0)=0\) is useful for your computation.

2.2.4.21. (✳).

Hint.
You’ll need to look at limits from the left and right. The fact that \(f(0)=0\) is useful for your computation.

2.2.4.22. (✳).

Hint.
You’ll need to look at limits from the left and right. The fact that \(f(1)=0\) is useful for your computation.

2.2.4.23.

Hint.
There’s lots of room between \(0\) and \(\frac{1}{8}\text{;}\) see what you can do with it.

2.2.4.24.

Hint.
Set up your usual limit, then split it into two pieces

2.2.4.25.

Hint.
You don’t need the definition of the derivative for a line.

2.2.4.26. (✳).

Hint.
A generic point on the curve has coordinates \((\alpha, \alpha^2)\text{.}\) In terms of \(\alpha\text{,}\) what is the equation of the tangent line to the curve at the point \((\alpha, \alpha^2)\text{?}\) What does it mean for \((1,-3)\) to be on that line?

2.2.4.27. (✳).

Hint.
Remember for a constant \(n\text{,}\)
\begin{equation*} \ds\lim_{h \to 0} h^{n} = \left\{\begin{array}{ll} 0&n \gt 0\\ 1&n=0\\ DNE&n \lt 0 \end{array}\right. \end{equation*}

2.3 Interpretations of the Derivative
2.3.3 Exercises

2.3.3.1.

Hint.
Think about units.

2.3.3.8.

Hint.
There are 360 degrees in one rotation.

2.3.3.9.

Hint.
\(P'(t)\) was discussed in Question 2.3.3.7.

2.4 Arithmetic of Derivatives - a Differentiation Toolbox
2.4.2 Exercises

2.4.2.1.

Hint.
Look at the Sum rule.

2.4.2.2.

Hint.
Try an example, like \(f(x)=g(x)=x\text{.}\)

2.4.2.3.

Hint.
Simplify.

2.4.2.4.

Hint.
\(g(x)=f(x)+f(x)+f(x)\)

2.4.2.5.

Hint.
Use linearity and the known derivatives of \(x^2\) and \(x^{1/2}\text{.}\)

2.4.2.6.

Hint.
You have already seen \(\diff{}{x}\{\sqrt{x}\}\text{.}\)

2.4.2.7. (✳).

Hint.
The equation of a line can be determined using a point, and the slope. The derivative of \(x^3\) can be found by writing \(x^3=(x)(x^2)\text{.}\)

2.4.2.8. (✳).

Hint.
Be careful to distinguish between speed and velocity. The speed is the absolute value of the velocity.

2.4.2.10.

Hint.
How do you take care of that power?

2.4.2.11.

Hint.
You know how to take the derivative of a reciprocal; this might be faster than using the quotient rule.

2.4.2.12.

Hint.
Population growth is rate of change of population.

2.4.2.14. (✳).

Hint.
Interpret it as a derivative that you know how to compute.

2.4.2.15.

Hint.
The answer is not 10 square metres per second.

2.4.2.16.

Hint.
You don’t need to know \(g(0)\) or \(g'(0)\text{.}\)

2.6 Using the Arithmetic of Derivatives – Examples
2.6.2 Exercises

2.6.2.1.

Hint.
Check signs.

2.6.2.2.

Hint.
Read Lemma 2.6.9 carefully.

2.6.2.3.

Hint.
First, factor an \(x\) out of the derivative. What’s left over looks like a quadratic equation, if you take \(x^2\) to be your variable, instead of \(x\text{.}\)

2.6.2.4.

Hint.
\(\frac{1}{t}=t^{-1}\)

2.6.2.5.

Hint.
First simplify. Don’t be confused by the role reversal of \(x\) and \(y\text{:}\) \(x\) is just the name of the function \(\big(2y+\tfrac{1}{y}\big)\cdot y^3\text{,}\) which is a function of the variable \(y\text{.}\) You are to differentiate with respect to \(y\text{.}\)

2.6.2.6.

Hint.
\(\sqrt{x}=x^{1/2}\)

2.6.2.8.

Hint.
You don’t need to multiply through.

2.6.2.9.

Hint.
You can use the quotient rule.

2.6.2.13. (✳).

Hint.
There are two pieces of the given function that could cause problems.

2.6.2.14.

Hint.
\(\sqrt[3]{x}=x^{1/3}\)

2.6.2.15.

Hint.
Simplify first.

2.6.2.17. (✳).

Hint.
Let \(m\) be the slope of such a tangent line, and let \(P_1\) and \(P_2\) be the points where the tangent line is tangent to the two curves, respectively. There are three equations \(m\) fulfils: it has the same slope as the curves at the given points, and it is the slope of the line passing through the points \(P_1\) and \(P_2\text{.}\)

2.6.2.18.

Hint.
A line has equation \(y=mx+b\text{,}\) for some constants \(m\) and \(b\text{.}\) What has to be true for \(y=mb+x\) to be tangent to the first curve at the point \(x=\alpha\text{,}\) and to the second at the point \(x=\beta\text{?}\)

2.6.2.19. (✳).

Hint.
Compare this to one of the forms given in the text for the definition of the derivative.

2.7 Derivatives of Exponential Functions
2.7.3 Exercises

2.7.3.1.

Hint.
Two of the functions are the same.

2.7.3.3.

Hint.
When can you use the power rule?

2.7.3.4.

Hint.
What is the shape of the curve \(e^{ax}\text{,}\) when \(a\) is a positive consant?

2.7.3.5.

Hint.
Quotient rule

2.7.3.6.

Hint.
\(e^{2x}=\left(e^x\right)^2\)

2.7.3.7.

Hint.
\(e^{a+x}=e^ae^x\)

2.7.3.8.

Hint.
Figure out where the derivative is positive.

2.7.3.9.

Hint.
\(e^{-x}=\frac{1}{e^x}\)

2.7.3.10.

Hint.
Product rule will work nicely here. Alternately, review the result of Question 2.7.3.6.

2.7.3.11.

Hint.
To find the sign of a product, compare the signs of each factor. The function \(e^t\) is always positive.

2.7.3.12.

Hint.
After you differentiate, factor out \(e^x\text{.}\)

2.7.3.13.

Hint.
Simplify.

2.7.3.14. (✳).

Hint.
In order to be differentiable, a function should be continuous. To determine the differentiability of the function at \(x=1\text{,}\) use the definition of the derivative.

2.8 Derivatives of Trigonometric Functions
2.8.8 Exercises

2.8.8.1.

Hint.
A horizontal tangent line is where the graph appears to “level off.”

2.8.8.2.

Hint.
You are going to mark there points on the sine graph where the graph is the steepest, going up.

2.8.8.3.

Hint.
You need to memorize the derivatives of sine, cosine, and tangent.

2.8.8.4.

Hint.
There are infinitely many values. You need to describe them all.

2.8.8.5.

Hint.
Simplify first.

2.8.8.6.

Hint.
The identity won’t help you.

2.8.8.8.

Hint.
Quotient rule

2.8.8.11.

Hint.
Use an identity.

2.8.8.12.

Hint.
How can you move the negative signs to a location that you can more easily deal with?

2.8.8.13.

Hint.
Apply the quotient rule.

2.8.8.14. (✳).

Hint.
The only spot to worry about is when \(x=0\text{.}\) For \(f(x)\) to be differentiable, it must be continuous, so first find the value of \(b\) that makes \(f\) continuous at \(x=0\text{.}\) Then, find the value of \(a\) that makes the derivatives from the left and right of \(x=0\) equal to each other.

2.8.8.16. (✳).

Hint.
Compare this to one of the forms given in the text for the definition of the derivative.

2.8.8.17. (✳).

Hint.
Compare this to one of the forms given in the text for the definition of the derivative.

2.8.8.18. (✳).

Hint.
Compare this to one of the forms given in the text for the definition of the derivative.

2.8.8.19.

Hint.
\(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\)

2.8.8.20. (✳).

Hint.
In order for a derivative to exist, the function must be continuous, and the derivative from the left must equal the derivative from the right.

2.8.8.21. (✳).

Hint.
There are infinitely many places where it does not exist.

2.8.8.27.

Hint.
You can set up the derivative using the limit definition: \(f'(0)=\ds\lim_{h \to 0}\dfrac{f(h)-f(0)}{h}\text{.}\) If the limit exists, it gives you \(f'(0)\text{;}\) if the limit does not exist, you conclude \(f'(0)\) does not exist.
To evaluate the limit, recall that when we differentiated sine, we learned that for \(h\) near 0,
\begin{equation*} \cos h \leq \frac{\sin h}{h}\leq 1 \end{equation*}

2.8.8.28. (✳).

Hint.
Recall \(|x|=\left\{\begin{array}{rl} x&x\ge 0\\ -x&x \lt 0 \end{array}\right.\text{.}\) To determine whether \(h(x)\) is differentiable at \(x=0\text{,}\) use the definition of the derivative.

2.8.8.29. (✳).

Hint.
To decide whether the function is differentiable, use the definition of the derivative.

2.8.8.30. (✳).

Hint.
In this chapter, we learned \(\ds\lim_{x \to 0}\dfrac{\sin x}{x}=1\text{.}\) If you divide the numerator and denominator by \(x^5\text{,}\) you can make use of this knowledge.

2.9 One More Tool – the Chain Rule
2.9.4 Exercises

2.9.4.1.

Hint.
For parts 2.9.4.1.a and 2.9.4.1.b, remember the definition of a derivative:
\begin{equation*} \ds\diff{K}{U}=\ds\lim_{h \rightarrow 0}\dfrac{K(U+h)-K(U)}{h}. \end{equation*}
When \(h\) is positive, \(U+h\) is an increased urchin population; what is the sign of \(K(U+h)-K(U)\text{?}\)
For part 2.9.4.1.c, use the chain rule!

2.9.4.2.

Hint.
Remember that Leibniz notation suggests fractional cancellation.

2.9.4.3.

Hint.
If \(g(x)=\cos x\) and \(h(x)=5x+3\text{,}\) then \(f(x)=g(h(x))\text{.}\) So we apply the chain rule, with “outside” function \(\cos x\) and “inside” function \(5x+3\text{.}\)

2.9.4.4.

Hint.
You can expand this into a polynomial, but it’s easier to use the chain rule. If \(g(x)=x^5\text{,}\) and \(h(x)=x^2+2\text{,}\) then \(f(x)=g(h(x))\text{.}\)

2.9.4.5.

Hint.
You can expand this into a polynomial, but it’s easier to use the chain rule. If \(g(k)=k^{17}\text{,}\) and \(h(k)=4k^4+2k^2+1\text{,}\) then \(T(k)=g(h(k))\text{.}\)

2.9.4.6.

Hint.
If we define \(g(x)=\sqrt{x}\) and \(h(x)=\dfrac{x^2+1}{x^2-1}\text{,}\) then \(f(x)=g(h(x))\text{.}\) To differentiate the square root function: \(\ds\diff{}{x}\{\sqrt{x}\}=\ds\diff{}{x}\left\{x^{1/2}\right\}=\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}\text{.}\)

2.9.4.7.

Hint.
You’ll need to use the chain rule twice.

2.9.4.8. (✳).

Hint.
Use the chain rule.

2.9.4.9. (✳).

Hint.
Use the chain rule.

2.9.4.10. (✳).

Hint.
Use the chain rule.

2.9.4.11. (✳).

Hint.
Use the chain rule.

2.9.4.12. (✳).

Hint.
Recall \(\dfrac{1}{x^2}=x^{-2}\) and \(\sqrt{x^2-1}=(x^2-1)^{1/2}\text{.}\)

2.9.4.14.

Hint.
If we let \(g(x)=\sec x\) and \({h(x)}=e^{2x+7}\text{,}\) then \(f(x)=g(h(x))\text{,}\) so by the chain rule, \(f'(x)=g'(h(x))\cdot h'(x)\text{.}\) However, in order to evaluate \(h'(x)\text{,}\) we’ll need to use the chain rule again.

2.9.4.15.

Hint.
What trig identity can you use to simplify the first factor in the equation?

2.9.4.16.

Hint.
Velocity is the derivative of position with respect to time. In this case, the velocity of the particle is given by \(s'(t)\text{.}\)

2.9.4.17.

Hint.
The slope of the tangent line is the derivative. You’ll need to use the chain rule twice.

2.9.4.18. (✳).

Hint.
Start with the product rule, then use the chain rule to differentiate \(e^{4x}\text{.}\)

2.9.4.19. (✳).

Hint.
Start with the quotient rule; you’ll need the chain rule only to differentiate \(e^{3x}\text{.}\)

2.9.4.20. (✳).

Hint.
More than one chain rule needed here.

2.9.4.21. (✳).

Hint.
More than one chain rule application is needed here.

2.9.4.22. (✳).

Hint.
More than one chain rule application is needed here.

2.9.4.23. (✳).

Hint.
More than one chain rule application is needed here.

2.9.4.24. (✳).

Hint.
What rule do you need, besides chain? Also, remember that \(\cos^2x = [\cos x]^2\text{.}\)

2.9.4.27. (✳).

Hint.
The product of two functions is zero exactly when at least one of the functions is zero.

2.9.4.28.

Hint.
If \(t \ge 1\text{,}\) then \(0 \lt \frac{1}{t} \leq 1\text{.}\)

2.9.4.29.

Hint.
The notation \(\cos^3(5x-7)\) means \(\left[\cos(5x-7)\right]^3\text{.}\) So, if \(g(x)=x^3\) and \(h(x)=\cos(5x-7)\text{,}\) then \(g(h(x))=\left[\cos(5x+7)\right]^3=\cos^3(5x+7)\text{.}\)

2.9.4.30. (✳).

Hint.
In Example 2.6.6, we generalized the product rule to three factors:
\begin{align*} \diff{}{x}\{f(x)g(x)h(x)\} \amp=f'(x)g(x)h(x)+f(x)g'(x)h(x)\\ \amp\hskip1in+f(x)g(x)h'(x) \end{align*}
This isn’t strictly necessary, but it will simplify your computations.

2.9.4.31.

Hint.
At time \(t\text{,}\) the particle is at the point \(\big(x(t),y(t)\big)\text{,}\) with \(x(t)=\cos t\) and \(y(t)=\sin t\text{.}\) Over time, the particle traces out a curve; let’s call that curve \(y=f(x)\text{.}\) Then \(y(t) = f\big(x(t)\big)\text{,}\) so the slope of the curve at the point \(\big(x(t),y(t)\big)\) is \(f'\big(x(t)\big)\text{.}\) You are to determine the values of \(t\) for which \(f'\big(x(t)\big)=-1\text{.}\)

2.9.4.32. (✳).

Hint.
Set \(f(x) = e^{x+x^2}\) and \(g(x)=1+x\text{.}\) Compare \(f(0)\) and \(g(0)\text{,}\) and compare \(f'(x)\) and \(g'(x)\text{.}\)

2.9.4.33.

Hint.
If \(\sin 2x\) and \(2\sin x \cos x\) are the same, then they also have the same derivatives.

2.9.4.34.

Hint.
This is a long, nasty problem, but it doesn’t use anything you haven’t seen before. Be methodical, and break the question into as many parts as you have to. At the end, be proud of yourself for your problem-solving abilities and tenaciousness!

2.9.4.35.

Hint.
To sketch the curve, you can start by plotting points. Alternately, consider \(x^2+y\text{.}\)

2.10 The Natural Logarithm
2.10.3 Exercises

2.10.3.1.

Hint.
Each speaker produces 3dB of noise, so if \(P\) is the power of one speaker, \(3=V(P)=10\log_{10}\left(\frac{P}{S}\right)\text{.}\) Use this to find \(V(10P)\) and \(V(100P)\text{.}\)

2.10.3.2.

Hint.
The question asks you when \(A(t)=2000\text{.}\) So, solve \(2000=1000e^{t/20}\) for \(t\text{.}\)

2.10.3.3.

Hint.
What happens when \(\cos x\) is a negative number?

2.10.3.4.

Hint.
There are two easy ways: use the chain rule, or simplify first.

2.10.3.5.

Hint.
There are two easy ways: use the chain rule, or simplify first.

2.10.3.6.

Hint.
Don’t be fooled by a common mistake: \(\log(x^2+x)\) is not the same as \(\log(x^2)+\log x\text{.}\)

2.10.3.7.

Hint.
Use the base-change formula to convert this to natural logarithm (base \(e\)).

2.10.3.9.

Hint.
Use the chain rule.

2.10.3.10.

Hint.
Use the chain rule twice.

2.10.3.11. (✳).

Hint.
You’ll need to use the chain rule twice.

2.10.3.12. (✳).

Hint.
Use the chain rule.

2.10.3.13. (✳).

Hint.
Use the chain rule to differentiate.

2.10.3.14. (✳).

Hint.
You can differentiate this by using the chain rule several times.

2.10.3.15. (✳).

Hint.
Using logarithm rules before you differentiate will make this easier.

2.10.3.16.

Hint.
Using logarithm rules before you differentiate will make this easier.

2.10.3.17.

Hint.
First, differentiate using the chain rule and any other necessary rules. Then, plug in \(x=2\text{.}\)

2.10.3.18. (✳).

Hint.
In the text, you are given the derivative \(\ds\diff{}{x} a^x\text{,}\) where \(a\) is a constant.

2.10.3.19.

Hint.
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)

2.10.3.20. (✳).

Hint.
Use Question 2.10.3.19 and the base-change formula, \(\log_b(a)=\dfrac{\log a}{\log b}\text{.}\)

2.10.3.21.

Hint.
To make this easier, use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.

2.10.3.22.

Hint.
To make this easier, use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.

2.10.3.23.

Hint.
It’s not going to come out nicely, but there’s a better way than blindly applying quotient and product rules, or expanding giant polynomials.

2.10.3.24. (✳).

Hint.
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.

2.10.3.25. (✳).

Hint.
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.

2.10.3.26. (✳).

Hint.
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.

2.10.3.27. (✳).

Hint.
You’ll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)

2.10.3.28. (✳).

Hint.
You’ll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)

2.10.3.29. (✳).

Hint.
You’ll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)

2.10.3.30.

Hint.
Evaluate \(\ds\diff{}{x}\left\{\log\left(\left[f(x)\right]^{g(x)}\right)\right\}\text{.}\)

2.10.3.31.

Hint.
Differentiate \(y=\log(f(x))\text{.}\) When is the derivative equal to zero?

2.11 Implicit Differentiation
2.11.2 Exercises

2.11.2.1.

Hint.
Where did the \(y'\) come from?

2.11.2.2.

Hint.
The three points to look at are \((0,-4)\text{,}\) \((0,0)\text{,}\) and \((0,4)\text{.}\) What does the slope of the tangent line look like there?

2.11.2.3.

Hint.
A function must pass the vertical line test: one input cannot result in two different outputs.

2.11.2.4. (✳).

Hint.
Remember that \(y\) is a function of \(x\text{.}\) Use implicit differentiation, then collect all the terms containing \(\ds\diff{y}{x}\) on one side of the equation to solve for \(\ds\diff{y}{x}\text{.}\)

2.11.2.5. (✳).

Hint.
Differentiate implicitly, then solve for \(y'\text{.}\)

2.11.2.6. (✳).

Hint.
Remember that \(y\) is a function of \(x\text{.}\) You can determine explicitly the values of \(x\) for which \(y(x)=1\text{.}\)

2.11.2.8. (✳).

Hint.
Plug in \(y=0\) at a strategic point in your work to simplify your computation.

2.11.2.10. (✳).

Hint.
Plug in \(y = 0\) at a strategic point in your work to simplify your computation.

2.11.2.11.

Hint.
If the tangent line has slope \(y'\text{,}\) and it is parallel to \(y=x\text{,}\) then \(y'=1\text{.}\)

2.11.2.12. (✳).

Hint.
You don’t need to solve for \(y'\) in general: only at a single point.

2.11.2.13. (✳).

Hint.
After you differentiate implicitly, get all the terms containing \(y'\) onto one side so you can solve for \(y'\text{.}\)

2.11.2.14. (✳).

Hint.
You don’t need to solve for \(\diff{y}{x}\) for all values of \(x\)--only when \(y=0\text{.}\)

2.12 Inverse Trigonometric Functions
2.12.2 Exercises

2.12.2.1.

Hint.
Remember that only certain numbers can come out of sine and cosine, but any numbers can go in.

2.12.2.2.

Hint.
What is the range of the arccosine function?

2.12.2.3.

Hint.
A one-to-one function passes the horizontal line test. To graph the inverse of a function, reflect it across the line \(y=x\text{.}\)

2.12.2.4.

Hint.
Your answer will depend on \(a\text{.}\) The arcsine function alone won’t give you every value.

2.12.2.5.

Hint.
In order for \(x\) to be in the domain of \(f\text{,}\) you must be able to plug \(x\) into both arcsine and arccosecant.

2.12.2.6.

Hint.
For the domain of \(f\text{,}\) remember the domain of arcsine is \([-1,1]\text{.}\)

2.12.2.7.

Hint.
The domain of \(\arccos(t)\) is \([-1,1]\text{,}\) but you also have to make sure you aren’t dividing by zero.

2.12.2.8.

Hint.
\(\ds\diff{}{x}\left\{\arcsec x\right\} = \dfrac{1}{|x|\sqrt{x^2-1}}\text{,}\) and the domain of \(\arcsec x\) is \(|x|\ge1\text{.}\)

2.12.2.9.

Hint.
The domain of \(\arctan(x)\) is all real numbers.

2.12.2.10.

Hint.
The domain of \(\arcsin x\) is \([-1,1]\text{,}\) and the domain of \(\sqrt{x}\) is \(x \geq 0\text{.}\)

2.12.2.11.

Hint.
This occurs only once.

2.12.2.12.

Hint.
The answer is a very simple expression.

2.12.2.13. (✳).

Hint.
chain rule

2.12.2.16.

Hint.
You can simplify the expression before you differentiate to remove the trigonometric functions. If \(\arctan x =\theta\text{,}\) then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:
With the sides labeled, you can figure out \(\sin\left(\arctan x\right)=\sin\left(\theta\right)\text{.}\)

2.12.2.17.

Hint.
You can simplify the expression before you differentiate to remove the trigonometric functions. If \(\arcsin x =\theta\text{,}\) then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:
With the sides labeled, you can figure out \(\cot\left(\arcsin x\right)=\cot\left(\theta\right)\text{.}\)

2.12.2.18. (✳).

Hint.
What is the slope of the line \(y=2x+9\text{?}\)

2.12.2.19.

Hint.
Differentiate using the chain rule.

2.12.2.20. (✳).

Hint.
If \(g(y)=f^{-1}(y)\text{,}\) then \(f(g(y))=f\left(f^{-1}(y)\right)=y\text{.}\) Differentiate this last equality using the chain rule.

2.12.2.21. (✳).

Hint.
To simplify notation, let \(g(y)=f^{-1}(y)\text{.}\) Simplify and differentiate \(g(f(x))\text{.}\)

2.12.2.22. (✳).

Hint.
To simplify notation, let \(g(y)=f^{-1}(y)\text{.}\) Simplify and differentiate \(g(f(x))\text{.}\)

2.12.2.23.

Hint.
Use logarithmic differentiation.

2.12.2.24.

Hint.
Where are those functions defined?

2.12.2.25.

Hint.
Compare this to one of the forms given in the text for the definition of the derivative.

2.12.2.26.

Hint.
\(f^{-1}(7)\) is the number \(y\) that satisfies \(f(y)=7\text{.}\)

2.12.2.27.

Hint.
If \(f^{-1}(y)=0\text{,}\) that means \(f(0)=y\text{.}\) So, we’re looking for the number that we plug into \(f^{-1}\) to get 0.

2.12.2.28.

Hint.
As usual, after you differentiate implicitly, get all the terms containing \(y'\) onto one side of the equation, so you can factor out \(y'\text{.}\)

2.13 The Mean Value Theorem
2.13.5 Exercises

2.13.5.1.

Hint.
How long would it take the caribou to travel 5000 km, travelling at its top speed?

2.13.5.2.

Hint.
Let \(f(x)\) be the position of the crane, where \(x\) is the hour of the day.

2.13.5.3.

Hint.
For an example, look at Figure 2.13.4.

2.13.5.4.

Hint.
How does this question differ from the statement of the mean value theorem?

2.13.5.6.

Hint.
Where is \(f(x)\) differentiable?

2.13.5.7. (✳).

Hint.
To use Rolle’s Theorem, you will want two values where the function is zero. If you’re stuck finding one of them, think about when \(x^2-2\pi x\) is equal to zero.

2.13.5.11.

Hint.
To show that there are exactly \(n\) roots, you need to not only show that \(n\) exist, but also that there are not more than \(n\text{.}\)

2.13.5.12.

Hint.
To show that there are exactly \(n\) roots, you need to not only show that \(n\) exist, but also that there are not more than \(n\text{.}\) If you can’t explicitly find the root(s), you can use the intermediate value theorem to show they exist.

2.13.5.13.

Hint.
If \(f(x)=0\text{,}\) then \(|x^3|=\left|\sin\left(x^5\right)\right| \leq 1\text{.}\) When \(|x| \lt 1\text{,}\) is \(\cos(x^5)\) positive or negative?

2.13.5.14.

Hint.
Let \(f(x)=e^x-4\cos(2x)\text{,}\) and use Rolle’s Theorem. What is the interval where \(f(x)\) can have a positive root?

2.13.5.15. (✳).

Hint.
For 2.13.5.15.b, what does Rolle’s Theorem tell you has to happen in order for \(f(x)\) to have more than one root in \([-1,1]\text{?}\)

2.13.5.16. (✳).

Hint.
Since \(f(x)=e^x\) is a continuous and differentiable function, the MVT promises that there exists some number \(c\) such that
\begin{equation*} f'(c)=\frac{f(T)-f(0)}{T}. \end{equation*}
Find that \(c\text{,}\) in terms of \(T\text{.}\)

2.13.5.17.

Hint.
Let \(f(x)=\arcsec x + \arccsc c -C\text{.}\) What is \(f'(x)\text{?}\)

2.13.5.18. (✳).

Hint.
Show that \(f\) is differentiable by showing that \(f'(x)\) exists for every \(x\text{.}\) Then, the Mean Value Theorem applies. What is the largest \(f'(x)\) can be, for any \(x\text{?}\) If \(f(100) \lt 100\text{,}\) what does the MVT tell you must be true of \(f'(c)\) for some \(c\text{?}\)

2.13.5.19.

Hint.
In order for \(f^{-1}(x)\) to be defined over an interval, \(f(x)\) must be one--to--one over that interval.

2.13.5.20.

Hint.
In order for \(f^{-1}(x)\) to be defined over an interval, \(f(x)\) must be one--to--one over that interval.

2.13.5.21.

Hint.
Let \(h(x)=f(x)-g(x)\text{.}\) What does the Mean Value Theorem tell you about the derivative of \(h\text{?}\)

2.13.5.22.

Hint.
Rolle’s Theorem relates the roots of a function to the roots of its derivative.

2.13.5.23.

Hint.
To show that there are exactly \(n\) distinct roots, you need to not only show that \(n\) exist, but also that there are not more than \(n\text{.}\)

2.14 Higher Order Derivatives
2.14.2 Exercises

2.14.2.1.

Hint.
If you know the first derivative, this should be easy.

2.14.2.2.

Hint.
Exactly two of the statements must be true.

2.14.2.3.

Hint.
Use factorials, as in Example 2.14.2.

2.14.2.4.

Hint.
The problem isn’t with any of the algebra.

2.14.2.5.

Hint.
Recall \(\ds\diff{}{x}\log x =\ds\frac{1}{x}\text{.}\)

2.14.2.6.

Hint.
Recall \(\ds\diff{}{x}\{\arctan x\}=\dfrac{1}{1+x^2}=(1+x^2)^{-1}\text{.}\)

2.14.2.7.

Hint.
Use implicit differentiation.

2.14.2.8.

Hint.
The acceleration is given by \(s''(t)\text{.}\)

2.14.2.9.

Hint.
Remember to use the chain rule.

2.14.2.10.

Hint.
\(h'(t)\) gives the velocity of the particle, and \(h''(t)\) gives its acceleration--the rate the velocity is changing.

2.14.2.11.

Hint.
\(h'(t)\) gives the velocity of the particle, and \(h''(t)\) gives its acceleration--the rate the velocity is changing. Be wary of signs--as in legends, they may be misleading.

2.14.2.12.

Hint.
You don’t need to solve for \(y''\) in general--only when \(x=y=0\text{.}\) To do this, you also need to find \(y'\) at the point \((0,0)\text{.}\)

2.14.2.13.

Hint.
To show that two functions are unequal, you can show that one input results in different outputs.

2.14.2.14.

Hint.
Only one of the curves could possibly represent \(y=f(x)\text{.}\)

2.14.2.15.

Hint.
Remember \(\ds\diff{}{x}\{2^x\}=2^x\log2\text{.}\)

2.14.2.16.

Hint.
Differentiate a few times until you get zero, remembering that \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are all constants.

2.14.2.17. (✳).

Hint.
Use a similar method to Question 2.9.4.32, Section 2.9.

2.14.2.18. (✳).

Hint.
For 2.14.2.18.b, you know a point where the curve and tangent line intersect, and you know what the tangent line looks like. What do the derivatives tell you about the shape of the curve?

2.14.2.19.

Hint.
Review Pascal’s Triangle.

2.14.2.20.

Hint.
Rolle’s Theorem relates the roots of a function to the roots of its derivative. So, the fifth derivative tells us something about the fourth, the fourth derivative tells us something about the third, and so on.

2.14.2.21.

Hint.
You’ll want to use Rolle’s Theorem, but the first derivative won’t be very tractable--use the idea behind Question 2.14.2.20.

2.14.2.22. (✳).

Hint.
You can re-write this function as a piecewise function, with branches \(x \ge 0\) and \(x \lt 0\text{.}\) To figure out the derivatives at \(x = 0\text{,}\) use the definition of a derivative.

3 Applications of derivatives
3.1 Velocity and Acceleration
3.1.2 Exercises

3.1.2.1.

Hint.
Is the velocity changing at \(t=2\text{?}\)

3.1.2.2.

Hint.
The acceleration (rate of change of velocity) is constant.

3.1.2.3.

Hint.
Remember the difference between speed and velocity.

3.1.2.4.

Hint.
How is this different from the wording of Question 3.1.2.3?

3.1.2.5.

Hint.
The equation of an object falling from rest on the earth is derived in Example 3.1.2. It would be difficult to use exactly the version given for \(s(t)\text{,}\) but using the same logic, you can find an equation for the height of the flower pot at time \(t\text{.}\)

3.1.2.6.

Hint.
Remember that a falling object has an acceleration of \(9.8\;\frac{\mathrm{m}}{\mathrm{s}^2}\text{.}\)

3.1.2.7.

Hint.
Acceleration is constant, so finding a formula for the distance your keys have travelled is a similar problem to finding a formula for something falling.

3.1.2.8.

Hint.
See Example 3.1.3.

3.1.2.9.

Hint.
Be careful to match up the units.

3.1.2.10.

Hint.
Think about what it means for the car to decelerate at a constant rate. You might also review Question 3.1.2.2.

3.1.2.11.

Hint.
Let \(a\) be the acceleration of the shuttle. Start by finding \(a\text{,}\) then find the position function of the shuttle.

3.1.2.12.

Hint.
Review Example 3.1.2, but account for the fact that your initial velocity is not zero.

3.1.2.13.

Hint.
Be very careful with units. The acceleration of gravity you’re used to is \(9.8\) metres per second squared, so you might want to convert \(325\) kpm to metres per second.

3.1.2.14.

Hint.
Since gravity alone brings it down, its acceleration is a constant \(-9.8\;\frac{\mathrm{m}}{\mathrm{s}^2}\text{.}\)

3.1.2.15.

Hint.
First, find an equation for \(a(t)\text{,}\) the acceleration of the car, noting that \(a'(t)\) is constant. Then, use this to find an equation for the velocity of the car. Be careful about seconds versus hours.

3.1.2.16.

Hint.
We recommend using two different functions to describe your height: \(h_1(t)\) while you are in the air, not yet touching the trampoline, and \(h_2(t)\) while you are in the trampoline, going down.
Both \(h_1(t)\) and \(h_2(t)\) are quadratic equations, since your acceleration is constant over both intervals, but be very careful about signs.

3.1.2.17.

Hint.
First, find an expression for the speed of the object. You can let \(v_0\) be its velocity at time \(t=0\text{.}\)

3.2 Related Rates
3.2.2 Exercises

3.2.2.1.

Hint.
If you know \(P\text{,}\) you can figure out \(Q\text{.}\)

3.2.2.2. (✳).

Hint.
Since the point moves along the unit circle, we know that \(x^2+y^2=1\text{,}\) where \(x\) and \(y\) are functions of time.

3.2.2.3. (✳).

Hint.
You’ll need some implicit differentiation: what should your variable be? Example 3.2.3 shows how to work with percentage rate of change.

3.2.2.4. (✳).

Hint.
For 3.2.2.4.b, refer to Example 3.2.3 for percentage rate of change.

3.2.2.5. (✳).

Hint.
Pay attention to direction, and what it means for the sign (plus/minus) of the velocities of the particles.

3.2.2.6. (✳).

Hint.
You’ll want to think about the difference in the \(y\)-coordinates of the two particles.

3.2.2.7. (✳).

Hint.
Draw a picture, and be careful about signs.

3.2.2.8. (✳).

Hint.
You’ll want to think about the difference in height of the two snails.

3.2.2.9. (✳).

Hint.
The length of the ladder is changing.

3.2.2.10.

Hint.
If a trapezoid has height \(h\) and (parallel) bases \(b_1\) and \(b_2\text{,}\) then its area is \(h\left(\frac{b_1+b_2}{2}\right)\text{.}\) To figure out how wide the top of the water is when the water is at height \(h\text{,}\) you can cut the trapezoid up into a rectangle and two triangles, and make use of similar triangles.

3.2.2.11.

Hint.
Be careful with units. One litre is 1000 cm\(^3\text{,}\) which is not the same as \(10\) m\(^3\text{.}\)

3.2.2.12.

Hint.
You, the rocket, and the rocket’s original position form a right triangle.

3.2.2.13. (✳).

Hint.
Your picture should be a triangle.

3.2.2.14.

Hint.
Let \(\theta\) be the angle between the two hands. Using the Law of Cosines, you can get an expression for \(D\) in terms of \(\theta\text{.}\) To find \(\ds\diff{\theta}{t}\text{,}\) use what you know about how fast clock hands move.

3.2.2.15. (✳).

Hint.
The area in the annulus is the area of the outer circle minus the area of the inner circle.

3.2.2.16.

Hint.
The volume of a sphere with radius \(R\) is \(\dfrac{4}{3}\pi r^3\text{.}\)

3.2.2.17.

Hint.
The area of a triangle is half its base times its height. To find the base, split the triangle into two right triangles.

3.2.2.18.

Hint.
The easiest way to figure out the area of the sector of an annulus (or a circle) is to figure out the area of the entire annulus, then multiply by what proportion of the entire annulus the sector is. For example, if your sector is \(\frac{1}{10}\) of the entire annulus, then its area is \(\frac{1}{10}\) of the area of the entire annulus. (See Section A.4 to see how this works out for circles.)

3.2.2.19.

Hint.
Think about the ways in which this problem is similar to and different from Example 3.2.6 and Question 3.2.2.18.

3.2.2.20.

Hint.
The volume of a cone with height \(h\) and radius \(r\) is \(\frac{1}{3}\pi r^2h\text{.}\) Also, one millilitre is the same as one cubic centimetre.

3.2.2.21.

Hint.
If you were to install the buoy, how would you choose the length of rope? For which values of \(\theta\) do \(\sin\theta\) and \(\cos\theta\) have different signs? How would those values of \(\theta\) look on the diagram?

3.2.2.22.

Hint.
At both points of interest, the point is moving along a straight line. From the diagram, you can figure out the equation of that line.
For the question “How fast is the point moving?” in part (b), remember that the velocity of an object can be found by differentiating (with respect to time) the equation that gives the position of the object. The complicating factors in this case are that (1) the position of our object is not given as a function of time, and (2) the position of our object is given in two dimensions, not one.

3.2.2.23.

Hint.
(a) Since the perimeter of the cross section of the bottle does not change, \(p\) (the perimeter of the ellipse) is the same as the perimeter of the circle of radius 5.
(b) The volume of the bottle will be the area of its cross section times its height. This is always the case when you have some two-dimensional shape, and turn it into a three-dimensional object by “pulling” the shape straight up. (For example, you can think of a cylinder as a circle that has been “pulled” straight up. To understand why this formula works, think about what is means to measure the area of a shape in square centimetres, and the volume of an object in cubic centimetres.)
(c) You can use what you know about \(a\) and the formula from (a) to find \(b\) and \(\ds\diff{b}{t}\text{.}\) Then use the formula from \((b)\text{.}\)

3.2.2.24.

Hint.
If \(A=0\text{,}\) you can figure out \(C\) and \(D\) from the relationship given.

3.3 Exponential Growth and Decay — a First Look at Differential Equations
3.3.4 Exercises

Exercises for § 3.3.1

3.3.4.1.
Hint.
Review the definition of a differential equation at the beginning of this section.
3.3.4.2.
Hint.
You can test whether a given function solves a differential equation by substituting the function into the equation.
3.3.4.3.
Hint.
Solve \(0=Ce^{-kt}\) for \(t\text{.}\)
3.3.4.4. (✳).
Hint.
No calculus here--just a review of the algebra of exponentials.
3.3.4.5. (✳).
Hint.
Use Theorem 3.3.2.
3.3.4.6.
Hint.
From the text, we see the half-life of Carbon-14 is 5730 years. A microgram (\(\mu\)g) is one-millionth of a gram, but you don’t need to know that to solve this problem.
3.3.4.7.
Hint.
The quantity of Radium-226 in the sample at time \(t\) will be \(Q(t)=Ce^{-kt}\) for some positive constants \(C\) and \(k\text{.}\) You can use the given information to find \(C\) and \(e^{-k}\text{.}\)
In the following work, remember we use \(\log\) to mean natural logarithm, \(\log_e\text{.}\)
3.3.4.8. (✳).
Hint.
The fact that the mass of the sample decreases at a rate proportional to its mass tells us that, if \(Q(t)\) is the mass of Polonium-201, the following differential equation holds:
\begin{equation*} \diff{Q}{t}=-kQ(t) \end{equation*}
where \(k\) is some positive constant. Compare this to Theorem 3.3.2.
3.3.4.9.
Hint.
The amount of Radium-221 in a sample at time \(t\) will be \(Q(t)=Ce^{-kt}\) for some positive constants \(C\) and \(k\text{.}\) You can leave \(C\) as a variable--it’s the original amount in the sample, which isn’t specified. What you want to find is the value of \(t\) such that \(Q(t)=0.0001Q(0)=0.0001C\text{.}\)
3.3.4.10.
Hint.
You don’t need to know the original amount of Polonium-210 in order to answer this question: you can leave it as some constant \(C\text{,}\) or you can call it 100%.
3.3.4.11.
Hint.
Try to find the most possible and least possible remaining Uranium-232, given the bounds in the problem.

Exercises for § 3.3.2

3.3.4.1.
Hint.
You can refer to Corollary 3.3.8, but you can also just differentiate the various proposed functions and see whether, in fact, \(\ds\diff{T}{t}\) is the same as \(5[T-20]\text{.}\)
3.3.4.2.
Hint.
From Newton’s Law of Cooling and Corollary 3.3.8, the temperature of the object will be
\begin{equation*} T(t)=[T(0)-A]e^{Kt}+A \end{equation*}
where \(A\) is the ambient temperature, \(T(0)\) is the initial temperature of the copper, and \(K\) is some constant.
3.3.4.3.
Hint.
What is \(\ds\lim_{t \to \infty}e^{Kt}\) when \(K\) is positive, negative, or zero?
3.3.4.4.
Hint.
Solve \(A=[T(0)-A]e^{kt}\) for \(t\text{.}\)
3.3.4.5.
Hint.
From Newton’s Law of Cooling and Corollary 3.3.8, we know the temperature of the copper will be
\begin{equation*} T(t)=[T(0)-A]e^{Kt}+A \end{equation*}
where \(A\) is the ambient temperature, \(T(0)\) is the initial temperature of the copper, and \(K\) is some constant. Use the given information to find an expression for \(T(t)\) not involving any unknown constants.
3.3.4.6.
Hint.
From Newton’s Law of Cooling and Corollary 3.3.8, we know the temperature of the stone \(t\) minutes after it leaves the fire is
\begin{equation*} T(t)=[T(0)-A]e^{Kt}+A \end{equation*}
where \(A\) is the ambient temperature, \(T(0)\) is the temperature of the stone the instant it left the fire, and \(K\) is some constant.
3.3.4.8. (✳).
Hint.
Newton’s Law of Cooling models the temperature of the tea after \(t\) minutes as
\begin{equation*} T(t)=[T(0)-A]e^{Kt}+A \end{equation*}
where \(A\) is the ambient temperature, \(T(0)\) is the initial temperature of the tea, and \(K\) is some constant.
3.3.4.9.
Hint.
What is \(\ds\lim_{t \to \infty}T(t)\text{?}\)

Exercises for § 3.3.3

3.3.4.1.
Hint.
\(P(0)\) is also (probably) a positive constant.
3.3.4.2.
Hint.
The assumption that the animals grow according to the Malthusian model tells us that their population \(t\) years after 2015 is given by \(P(t)=121e^{bt}\) for some constant \(b\text{.}\)
3.3.4.3.
Hint.
The Malthusian model says that the population of bacteria \(t\) hours after being placed in the dish will be \(P(t)=1000e^{bt}\) for some constant \(b\text{.}\)
3.3.4.4.
Hint.
If 1928 is \(a\) years after the shipwreck, you might want to make use of the fact that \(e^{b(a+1)}=e^{ba}e^b\text{.}\)
3.3.4.5.
Hint.
If the population has a net birthrate per individual per unit time of \(b\text{,}\) then the Malthusian model predicts that the number of individuals at time \(t\) will be \(P(t)=P(0)e^{bt}\text{.}\) You can use the test population to find \(e^b\text{.}\)
3.3.4.6.
Hint.
One way to investigate the sign of \(k\) is to think about \(f'(t)\text{:}\) is it positive or negative?

Further problems for § 3.3

3.3.4.1. (✳).
Hint.
Use Theorem 3.3.2 to figure out what \(f(x)\) looks like.
3.3.4.2.
Hint.
To use Corollary 3.3.8, you need to re-write the differential equation as
\begin{equation*} \diff{T}{t}=7\left[T-\left(-\frac{9}{7}\right)\right]. \end{equation*}
3.3.4.3. (✳).
Hint.
The amount of the material at time \(t\) will be \(Q(t)=Ce^{-kt}\) for some constants \(C\) and \(k\text{.}\)
3.3.4.4.
Hint.
In your calculations, it might come in handy that \(e^{30K}=\left(e^{15K}\right)^2\text{.}\)
3.3.4.5. (✳).
Hint.
The differential equation in the problem has the same form as the differential equation from Newton’s Law of Cooling.
3.3.4.6. (✳).
Hint.
We know the form of the solution \(A(t)\) from Corollary 3.3.8.
3.3.4.7. (✳).
Hint.
If a function’s rate of change is proportional to the function itself, what does the function looks like?
3.3.4.8. (✳).
Hint.
The equation from Newton’s Law of Cooling, in Corollary 3.3.8, has a similar form to the differential equation in this question.

3.4 Approximating Functions Near a Specified Point — Taylor Polynomials
3.4.11 Exercises

Exercises for § 3.4.1

3.4.11.1.
Hint.
An approximation should be something you can actually figure out--otherwise it’s no use.
3.4.11.2.
Hint.
You’ll need some constant \(a\) to approximation \(\log(0.93) \approx \log(a)\text{.}\) This \(a\) should have two properties: it should be close to 0.93, and you should be able to easily evaluate \(\log(a)\text{.}\)
3.4.11.3.
Hint.
You’ll need some constant \(a\) to approximate \(\arcsin(0.1) \approx \arcsin(a)\text{.}\) This \(a\) should have two properties: it should be close to 0.1, and you should be able to easily evaluate \(\arcsin(a)\text{.}\)
3.4.11.4.
Hint.
You’ll need some constant \(a\) to approximate \(\sqrt{3}\tan(1) \approx \sqrt{3}\tan(a)\text{.}\) This \(a\) should have two properties: it should be close to 1, and you should be able to easily evaluate \(\sqrt{3}\tan(a)\text{.}\)
3.4.11.5.
Hint.
We could figure out \(10.1^3\) exactly, if we wanted, with pen and paper. Since we’re asking for an approximation, we aren’t after perfect accuracy. Rather, we’re after ease of calculation.

Exercises for § 3.4.2

3.4.11.1.
Hint.
The linear approximation \(L(x)\) is chosen so that \(f(5)=L(5)\) and \(f'(5)=L'(5)\text{.}\)
3.4.11.2.
Hint.
The graph of the linear approximation is a line, passing through \((2,f(2))\text{,}\) with slope \(f'(2)\text{.}\)
3.4.11.3.
Hint.
It’s an extremely accurate approximation.
3.4.11.4.
Hint.
You’ll need to centre your approximation about some \(x=a\text{,}\) which should have two properties: you can easily compute \(\log(a)\text{,}\) and \(a\) is close to \(0.93\text{.}\)
3.4.11.5.
Hint.
Approximate the function \(f(x) = \sqrt{x}\text{.}\)
3.4.11.6.
Hint.
Approximate the function \(f(x)=\sqrt[5]{x}\text{.}\)
3.4.11.7.
Hint.
Approximate the function \(f(x)=x^3\text{.}\)
3.4.11.8.
Hint.
One possible choice of \(f(x)\) is \(f(x)=\sin x\text{.}\)
3.4.11.9.
Hint.
Compare the derivatives.

Exercises for § 3.4.3

3.4.11.1.
Hint.
If \(Q(x)\) is the quadratic approximation of \(f\) about \(3\text{,}\) then \(Q(3)=f(3)\text{,}\) \(Q'(3)=f'(3)\text{,}\) and \(Q''(3)=f''(3)\text{.}\)
3.4.11.2.
Hint.
It is a very good approximation.
3.4.11.3.
Hint.
Approximate \(f(x)=\log x\text{.}\)
3.4.11.4.
Hint.
You’ll probably want to centre your approximation about \(x=0\text{.}\)
3.4.11.5.
Hint.
The quadratic approximation of a function \(f(x)\) about \(x=a\) is
\begin{equation*} f(x) \approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2 \end{equation*}
3.4.11.6.
Hint.
One way to go about this is to approximate the function \(f(x) = 5 \cdot x^{1/3}\) , because then \(5^{4/3} = 5 \cdot 5^{1/3} =f(5)\text{.}\)
3.4.11.7.
Hint.
For 3.4.11.7.c, look for cancellations.
3.4.11.8.
Hint.
Compare (c) to (b).
Compare (e) and (f) to (d).
To get an alternating sign, consider powers of \((-1)\text{.}\)
3.4.11.9.
Hint.
You can evaluate \(f(1)\) exactly. Recall \(\ds\diff{}{x}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}\text{.}\)
3.4.11.10.
Hint.
Let \(f(x)=e^x\text{,}\) and use the quadratic approximation of \(f(x)\) about \(x=0\) (given in your text, or you can reproduce it) to approximate \(f(1)\text{.}\)
3.4.11.11.
Hint.
Be wary of indices: for example \(\ds\sum_{n=1}^3 n = \ds\sum_{n=5}^7 (n-4)\text{.}\)

Exercises for § 3.4.4

3.4.11.1.
Hint.
\(T_3''(x)\) and \(f''(x)\) agree when \(x=1\text{.}\)
3.4.11.2.
Hint.
The \(n\)th order Taylor polynomial for \(f(x)\) about \(x=5\) is
\begin{equation*} T_n(x)=\sum_{k=0}^{n} \frac{f^{(k)}(5)}{k!}(x-5)^k \end{equation*}
Match up the terms.
3.4.11.3.
Hint.
The fourth order Maclaurin polynomial for \(f(x)\) is
\begin{align*} T_4(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f^{(4)}(0)x^4\\ \end{align*}

while the third order Maclaurin polynomial for \(f(x)\) is

\begin{align*} T_3(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3 \end{align*}
3.4.11.4.
Hint.
The third order Taylor polynomial for \(f(x)\) about \(x=1\) is
\begin{equation*} T_3(x)=f(1)+f'(1)(x-1)+\frac{1}{2}f''(1)(x-1)^2+\frac{1}{3!}f'''(1)(x-1)^3 \end{equation*}
How can you recover \(f(1)\text{,}\) \(f'(1)\text{,}\) \(f''(1)\text{,}\) and \(f'''(1)\) from \(T_4(x)\text{?}\)
3.4.11.5.
Hint.
Compare the given polynomial to the more standard form of the \(n\)th order Taylor polynomial,
\begin{equation*} \sum_{k=0}^{n} \frac{1}{k!}f^{(k)}(5)(x-5)^{k} \end{equation*}
and notice that the term you want (containing \(f^{(10)}(5)\)) corresponds to \(k=10\) in the standard form, but is not the term corresponding to \(k=10\) in the polynomial given in the question.
3.4.11.6.
Hint.
\(T_3'''(a)=f'''(a)\)

Exercises for § 3.4.5

3.4.11.1.
Hint.
The derivatives of \(f(x)\) repeat themselves.
3.4.11.2.
Hint.
You are approximating a polynomial with a polynomial.
3.4.11.3.
Hint.
Recall \(\ds\diff{}{x}\left\{2^x\right\}=2^x\log 2\text{,}\) where \(\log 2\) is the constant \(\log_e2\text{.}\)
3.4.11.4.
Hint.
Just keep differentiating--it gets easier!
3.4.11.5.
Hint.
Start by differentiating, and finding the pattern for \(f^{(k)}(0)\text{.}\) Remember the chain rule!
3.4.11.6.
Hint.
You’ll need to differentiate \(x^x\text{.}\) This is accomplished using logarithmic differentiation, covered in Section 2.10.
3.4.11.7.
Hint.
What is \(6\arctan \left(\dfrac{1}{\sqrt{3}}\right)\text{?}\)
3.4.11.8.
Hint.
After a few derivatives, this will be very similar to Example 3.4.13.
3.4.11.9.
Hint.
Treat the even and odd powers separately.
3.4.11.10.
Hint.
Compare this to the Maclaurin polynomial for \(e^x\text{.}\)
3.4.11.11.
Hint.
Compare this to the Maclaurin polynomial for cosine.

Exercises for § 3.4.6

3.4.11.1.
Hint.
\(\Delta x\) and \(\Delta y \) represent changes in \(x\) and \(y\text{,}\) respectively, while \(f(x)\) and \(f\left(x+\Delta x \right)\) are the \(y\)-values the function takes.
3.4.11.2.
Hint.
Let \(f(x)\) be the number of problems finished after \(x\) minutes of work.
3.4.11.3.
Hint.
\(\Delta y = f(5.1)-f(5)\)
3.4.11.4.
Hint.
Use the approximation \(\Delta y \approx s'(4)\Delta x\) when \(x\) is near 4.

Exercises for § 3.4.7

3.4.11.1.
Hint.
Is the linear approximation exact, or approximate?
3.4.11.2.
Hint.
When an exact value \(Q_0\) is measured as \(Q_0+\Delta Q\text{,}\) Definition 3.4.25
gives us the absolute error as \(|\Delta Q|\text{,}\) and the percentage error as \(100\dfrac{|\Delta Q|}{Q_0}\text{.}\)
3.4.11.3.
Hint.
Let \(\Delta y\) is the change in \(f(x)\) associated to a change in \(x\) from \(a\) to \(a+\Delta x\text{.}\) The linear approximation tells us
\begin{equation*} \Delta y \approx f'(a)\Delta x \end{equation*}
while the quadratic approximation tells us
\begin{equation*} \Delta y \approx f'(a)\Delta x+\frac{1}{2}f''(a)\left(\Delta x\right)^2 \end{equation*}
3.4.11.4.
Hint.
The exact area desired is \(A_0\text{.}\) Let the corresponding exact radius desired be \(r_0\text{.}\) The linear approximation tells us \(\Delta A \approx A'(r_0) \Delta r\text{.}\) Use this relationship, and what you know about the error allowable in \(A\text{,}\) to find the error allowable in \(r\text{.}\)
3.4.11.5.
Hint.
For part (b), cut the triangle (with angle \(\theta\) and side \(d\)) into two right triangles.
3.4.11.6.
Hint.
The volume of a cone of height \(h\) and radius \(r\) is \(\frac{1}{3}\pi r^2 h\text{.}\)
3.4.11.7.
Hint.
Remember that the amount of the isotope present at time \(t\) is \(Q(t)=Q(0)e^{-kt}\) for some constant \(k\text{.}\) The measured quantity after 3 years will allow you to replace \(k\) in the equation, then solving \(Q(t)=\frac{1}{2}Q(0)\) for \(t\) will give you the half-life of the isotope.

Exercises for § 3.4.8

3.4.11.1.
Hint.
\(R(10)=f(10)-F(10)=-3-5\)
3.4.11.2.
Hint.
Equation 3.4.33 tells us
\begin{equation*} |f(2)-T_3(2)| = \left|\frac{f^{(4)}(c)}{4!}(2-0)^4\right| \end{equation*}
for some \(c\) strictly between 0 and 2.
3.4.11.3.
Hint.
You are approximating a third order polynomial with a fifth order Taylor polynomial. You should be able to tell how good your approximation will be without a long calculation.
3.4.11.4.
Hint.
Draw a picture — it should be clear how the two approximations behave.
3.4.11.5.
Hint.
In this case, Equation 3.4.33 tells us that
\begin{equation*} \left|f(11.5)-T_5(11.5)\right| = \left|\dfrac{f^{(6)}(c)}{6!}(11.5-11)^{6}\right| \end{equation*}
for some \(c\) strictly between 11 and 11.5.
3.4.11.6.
Hint.
In this case, Equation 3.4.33 tells us that \(\left|f(0.1)-T_2(0.1)\right| = \left|\dfrac{f^{(3)}(c)}{3!}(0.1-0)^{3}\right|\) for some \(c\) strictly between 0 and 0.1.
3.4.11.7.
Hint.
In our case, Equation 3.4.33 tells us
\begin{equation*} \left|f\left(-\dfrac{1}{4}\right)-T_5\left(-\dfrac{1}{4}\right)\right| = \left|\dfrac{f^{(6)}(c)}{6!}\left(-\dfrac{1}{4}-0\right)^6\right| \end{equation*}
for some \(c\) between \(-\dfrac{1}{4}\) and 0.
3.4.11.8.
Hint.
In this case, Equation 3.4.33 tells us that \(\left|f(30)-T_3(30)\right| = \left|\dfrac{f^{(4)}(c)}{4!}(30-32)^{4}\right|\) for some \(c\) strictly between 30 and 32.
3.4.11.9.
Hint.
In our case, Equation 3.4.33 tells us \(\left|f\left(0.01\right)-T_1\left(0.01\right)\right| = \left|\dfrac{f^{(2)}(c)}{2!}\left(0.01-\frac{1}{\pi}\right)^2\right|\) for some \(c\) between 0.01 and \(\dfrac{1}{\pi}\text{.}\)
3.4.11.10.
Hint.
Using Equation 3.4.33, \(\left|f\left(\dfrac{1}{2}\right)-T_2\left(\dfrac{1}{2}\right)\right| = \left|\dfrac{f^{(3)}(c)}{3!}\left(\dfrac{1}{2}-0\right)^{3}\right|\) for some \(c\) in \(\left(0,\dfrac{1}{2}\right)\text{.}\)
3.4.11.11.
Hint.
It helps to have a formula for \(f^{(n)}(x)\text{.}\) You can figure it out by taking several derivatives and noticing the pattern, but also this has been given previously in the text.
3.4.11.12.
Hint.
You can approximate the function \(f(x)=x^{\tfrac{1}{7}}\text{.}\)
It’s a good bit of trivia to know \(3^7=2187\text{.}\)
A low order Taylor approximation will give you a good enough estimation. If you guess an order, and take that Taylor polynomial, the error will probably be less than 0.001 (but you still need to check).
3.4.11.13.
Hint.
Use the 6th order Maclaurin approximation for \(f(x)=\sin x\text{.}\)
3.4.11.14.
Hint.
For part 3.4.11.14.c, after you plug in the appropriate values to Equation 3.4.33, simplify the upper and lower bounds for \(e\) separately. In particular, for the upper bound, you’ll have to solve for \(e\text{.}\)

Further problems for § 3.4

3.4.11.1. (✳).
Hint.
Compare the given polynomial to the definition of a Maclaurin polynomial.
3.4.11.3. (✳).
Hint.
Compare the given polynomial to the definition of a Taylor polynomial.
3.4.11.5. (✳).
Hint.
You can use the error formula to determine whether the approximation is too large or too small.
3.4.11.6. (✳).
Hint.
Use the function \(f(x)=\sqrt{x}\text{.}\)
3.4.11.7. (✳).
Hint.
Use the function \(f(x)=x^{1/3}\text{.}\) What is a good choice of centre?
3.4.11.8. (✳).
Hint.
Try using the function \(f(x)=x^5\text{.}\)
3.4.11.9. (✳).
Hint.
If you use the function \(f(x)=\sin(x)\text{,}\) what is a good centre?
3.4.11.10. (✳).
Hint.
Recall \(\ds\diff{}{x}\left\{\arctan x\right\} = \dfrac{1}{1+x^2}\text{.}\)
3.4.11.11. (✳).
Hint.
Try using the function \(f(x)=(2+x)^3\text{.}\)
3.4.11.12. (✳).
Hint.
You can try using \(f(x)=(8+x)^{1/3}\text{.}\) What is a suitable centre for your approximation?
3.4.11.13. (✳).
Hint.
This is the same as the Maclaurin polynomial.
3.4.11.14. (✳).
Hint.
This is a straightforward application of Equation 3.4.33.
3.4.11.17. (✳).
Hint.
\(5^{2/3}=f(5^2)=f(25)\)
3.4.11.18.
Hint.
The fourth order Maclaurin polynomial for \(f(x)\) is
\begin{align*} T_4(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f^{(4)}(0)x^4\\ \end{align*}

while the third order Maclaurin polynomial for \(f(x)\) is

\begin{align*} T_3(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3 \end{align*}
3.4.11.19. (✳).
Hint.
For part 3.4.11.19.c, think about what the quadratic approximation looks like — is it pointing up or down?
3.4.11.22. (✳).
Hint.
For (c), you can write \(f(x)\) as the sum of \(Q(x)\) and its error term.
For (d), you can use the linear approximation of \(e^x\) centred at \(0\text{,}\) with its error term when \(x=0.1\text{.}\)

3.5 Optimisation
3.5.4 Exercises

Exercises for § 3.5.1

3.5.4.1.
Hint.
Estimate \(f'(0)\text{.}\)
3.5.4.2.
Hint.
If the graph is discontinuous at a point, it is not differentiable at that point.
3.5.4.3.
Hint.
Try making a little bump at \(x=2\text{,}\) the letting the function get quite large somewhere else.
3.5.4.4.
Hint.
Critical points are those values of \(x\) for which \(f'(x)=0\text{.}\)
Singular points are those values of \(x\) for which \(f(x)\) is not differentiable.
3.5.4.5.
Hint.
We’re only after local extrema, not global. Let \(f(x)\) be our function. If there is some interval around \(x=2\) where nothing is bigger than \(f(2)\text{,}\) then \(f(2)\) is a local maximum, whether or not it is a maximum overall.
3.5.4.6.
Hint.
By Theorem 3.5.4, if \(x=2\) not a critical point, then it must be a singular point.
3.5.4.7.
Hint.
You should be able to figure out the global minima of \(f(x)\) in your head.
Remember with absolute values, \(|X|=\left\{\begin{array}{ll} X&X\ge0\\ -X&X \lt 0 \end{array}\right.\text{.}\)
3.5.4.8.
Hint.
Review the definitions of critical points and extrema: Definition 3.5.6 and Definition 3.5.3.

Exercises for § 3.5.2

3.5.4.1.
Hint.
One way to avoid a global minimum is to have \(\ds\lim_{x \to \infty}f(x)=-\infty\text{.}\) Since \(f(x)\) keeps getting lower and lower, there is no one value that is the lowest.
3.5.4.2.
Hint.
Try allowing the function to approach the \(x\)-axis without ever touching it.
3.5.4.3.
Hint.
Since the global minimum value occurs at \(x=5\) and \(x=-5\text{,}\) it must be true that \(f(5)=f(-5)\text{.}\)
3.5.4.4.
Hint.
Global extrema will either occur at critical points in the interval \((-5,5)\) or at the endpoints \(x=5,\,x=-5\text{.}\)
3.5.4.5.
Hint.
You only need to consider critical points that are in the interval \((-4,0).\)

Exercises for § 3.5.3

3.5.4.1. (✳).
Hint.
Factor the derivative.
3.5.4.2. (✳).
Hint.
Remember to test endpoints.
3.5.4.4. (✳).
Hint.
One way to decide whether a critical point \(x=c\) is a local extremum is to consider the first derivative. For example: if \(f'(x)\) is negative for all \(x\) just to the left of \(c\text{,}\) and positive for all \(x\) just to the right of \(c\text{,}\) then \(f(x)\) decreases up till \(c\text{,}\) then increases after \(c\text{,}\) so \(f(x)\) has a local minimum at \(c\text{.}\)
3.5.4.5. (✳).
Hint.
One way to decide whether a critical point \(x=c\) is a local extremum is to consider the first derivative. For example: if \(f'(x)\) is negative for all \(x\) just to the left of \(c\text{,}\) and positive for all \(x\) just to the right of \(c\text{,}\) then \(f(x)\) decreases up till \(c\text{,}\) then increases after \(c\text{,}\) so \(f(x)\) has a local minimum at \(c\text{.}\)
3.5.4.6. (✳).
Hint.
Start with a formula for travel time from \(P\) to \(B\text{.}\) You might want to assign a variable to the distance from \(A\) where your buggy first reaches the road.
3.5.4.7. (✳).
Hint.
A box has three dimensions; make variables for them, and write the relations given in the problem in terms of these variables.
3.5.4.8. (✳).
Hint.
Find a formula for the cost of the base, and another formula for the cost of the other sides. The total cost is the sum of these two formulas.
3.5.4.9. (✳).
Hint.
The setup is this:
3.5.4.10. (✳).
Hint.
Put the whole system on \(xy\)-axes, so that you can easily describe the pieces using \((x,y)\)-coordinates.
3.5.4.11. (✳).
Hint.
The surface area consists of two discs and a strip. Find the areas of these pieces.
The volume of a cylinder with radius \(r\) and height \(h\) is \(\pi r^2 h\text{.}\)
3.5.4.12. (✳).
Hint.
If the circle has radius \(r\text{,}\) and the entire window has perimeter \(P\text{,}\) what is the height of the rectangle?
3.5.4.14. (✳).
Hint.
Use logarithmic differentiation to find \(f'(x)\text{.}\)
3.5.4.15. (✳).
Hint.
When you are finding the global extrema of a function, remember to check endpoints as well as critical points.

3.6 Sketching Graphs
3.6.7 Exercises

Exercises for § 3.6.1

3.6.7.1.
Hint.
What happens if \(g(x)=x+3\text{?}\)
3.6.7.2.
Hint.
Use domains and intercepts to distinguish between the functions.
3.6.7.3.
Hint.
To find \(p\text{,}\) the equation \(f(0)=2\) gives you two possible values of \(p\text{.}\) Consider the domain of \(f(x)\) to decide between them.
3.6.7.4.
Hint.
Check for horizontal asymptotes by evaluating \(\ds\lim_{x \to \pm \infty}f(x)\text{,}\) and check for vertical asymptotes by finding any value of \(x\) near which \(f(x)\) blows up.
3.6.7.5.
Hint.
Check for horizontal asymptotes by evaluating \(\ds\lim_{x \to \pm \infty}f(x)\text{,}\) and check for vertical asymptotes by finding any value of \(x\) near which \(f(x)\) blows up.

Exercises for § 3.6.2

3.6.7.1.
Hint.
For each of the graphs, consider where the derivative is positive, negative, and zero.
3.6.7.2. (✳).
Hint.
Where is \(f'(x) \gt 0\text{?}\)
3.6.7.3. (✳).
Hint.
Consider the signs of the numerator and the denominator of \(f'(x)\text{.}\)
3.6.7.4. (✳).
Hint.
Remember \(\ds\diff{}{x}\{\arctan x\}=\dfrac{1}{1+x^2}\text{.}\)

Exercises for § 3.6.3

3.6.7.1.
Hint.
There are two intervals where the function is concave up, and two where it is concave down.
3.6.7.2.
Hint.
Try allowing your graph to have horizontal asymptotes. For example, let the function get closer and closer to the \(x\)-axis (or another horizontal line) without touching it.
3.6.7.3.
Hint.
Consider \(f(x)=(x-3)^4\text{.}\)
3.6.7.5. (✳).
Hint.
You must show it has at least one inflection point (try the Intermediate Value Theorem), and at most one inflection point (consider whether the second derivative is increasing or decreasing).
3.6.7.7.
Hint.
Since \(x=3\) is an inflection point, we know the concavity of \(f(x)\) changes at \(x=3\text{.}\) That is, there is some interval around 3, with endpoints \(a\) and \(b\text{,}\) such that
  • \(f''(a) \lt 0\) and \(f''(x) \lt 0\) for every \(x\) between \(a\) and 3, and
  • \(f''(b) \gt 0\) and \(f''(x) \gt 0\) for every \(x\) between \(b\) and 3.
Use the IVT to show that \(f''(x)=3\) for some \(x\) between \(a\) and \(b\text{;}\) then show that this value of \(x\) can’t be anything except \(x=3\text{.}\)

Exercises for § 3.6.4

3.6.7.1.
Hint.
This function is symmetric across the \(y\)-axis.
3.6.7.2.
Hint.
There are two.
3.6.7.3.
Hint.
Since the function is even, you only have to reflect the portion shown across the \(y\)-axis to complete the sketch.
3.6.7.4.
Hint.
Since the function is odd, to complete the sketch, reflect the portion shown across the \(y\)-axis, then the \(x\)-axis.
3.6.7.5.
Hint.
A function is even if \(f(-x)=f(x)\text{.}\)
3.6.7.6.
Hint.
Its period is not \(2\pi\text{.}\)
3.6.7.7.
Hint.
Simplify \(f(-x)\) to see whether it is the same as \(f(x)\text{,}\) \(-f(x)\text{,}\) or neither.
3.6.7.8.
Hint.
Simplify \(f(-x)\) to see whether it is the same as \(f(x)\text{,}\) \(-f(x)\text{,}\) or neither.
3.6.7.9.
Hint.
Find the smallest value \(k\) such that \(f(x+k)=f(x)\) for any \(x\) in the domain of \(f\text{.}\)
You may use the fact that the period of \(g(X)=\tan X\) is \(\pi\text{.}\)
3.6.7.10.
Hint.
It is true that \(f(x)=f(x+2\pi)\) for every \(x\) in the domain of \(f(x)\text{,}\) but the period is not \(2\pi\text{.}\)

Exercises for § 3.6.6

3.6.7.1. (✳).
Hint.
You’ll find the intervals of increase and decrease. These will give you a basic outline of the behaviour of the function. Use concavity to refine your picture.
3.6.7.2. (✳).
Hint.
The local maximum is also a global maximum.
3.6.7.3. (✳).
Hint.
The sign of the first derivative is determined entirely by the numerator, but the sign of the second derivative depends on both the numerator and the denominator.
3.6.7.4. (✳).
Hint.
The function is odd.
3.6.7.5. (✳).
Hint.
The function is continuous at \(x=0\text{,}\) but its derivative is not.
3.6.7.6. (✳).
Hint.
Since you aren’t asked to find the intervals of concavity exactly, sketch the intervals of increase and decrease, and turn them into a smooth curve. You might not get exactly the intervals of concavity that are given in the solution, but there should be the same number of intervals as the solution, and they should have the same positions relative to the local extrema.
3.6.7.7. (✳).
Hint.
Use intervals of increase and decrease, concavity, and asymptotes to sketch the curve.
3.6.7.8.
Hint.
Although the function exhibits a certain kind of repeating behaviour, it is not periodic.
3.6.7.9. (✳).
Hint.
The period of this function is \(2\pi\text{.}\) So, it’s enough to graph the curve \(y=f(x)\) over the interval \([-\pi,\pi]\text{,}\) because that figure will simply repeat.
Use trigonometric identities to write \(f''(x)=-4(4\sin^2 x + \sin x -2)\text{.}\) Then you can find where \(f''(x)=0\) by setting \(y=\sin x\) and solving \(0=4y^2+y-2\text{.}\)
3.6.7.10.
Hint.
There is one point where the curve is continuous but has a vertical tangent line.
3.6.7.11. (✳).
Hint.
Use \(\displaystyle\lim_{x \rightarrow - \infty}f'(x)\) to determine \(\displaystyle\lim_{x \rightarrow - \infty}f(x)\text{.}\)
3.6.7.12. (✳).
Hint.
Once you have the graph of a function, reflect it over the line \(y=x\) to graph its inverse. Be careful of the fact that \(f(x)\) is only defined in this problem for \(x \geq 0\text{.}\)
3.6.7.14. (✳).
Hint.
For (a), don’t be intimidated by the new names: we can graph these functions using the methods learned in this section.
For (b), remember that to define an inverse of a function, we need to restrict the domain of that function to an interval where it is one-to-one. Then to graph the inverse, we can simply reflect the original function over the line \(y=x\text{.}\)
For (c), set \(y(x)=\cosh^{-1}(x)\text{,}\) so \(\cosh(y(x))=x\text{.}\) The differentiate using the chain rule. To get your final answer in terms of \(x\) (instead of \(y\)), use the identity \(\cosh^2(y)-\sinh^2(y)=1\text{.}\)

3.7 L’Hôpital’s Rule, Indeterminate Forms
3.7.4 Exercises

3.7.4.1.

Hint.
Try making one function a multiple of the other.

3.7.4.2.

Hint.
Try making one function a multiple of the other, but not a constant multiple.

3.7.4.3.

Hint.
Try modifying the function from Example 3.7.20.

3.7.4.4. (✳).

Hint.
Plugging in \(x=1\) to the numerator and denominator makes both zero. This is exactly one of the indeterminate forms where l’Hôpital’s rule can be directly applied.

3.7.4.5. (✳).

Hint.
Is this an indeterminate form?

3.7.4.6. (✳).

Hint.
First, rearrange the expression to a more natural form (without a negative exponent).

3.7.4.7. (✳).

Hint.
If at first you don’t succeed, try, try again.

3.7.4.8. (✳).

Hint.
Keep at it!

3.7.4.9.

Hint.
Rather than use l’Hôpital, try factoring out \(x^2\) from the numerator and denominator.

3.7.4.10. (✳).

Hint.
Keep going!

3.7.4.12.

Hint.
Try plugging in \(x=0\text{.}\) Is this an indeterminate form?

3.7.4.13.

Hint.
Simplify the trigonometric part first.

3.7.4.14.

Hint.
Rationalize, then remember your training.

3.7.4.15. (✳).

Hint.
If it is too difficult to take a derivative for l’Hôpital’s Rule, try splitting up the function into smaller chunks and evaluating their limits independently.

3.7.4.17. (✳).

Hint.
Try manipulating the function to get it into a nicer form

3.7.4.19.

Hint.
\(\lim\limits_{x \to 0}\sqrt[x^2]{\sin^2 x}=(\sin^2 x)^{\frac{1}{x^2}}\text{;}\) what form is this?

3.7.4.20.

Hint.
\(\lim\limits_{x \to 0}\sqrt[x^2]{\cos x} = \ds\lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}\)

3.7.4.21.

Hint.
logarithms

3.7.4.22.

Hint.
Introduce yet another logarithm.

3.7.4.23. (✳).

Hint.
If the denominator tends to zero, and the limit exists, what must be the limit of the numerator?

3.7.4.24. (✳).

Hint.
Start with one application of l’Hôpital’s Rule. After that, you need to consider three distinct cases: \(k \gt 2\text{,}\) \(k \lt 2\text{,}\) and \(k=2\text{.}\)

3.7.4.25.

Hint.
Percentage error: \(100\left|\frac{\mbox{exact}-\mbox{approx}}{\mbox{exact}}\right|\text{.}\) Absolute error: \(|\mbox{exact}-\mbox{approx}|\text{.}\) (We’ll see these definitions again in 3.4.25.)

4 Towards Integral Calculus
4.1 Introduction to Antiderivatives
4.1.2 Exercises

4.1.2.1.

Hint.
The function \(f(x)\) is an antiderivative of \(f'(x)\text{,}\) but it is not the most general one.

4.1.2.2.

Hint.
When \(f(x)\) is positive, its antiderivative \(F(x)\) is increasing. When \(f(x)\) is negative, its antiderivative \(F(x)\) is decreasing. When \(f(x)=0\text{,}\) \(F(x)\) has a horizontal tangent line.

4.1.2.3.

Hint.
For any constant \(n \neq -1\text{,}\) an antiderivative of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\text{.}\)

4.1.2.4.

Hint.
For any constant \(n \neq -1\text{,}\) an antiderivative of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\text{.}\)

4.1.2.5.

Hint.
For any constant \(n\neq -1\text{,}\) an antiderivative of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\text{.}\) The constant \(n\) does not have to be an integer.

4.1.2.6.

Hint.
What is the derivative of \(\sqrt{x}\text{?}\)

4.1.2.7.

Hint.
The derivative of \(e^{5x+11}\) is close to, but not exactly the same as, \(f(x)\text{.}\) Don’t be afraid to just make a guess. But be sure to check by differentiating your guess. If the derivative isn’t what you want, you will often still learn enough to be able to then guess the correct antiderivative.

4.1.2.8.

Hint.
From the table in the text, an antiderivative of \(\sin x\) is \(-\cos x\text{,}\) and an antiderivative of \(\cos x\) is \(\sin x\text{.}\)

4.1.2.9.

Hint.
What is the derivative of tangent?

4.1.2.10.

Hint.
What is an antiderivative of \(\dfrac{1}{x}\text{?}\)

4.1.2.11.

Hint.
\(\dfrac{7}{\sqrt{3-3x^2}}=\dfrac{7}{\sqrt{3}}\left(\dfrac{1}{\sqrt{1-x^2}}\right)\)

4.1.2.12.

Hint.
How is this similar to the derivative of the arctangent function?

4.1.2.13.

Hint.
First, find the antiderivative of \(f'(x)\text{.}\) Your answer will have an unknown \(+C\) in it. Figure out which value of \(C\) gives \(f(1)=10\text{.}\)

4.1.2.14.

Hint.
Remember that one antiderivative of \(\cos x\) is \(\sin x\) (not   \(-\sin x\)).

4.1.2.15.

Hint.
An antiderivative of \(\dfrac{1}{x}\) is \(\log(x)+C\text{,}\) but only for \(x\gt 0\text{.}\)

4.1.2.16.

Hint.
What is the derivative of the arcsine function?

4.1.2.17.

Hint.
If \(P(t)\) is the population at time \(t\text{,}\) then the information given in the problem is \(P'(t)=100e^{2t}\text{.}\)

4.1.2.18.

Hint.
You can’t know exactly —there will be a constant involved.

4.1.2.19.

Hint.
If \(P(t)\) is the amount of power in kW-hours the house has used since time \(t=0\text{,}\) where \(t\) is measured in hours, then the information given is that \(P'(t)=0.5\sin\left(\dfrac{\pi}{24}t\right)+0.25\) kW.

4.1.2.20. (✳).

Hint.
The derivatives should match. Remember \(\sin^{-1}\) is another way of writing arcsine.

4.1.2.21.

Hint.
Think about the product rule.

4.1.2.22.

Hint.
Think about the quotient rule for derivatives.

4.1.2.23.

Hint.
Notice that the derivative of \(x^3\) is \(3x^2\text{.}\)

4.1.2.24.

Hint.
Think about the chain rule for derivatives. You might need to multiply your first guess by a constant.

4.1.2.25.

Hint.
Simplify.

4.1.2.26.

Hint.
This problem is similar to Question 4.1.2.11, but you’ll have to do some harder factoring. Try getting \(f(x)\) into the form \(a\left(\dfrac{1}{\sqrt{1-(bx)^2}}\right)\) for some constants \(a\) and \(b\text{.}\)

4.1.2.27.

Hint.
Following Example 4.1.7 let \(V(H)\) be the volume of the solid formed by rotating the segment of the parabola from \(x=-H\) to \(x=H\text{.}\) The plan is to evaluate
\begin{equation*} V'(H) = \ds\lim\limits_{h\rightarrow H}\frac{V(H)-V(h)}{H-h} \end{equation*}
and then antidifferentiate \(V'(H)\) to find \(V(H)\text{.}\) Since you don’t know \(V(H)-V(h)\) (yet), first find upper and lower bounds on it when \(h \lt H\text{.}\) These bounds can be the volumes of two cylinders, one with radius \(H\) (and what height?) and the other with radius \(h\text{.}\)

4.1.2.28.

Hint.
Treat \(x\lt 0\) and \(x\gt 0\) separately.
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