1.1.1.
Hint.
Draw sketches. Don’t forget the range that the parameter runs over.
Appendix B Hints for Exercises
1 Curves
1.1 Derivatives, Velocity, Etc.
Exercises
1.1.2.
Hint.
Find the value of \(t\) at which the three points occur on the curve.
1.1.3.
Hint.
The curve “crosses itself” when \((\sin t,t^2)\) gives the same coordinate for different values of \(t\text{.}\) When these crossings occur will depend on which crossing you’re referring to, so your answers should all depend on \(t\text{.}\)
1.1.4.
Hint.
For part (b), find the position of \(P\) relative to the centre of the circle. Then combine your answer with part (a).
1.1.5.
Hint.
We aren’t concerned with \(x\text{,}\) so we can eliminate it by solving one equation for \(x\) as a function of \(y\) and \(z\) and plugging the result into the other equation.
1.1.6.
Hint.
To determine whether the particle is rising or falling, we only need to consider its \(z\)-coordinate.
1.1.7.
Hint.
This is the setup from Lemma 1.1.4. The two quantities you’re labelling are related, but different.
1.1.8.
Hint.
See the note just before Example 1.1.6.
1.1.9. (✳).
Hint.
To simplify your answer, remember: the cross product of \(\va\) and \(\vb\) is a vector orthogonal to both \(\va\) and \(\vb\text{;}\) the cross product of a vector with itself is zero; and two orthogonal vectors have dot product 0.
1.1.10.
Hint.
Evaluate \(\diff{}{t} |\vr(t)|^2\text{.}\)
1.1.11. (✳).
Hint.
Just compute \(|\vv(t)|\text{.}\) Note that \(\big(e^{at}+e^{-at}\big)^2
=e^{2at} + 2 + e^{-2at}\text{.}\)
1.1.12.
Hint.
To figure out what the path looks like, first concentrate on the \(x\)- and \(y\)-coordinates.
1.1.13. (✳).
Hint.
Review §1.5. The arc length should be positive.
1.1.14.
Hint.
\begin{equation*}
\int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t
\end{equation*}
The notation looks a little confusing at first, but we can break it down piece by piece: \(\diff{\vr}{t}(t)\) is a vector, whose components are functions of \(t\text{.}\) If we take its magnitude, we’ll get one big function of \(t\text{.}\) That function is what we integrate. Before integrating it, however, we should simplify as much as possible.
1.1.16.
Hint.
\(\vr(t)\) is the position of the particle, so its acceleration is \(r''(t)\text{.}\)
1.1.17. (✳).
Hint.
Review §1.5.
1.1.18. (✳).
Hint.
Review §1.1.
1.1.19.
Hint.
(a) First parametrize \(x^2+y^2=9\text{.}\)
1.1.20. (✳).
Hint.
If you got the answer \(0\) in part (b), you dropped some absolute value signs.
1.1.22. (✳).
Hint.
The integral you get can be evaluated with a simple substitution. You may want to factor the integrand first.
1.1.23.
Hint.
(b) \(\frac{1}{4x}+1+x\) is a perfect square.
(c), (d) Let
- \(\vr(x)\) be the position of the particle when its first coordinate is \(x\text{,}\)
- \(\vR(t)\) be the position of the particle at time \(t\text{,}\) and
- \(x(t)\) be the \(x\)--coordinate of the particle at time \(t\text{.}\)
Then \(\vR(t) = \vr\big(x(t)\big)\text{.}\) We are told \(|\vR'(t)|=9\) for all \(t\text{.}\)
1.1.24.
Hint.
Given the position of a particle, you can find its velocity.
1.1.25. (✳).
Hint.
If \(\vr(u)\) is the parametrization of \(\cC\) by \(u\text{,}\) then the position of the particle at time \(t\) is \(\vR(t) = \vr\big(u(t)\big)\text{.}\)
1.1.26. (✳).
Hint.
By Newton’s law, \(\vF=m\va\text{.}\)
1.1.27. (✳).
Hint.
Denote by \(\vr(x)\) the parametrization of \(C\) by \(x\text{.}\) If the \(x\)-coordinate of the particle at time \(t\) is \(x(t)\text{,}\) then the position of the particle at time \(t\) is \(\vR(t)=\vr\big(x(t)\big)\text{.}\) Also, though the particle is moving at a constant speed, it doesn’t necessarily have a constant value of \(\diff{\vx}{t}\text{.}\)
1.1.28.
Hint.
The question is already set up as an \(xy\)-plane, with the camera at the origin, so the vector in the direction the camera is pointing is \((x(t),y(t))\text{.}\) Let \(\theta\) be the angle the camera makes with the positive \(x\)-axis (due east). The tangent function gives a clean-looking relation between \(\theta(t)\text{,}\) \(x(t)\text{,}\) and \(y(t)\text{.}\)
1.1.29.
Hint.
Usng the Theorem of Pappus, the surface area and volume of this pipe are the same as that of a straight pipe with the same length and radius.
1.1.30.
Hint.
A helix can be parametrized by \(\vr(\theta)=a\cos\theta\,\hi+a\sin\theta\,\hj+b\theta\,\hk\text{.}\)
1.1.31.
Hint.
Define \(\vu(t)=e^{\alpha t}\frac{d\vr}{dt}(t)\) and substitute \(\diff{\vr}{t}(t)=e^{-\alpha t}\vu(t)\) into the given differential equation to find a differential equation for \(\vu\text{.}\)
1.2 Reparametrization
Exercises
1.2.1.
Hint.
You’re asked to find the arclength of the curve from \(s=1\) to \(s=t\text{.}\)
1.2.2.
Hint.
The arclength will be 0 at \(P\text{.}\)
1.2.3.
Hint.
\(\va(t_0)\) and \(\vb(s_0)\) describe the same point on \(\vR\text{.}\)
1.2.4. (✳).
Hint.
On your way to finding the relationship between \(t\) and arclength, you should realize that the curve has constant speed (with respect to \(t\)), though not constant velocity.
1.2.5. (✳).
Hint.
For which values of \(t\) is \(|\vr(t)|\le 1\text{?}\) Check the domain of \(t\) — we’re not starting at zero.
1.2.6.
Hint.
Be careful with the domain.
1.2.7.
Hint.
Remember \(\sqrt{x^2}=|x|\text{.}\) You will need to consider cases for this one.
1.3 Curvature
Exercises
1.3.1.
Hint.
The curve is a circle, so you don’t need to do any calculus.
1.3.2.
Hint.
Because \(\vr\) is a circle, you can parametrize it with respect to arclength without using an integral. You found \(\ka\) in Question 1.3.1.
1.3.3.
Hint.
When \(t\) is large, does the spiral locally look like a circle of large radius, or small?
1.3.4.
Hint.
\(\diff{s}{t}=\left| \vv(t)\right|=\left| \vr'(t)\right|\)
1.3.5.
Hint.
\(\hT=\frac{\vv(t)}{|\vv(t)|}=\frac{\vr'(t)}{|\vr'(t)|}\)
1.3.7.
Hint.
You can find the last two quantities by making use of the first three. Looking ahead, the formula list in Section 1.5 might come in handy.
1.3.8.
Hint.
We can calculate \(\ka = \dfrac{|\vv(t) \times \va(t)|}{\left| \left(\diff{s}{t}\right)^3\right|}\text{.}\) We can also figure out what kind of a shape our curve is.
1.3.9.
Hint.
The maximum and minimum values of \(\ka(t)\) should be obvious from your formula for \(\ka(t)\text{.}\)
1.3.11. (✳).
Hint.
For part (a), determine \(\vr(0)\text{,}\) \(\vr(\pi)\text{,}\) \(\vr(2\pi)\text{,}\) \(\vr(3\pi)\text{,}\) and \(\vr(4\pi)\text{,}\) to help you map out the motion. Also visualize the thumbtack as the wheel moves.
For part (d), use the fact that you only care about \(t=\pi\text{:}\) where is this on your sketch? What does that mean about the direction of \(\hN\text{?}\)
1.3.12.
Hint.
You should find that \(s=\theta\text{!}\)
1.3.13. (✳).
Hint.
Since \(\ka(x)\) is never negative, \(\ka(x)\) is maximum when \(\ka^2(x)\) is maximum. The latter is easier to compute.
1.4 Curves in Three Dimensions
Exercises
1.4.1.
Hint.
Use the right-hand rule to figure out how \(\hB\) is oriented.
1.4.2.
Hint.
Speed is the norm of velocity. Does that fit this equation?
1.4.3.
Hint.
Review Example 1.4.4 and remember that positive torsion indicates “right-handed twisting.” You shouldn’t actually need to calculate anything.
1.4.4.
Hint.
(a) Show that the tangent vector \(\hT(s)\) is a constant.
(b) Guess the plane. To do so, first show that the binormal \(\hB(s)\) is a constant. Then show that \((\vr(s)-\vr(0))\cdot\hB \) is a constant.
(c) Guess the circle. To do so, first show that \(\vr_c(s)=\vr(s)+\frac{1}{\ka(s)}\hN(s)\) is a constant.
1.4.5. (✳).
Hint.
It is not necessary to compute anything.
1.4.6. (✳).
Hint.
Both parts of this question make use of the quantity \(\diff{s}{t}\text{.}\)
1.4.7.
Hint.
\(\tau(t)=\dfrac{(\vv(t) \times \va(t))\cdot\diff{\va}{t}}{|\vv(t) \times \va(t)|^2}\)
1.4.8.
Hint.
Review §1.5.
1.4.9.
Hint.
The vector perpendicular to the plane containing the osculating circle is the binormal vector, \(\hB\text{.}\)
1.4.10. (✳).
Hint.
(a) The tangent vector of the curve is also a normal vector for the specified plane.
(b) Review §1.5.
1.4.11. (✳).
Hint.
Remember \(\va(t)=\difftwo{s}{t}(t)\,\hat\vT(t)
+\ka(t)\big(\diff{s}{t}(t)\big)^2\hat\vN(t)\text{.}\) Remember also that \(\hB\) is orthogonal to \(\hT\) and \(\hN\text{,}\) which are in the plane of \(C\text{.}\)
1.4.12. (✳).
Hint.
By Theorem 1.3.3, the tangential component of acceleration is \(a_T(t) = \difftwo{s}{t}\)
1.4.13. (✳).
Hint.
Use your answers to previous parts to calculate (d). Tangential and normal components of acceleration are defined just before Example 1.3.4.
1.4.14. (✳).
Hint.
(a) All points on the curve obey an equation that contains \(x\)’s and \(y\)’s, but no \(z\)’s. There is a standard way to get a nice parametrization of this equation, that doesn’t involve using square roots.
(b) You don’t need to compute the constants for all points: only the given point.
1.4.15. (✳).
Hint.
For part (c), you only need to find \(\hN\) at a point, which is easier than finding it for all \(t\text{.}\)
1.4.16. (✳).
Hint.
First parametrize \(x^2+y^2=1\) in the standard way. You don’t need calculus for part (c).
1.4.17. (✳).
Hint.
Review §1.5.
1.4.18. (✳).
Hint.
Since \(0 \le t \le 1\text{,}\) you can simplify \(|t|=t\text{.}\)
1.4.19. (✳).
Hint.
For part (f), remember that you can write the equation of a plane easily once you know a point it passes through, and a vector normal to it. The plane should touch the curve when \(t=0\text{,}\) and the plane should contain \(\hT\) and \(\hN\text{.}\)
1.4.20. (✳).
Hint.
It might be easier to find \(\hB\) before you find \(\hN\text{,}\) then use the formula \(\hN(t)=\hB(t)\times\hT(t)\text{.}\)
1.4.21. (✳).
Hint.
The osculating plane at \(\vr(t_0)\) is the plane through \(\vr(t_0)\) with normal \(\hat\vB(t_0)\text{.}\) Also, notice the points for parts (a) and (b) are not the same.
1.4.22. (✳).
Hint.
Since \(t \gt 0\text{,}\) we can simplify \(\sqrt{t^2}=|t|=t\text{.}\)
1.4.23. (✳).
Hint.
In this context, “distance travelled” means “arclength.”
1.4.24. (✳).
Hint.
Use \(\hT\) and \(\hN\) to compute \(\hB\text{.}\)
1.4.25.
Hint.
(a) First find a parametrization \(\big(x(\theta),y(\theta)\big)\) for \(x^2+y^2=1\text{.}\)
1.4.26. (✳).
Hint.
You need to find the acceleration at \((1,1,1)\text{.}\) Think about what strategies are available for computing the acceleration.
1.4.27. (✳).
Hint.
For part (a), \(\vT(t)\) will be a vector of the form \(\vT(t) = \frac{(1,at,bt)}{\sqrt{1+4t^2}}\) where \(a\) and \(b\) are nonzero constant real numbers.
For part (b), \(\vN(t)\) will be a vector of the form \(\vN(t) = \frac{(-4t,\alpha, \beta)}{2\sqrt{1+4t^2}}\) where \(\alpha\) and \(\beta\) are nonzero constant real numbers.
For part (e), \(\kappa(t)\) will be a function of the form \(\kappa(t) = \frac{\gamma}{{(1+4t^2)}^{3/2}}\text{,}\) where \(\gamma\) is a positive constant real number.
1.4.28. (✳).
Hint.
Differentiate \(\hat\vN =\hat\vB\times\hat\vT\) with respect to \(s\text{.}\)
The vectors \(\hN, \hB,\) and \(\hT\) form a right-handed triple. Sketch them (the same way you might sketch the \(x\text{,}\) \(y\text{,}\) and \(z\) axes) to figure out the signs of their cross products.
1.4.29. (✳).
Hint.
In part (b), note that \(\va\) is the second derivative with respect to time (not \(\theta\)). Exploit \(\va = \diff{v}{t}\hat\vT + v^2\kappa\hat\vN\) to find what you’re asked for.
1.4.30. (✳).
Hint.
For part (d), what is the relationship between the \(y\)- and \(z\)-components of the particle’s position? How can you use that to find a plane containing the particle at all times \(t\text{?}\)
1.4.31. (✳).
Hint.
Rather than trying to wrangle trig identities, plug in \(\theta=\pi\) as soon as you can for part (a). For part (c), remember that you need the chain rule if you want to make use of your previous derivatives.
1.6 Integrating Along a Curve
Exercises
1.6.1.
Hint.
Your differential is \(\dee{s}\text{,}\) where \(s\) is arclength.
1.6.2.
Hint.
(a) You can parametrize the curve by \(\vr(\theta) = r(\theta)\,\cos\theta\,\hi +
r(\theta)\,\sin\theta\,\hj\text{,}\) \(\theta_1\le \theta\le\theta_2\text{.}\)
1.6.3.
Hint.
Following Definition 1.6.1, set \(f(x,y,z)=\frac{xy}{z}\text{,}\) \(x(t)=\frac23 t^3\text{,}\) \(y(t)=\sqrt3t^2\text{,}\) and \(z(t)=3t\text{.}\)
1.6.4.
Hint.
Parametrize the circle in the usual way.
1.6.5.
Hint.
\(C\) can be parametrized as \((1+t,2+2t,3+2t)\) for \(0 \le t \le 1\text{.}\)
1.6.7.
Hint.
Simplify! Also: \(\diff{}{t}\{\arcsec t\} = \frac{1}{|t|\sqrt{t^2-1}}\text{.}\)
1.6.8. (✳).
Hint.
Sketch \(C\) and determine the normal vectors from the sketch. You can use \(x\) or \(y\) as the integration variable in your integrals.
1.6.9. (✳).
Hint.
(c) How is \(x(t)^2+y(t)^2\) related to \(z(t)\text{?}\)
(d) First, sketch \(\big(x(t)\,,\,y(t)\big)\text{.}\)
1.6.10.
Hint.
Remember \(\bar x = \dfrac{\int_C x\rho\,\dee{s}}{\int_C \rho\,\dee{s}}\text{,}\) etc. The integrals you evaluate should all be straightforward applications of the power rule.
1.7 Sliding on a Curve
1.7.4 Exercises
1.7.4.1.
Hint.
Gravity pulls straight down, while the direction of the normal force depends on the curve of the wire. There is not enough information to know the magnitude of the forces, but you can approximate their directions.
1.7.4.2.
Hint.
This equation stems from \(\vF=m\va\text{.}\) In that equation, \(\va\) is what kind of derivative?
1.7.4.3.
Hint.
A thought experiment might help you avoid any calculations. If the wire were perfectly vertical or perfectly horizontal, what would \(W\hN\) be?
1.7.4.4.
Hint.
The skater reaches their highest point when \(|\vv|=0\text{.}\)
1.7.4.5.
Hint.
The highest vertical height occurs just as the skateboarder’s speed reduces to 0, at \(y_S=\frac{E}{mg}\text{.}\)
1.7.4.6.
Hint.
At the bottom of the culvert, all the skater’s energy is kinetic, not potential. That is, in the equation \(E=\frac12m|\vv|^2+mgy\text{,}\) we have \(y=0\text{.}\)
1.7.4.7.
1.7.4.8.
Hint.
When \(\theta=13\pi/3\text{,}\) \(\ddiff{2}{s}{\theta}=0\text{,}\) which is handy for a quicker calculation.
1.7.4.9.
Hint.
According to the equation in §1.7.2, the skiier will become airborne when:
\begin{equation*}
|\vv| \gt \sqrt{\frac{g}{\kappa}|\hj\cdot\hN|}
\end{equation*}
So, we need \(|\vv|\) to be greater than \(\sqrt{\frac{g}{\kappa}|\hj\cdot\hN|}\) for some point on the curve inside the range \(1/e \le t \le e\text{.}\)
Note that \(g\) is given in metres per second, while the other quantities are in kilometres and hours.
1.7.4.10.
Hint.
There are now three forces acting on the bead: one parallel to \(\hj\) (exerted by gravity), one parallel to \(\hN\) (exerted by the wire), and one parallel to \(\hT\) (exerted by the jet pack).
Follow the reasoning in the sliding bead section of the text, focusing on the tangential forces.
1.7.4.11.
Hint.
If the snowmachine is moving at a constant speed, the tangential component of its acceleration is zero. Part (a) is similar to Question 1.7.4.10.
1.7.4.12.
Hint.
Follow the discussion in §1.7.3.
It’s fine to leave part (b) pretty messy. Your answer for part (c) involves the root of a cubic function, but you don’t need a high degree of accuracy to decide between the three options given.
1.8 Optional — Polar Coordinates
Exercises
1.8.2.
Hint.
\(r\) is allowed to be negative.
1.8.3.
Hint.
Compute, for each angle \(\theta\text{,}\) the dot product \(\he_r(\theta)\cdot\he_\theta(\theta)\text{.}\)
1.8.5.
Hint.
The curve can be parametrized by \(\vr(\theta)= f(\theta)\big[\cos\theta\ \hi + \sin\theta\ \hj\big]\)
1.9 Optional — Central Forces
Exercises
2 Vector Fields
2.1 Definitions and First Examples
Exercises
2.1.1.
Hint.
Not all blanks represent a single interval.
2.1.2.
Hint.
Write down all coordinates where \(\vv(x,y)\cdot\hi=0\) or \(\vv(x,y)\cdot\hj=0\text{,}\) and look for a pattern.
2.1.3.
Hint.
If you know the speed and direction of an object, you can find its velocity.
2.1.5.
Hint.
When the twig is at \((x,y)\) it has velocity \(\vv(x,y)\text{.}\)
2.1.6.
Hint.
Whenever the twig is on the \(y\)-axis, its velocity is parallel to the \(y\)-axis. So it remains on the \(y\)-axis for all time.
2.1.7.
Hint.
Set your face to be at the origin, \((0,0,0)\text{.}\)
If \(A\) is “inversely proportional” to \(B\text{,}\) then there exists a constant \(\alpha\) such that \(AB=\alpha\text{.}\) That way when \(|B|\) goes up, \(|A|\) goes down, and vice-versa.
2.1.8.
Hint.
Start with the regions where \(\vv(x,y)\cdot\hi\) and \(\vv(x,y)\cdot\hj\) are positive and negative. As you move up/down/left/right, do the vectors get longer or shorter? More horizontal or more vertical?
2.1.9.
Hint.
\(\vv(x,y)\cdot \hi\) is the distance from \((x,y)\) to the origin, while \(\vv(x,y)\cdot\hj\) is the distance from \((x,y)\) to the point \((1,1)\text{.}\)
2.1.10.
Hint.
Factor \(x^2+xy=x(x+y)\) and \(y^2-xy=y(x-y)\text{.}\) Chop the plane up into eight regions using the two coordinate axes and the lines \(y=x\text{,}\) \(y=-x\text{.}\)
2.1.11.
Hint.
What is the geometric interpretation of each summand?
2.1.12.
Hint.
(a), (c) Intrepret the vector field geometrically.
2.1.13.
Hint.
The constant \(G\) is the same for all masses, but \(M\) differs. The net force is the sum of three force vectors.
2.1.14.
Hint.
For part a., make a triangle with \(P\) as one of its vertices that is similar to the triangle made by the pole, the wall, and the ground. Its hypotenuse has length \(p\text{;}\) let its base be \(b\) and its height be \(h\text{.}\) Find a way to translate between \((b,h)\) and \((x,y)\text{.}\)
For part b., use your answer from part a. Start by describing a point on a pole as its distance from the lower end of the pole, \(p\text{.}\) Then, consider \(\diff{z}{t}\) and \(\left(\diff{x}{t},\diff{y}{t}\right)\) separately. If you’re having a hard time simplifying your answer, note \(\sqrt{x^2+y^2}=\sqrt3(1-z)\) for any point \((x,y,z)\) on a pole when \(H=1\text{.}\)
2.2 Optional — Field Lines
2.2.2 Exercises
2.2.2.2.
Hint.
Express \(x'(t)\) and \(y'(t)\) purely in terms of \(x(t)\) and \(y(t)\text{.}\)
2.2.2.3. (✳).
Hint.
Review §2.2.
2.3 Conservative Vector Fields
Exercises
2.3.1.
Hint.
Carefully consider the context that lead to each of these equations.
2.3.2.
Hint.
One of the three options will NEVER be true, for any \(\vF\text{.}\)
2.3.3.
Hint.
Modify \(\varphi\text{,}\) the potential for \(\vF\text{.}\)
2.3.4.
Hint.
a. If \(\vF+\vG\) is conservative, what has to be true?
b. What if \(\vF\) and \(\vG\) are quite similar?
c. Find a potential for \(\vF+\vG\text{.}\)
2.3.5. (✳).
Hint.
Note that the domain is \(D=\Set{(x,y)}{x \gt 1}\text{.}\) Compare to Example 2.3.14.
2.3.6.
Hint.
A potential does exist.
2.3.7.
Hint.
Recall \(\diff{}{x} \ln |x| = \frac1x\text{.}\)
2.3.8.
Hint.
Try the screening test, Theorem 2.3.9.
2.3.9.
Hint.
\(\displaystyle\int\frac{x}{x^2+y^2+z^2}\,\dee{x}\) can be evaluated by inspection, or with the substitution \(u=x^2+y^2+z^2\text{.}\)
2.3.11.
Hint.
For what values of the constants \(A\) and \(B\) does the vector field \(\vF\) pass the screening test \(\vnabla\times\vF=\vZero\text{?}\)
2.3.12.
Hint.
Review Example 2.1.2.
2.3.13.
Hint.
Following Example 2.3.3, the particle can never escape the region
\begin{equation*}
\Set{(x,y,z)}{\varphi(x,y,z)\ge -E}
\end{equation*}
where \(E\) is the energy of the system.
2.3.14.
Hint.
Example 2.3.3 tells us \(\frac{1}{2} m |\vv(t)|^2 -\varphi\big(x(t),y(t),z(t)\big)=E\) is a constant quantity, provided \(\vF\) is conservative with potential \(\varphi(x,y,z)\text{.}\)
2.3.15.
Hint.
Find a potential \(\varphi\text{.}\) Notice \(f\text{,}\) \(g\text{,}\) and \(h\) are functions of one variable each — this simplifies things.
2.3.16.
Hint.
Write the points with curl \(\mathbf 0\) as multiples of a constant vector.
2.4 Line Integrals
2.4.2 Exercises
2.4.2.1.
Hint.
The top and bottom of the square can be easily paramerized using \(x\) as the parameter. The other two sides can be easily parameterized using \(y\) as the parameter.
2.4.2.2.
2.4.2.3.
Hint.
Please don’t do any computation, especially not to find \(C\text{!}\)
2.4.2.4.
Hint.
Review properties of conservative vector fields.
2.4.2.5. (✳).
2.4.2.6.
Hint.
Review Theorem 2.4.7.
2.4.2.7. (✳).
Hint.
Part (d) is a hint.
2.4.2.8.
Hint.
The last part of the question is a huge hint.
2.4.2.11. (✳).
Hint.
Parametrize the curve using \(y\) as a parameter.
2.4.2.12.
Hint.
(a) Use Theorem 2.4.8.
(c) You may parametrize the curve using \(x\) as the parameter. Exploit the fact that, for the value of \(\la\) found in part (a), \(\vF+\la \vG\) is conservative.
2.4.2.14. (✳).
Hint.
Parametrize the path using sines and cosines. The work done is \(\int_C \vF \cdot \dee{\vr}\)
2.4.2.15. (✳).
Hint.
Is \(\vF\) conservative?
2.4.2.16. (✳).
Hint.
Is \(\vF= xy\,\hj\) conservative? Sketch \(C\text{.}\)
2.4.2.17. (✳).
Hint.
That the line integral is to be independent of path is a huge hint.
2.4.2.18. (✳).
Hint.
Note that
- \(y=0\) on the line segment from \((1,0,0)\) to \((0, 0, 1)\) and
- \(x=0\) on the line segment from \((0,0,1)\) to \((0, 1, 0)\) and
- \(z=0\) on the line segment from \((0, 1, 0)\) to \((1, 0, 0)\)
2.4.2.19. (✳).
Hint.
That \(\vF\) is conservative should be a dead giveaway.
2.4.2.20. (✳).
Hint.
To calculate the integral, it might be easier to find a potential for \(\vF\) and use Theorem 2.4.2.
2.4.2.21. (✳).
Hint.
Your answer from (b) can help you in (c). Also, \(\cos(1)=\cos(-1)\text{,}\) because cosine is an even function.
2.4.2.22. (✳).
Hint.
Review §2.4.1 of the text.
2.4.2.23. (✳).
Hint.
Relate the integral of part (d) to the integral of part (c).
2.4.2.24. (✳).
Hint.
Write the integral of part (c) as \(\int_C\vG\cdot\dee{\vr}\text{.}\) What is the difference between \(\vG\) and \(\vF\text{?}\)
2.4.2.25. (✳).
Hint.
(d) How are \(\vG\) and \(\vF\) related?
2.4.2.26. (✳).
Hint.
(a) Start with \(\pdiff{f}{z} = y^2e^{yz}\text{.}\)
(b) Use the result of part (a) to do part (b).
2.4.2.27. (✳).
Hint.
The integral in part (b) is path independent. That’s a big hint.
2.4.2.28. (✳).
Hint.
Part (a) is a hint for part (b).
2.4.2.29. (✳).
Hint.
The three parts of this problem are closely related.
2.4.2.30. (✳).
Hint.
We can rewrite \(x^2 + y^2 + z^2 = 2z\) as \(x^2+y^2+(z-1)^2=1\text{.}\)
2.4.2.31. (✳).
Hint.
Newton’s law of motion is \(\vF=m\va\text{.}\) The work done over a displacement \(\dee{\vr}\) is \(W=\vF \cdot \dee{\vr}\text{.}\)
2.4.2.32.
Hint.
(a) The curve can be easily parametrized by using \(x\) as a parameter.
(b) Don’t evaluate the integral directly.
2.4.2.33. (✳).
Hint.
Refer to Example 1.4.4 for a parametrization of a helix.
2.4.2.34. (✳).
Hint.
(b) Parametrize each side of the square by arc length, and make use of the plentiful zeroes that arise.
2.4.2.35. (✳).
Hint.
Force is mass times acceleration, where acceleration is the second derivative of position, \(\vr(t)\text{,}\) with respect to time, \(t\text{.}\) The work done by \(\vF\) between time \(a\) and time \(b\) is \(\int_a^b \vF\cdot\dee{\vr}\text{.}\)
2.4.2.36. (✳).
Hint.
Note that the curve goes from \((2,2)\) to \((1,1)\) — not the other way around.
For part (b), one possibility is to look for a path consisting of the line segment from \((2,2)\) to \((2,Y)\text{,}\) followed by the line segment from \((2,Y)\) to \((1,Y)\text{,}\) followed by the line segment from \((1,Y)\) to \((1,1)\text{,}\) with \(Y\) being a parameter to be determined.
2.4.2.37. (✳).
Hint.
One possibility is to look for a path consisting of the line segment from \((0,0)\) to \((0,Y)\text{,}\) followed by the line segment from \((0,Y)\) to \((2,Y)\text{,}\) followed by the line segment from \((2,Y)\) to \((2,0)\text{,}\) with \(Y\) being a parameter to be determined.
2.4.2.38. (✳).
Hint.
Is \(\vF\) conservative?
2.4.2.39. (✳).
Hint.
On \(S\text{,}\) note \(z = 2 + x^2 - 3 y^2\text{.}\) Further, the vector field \(\tilde\vF(x,y,z) = z^2\,\hk\) is conservative (with potential \(\frac{1}{3} z^3\)), so \(\int_{C_1} \tilde\vF \cdot \dee{\vr}= \int_{C_2} \tilde\vF \cdot \dee{\vr}\) for any two curves \(C_1\) and \(C_2\) from \(P_1\) to \(P_2\text{.}\) Compare this to Questions 2.4.2.24 through 2.4.2.25.
2.4.2.40. (✳).
Hint.
Simplify the answer in part (a) as much as possible.
For part (c), start with \(\pdiff{f}{y} = xe^{3x^2}\) and \(\pdiff{f}{z} = x^2 \cos(x^2 z) \text{.}\)
For part (d), notice the difference between the given vector field and the conservative vector field of part (c). The resulting integral can be directly evaluated using methods from integral calculus.
2.4.2.41. (✳).
Hint.
For (b), remember \(\diff{s}{t}=\left| \diff{\vr}{t}\right|\)\\ Is the vector field of part (c) conservative?
2.4.2.42. (✳).
Hint.
For part (d), what is the difference between \(J\) and \(\int_\cC\vF\cdot\dee{\vr}\text{?}\)
For part (e), many parts of the integral are zero: find as many as you can.
2.4.2.43. (✳).
Hint.
By Newton’s law of motion, \(m\vr''(t)=\vF(t)\text{.}\)
Recall \(\ka(t) = \frac{|\vr'(t)\times\vr''(t)|}{|\vr'(t)|^3}\text{.}\)
2.4.2.44. (✳).
Hint.
(a) Remember the arclength of the parametrized path \(\vr(t)\) from \(t=a\) to \(t=b\) is given by \(\int_a^b |\vr'(t)|\ \dee{t}\text{.}\) In this case, \(|\vr'(t)|\) can be simplified considerably.
(b) Remember \(\ka(t) = \frac{|\vr'(t)\times\vr''(t)|}{|\vr'(t)|^3}\text{.}\)
(c) Gravity is conservative. Friction is not conservative.
(d) What are the tangential and normal components of acceleration?
3 Surface Integrals
3.1 Parametrized Surfaces
Exercises
3.1.1.
Hint.
Your answer will have the form \(\vr(x,y)= \psi_1(x,y)\hi+ \psi_2(x,y)\hj+ \psi_3(x,y)\hk\text{.}\)
3.1.3. (✳).
Hint.
First think about what properties \(\vr(u,v)\) has to have in order to be a parametrization.
3.1.4. (✳).
Hint.
First think about what properties \(\vr\) has to have in order to be a parametrization.
3.1.5. (✳).
Hint.
First think about what properties \(\vr(u,v)\) has to have in order to be a parametrization.
3.2 Tangent Planes
Exercises
3.2.1.
Hint.
What are the tangent planes to the two surfaces at \((0,0,0)\text{?}\)
3.2.2.
Hint.
Apply the chain rule to \(G\big(\vr(t)\big)=0\text{.}\)
3.2.4.
Hint.
To find a tangent vector to the curve of intersection of the surfaces \(F(x,y,z)=0\) and \(G(x,y,z)=0\) at \((x_0,y_0,z_0)\text{,}\) use Q[3.2.2] twice, once for the surface \(F(x,y,z)=0\) and once for the surface \(G(x,y,z)=0\text{.}\)
3.2.5.
Hint.
To find a tangent vector to the curve of intersection of the surfaces \(z=f(x,y)\) and \(z=g(x,y)\) at \((x_0,y_0,z_0)\text{,}\) use Q[3.2.2] twice, once for the surface \(z=f(x,y)\) and once for the surface \(z=g(x,y)\text{.}\)
3.2.10. (✳).
Hint.
Review §3.2.
3.2.11. (✳).
Hint.
Review §3.2.
3.2.13. (✳).
Hint.
Let \((x,y,z)\) be a desired point. Then
- \((x,y,z)\) must be on the surface and
- the normal vector to the surface at \((x,y,z)\) must be parallel to the plane’s normal vector.
3.2.14. (✳).
Hint.
First find a parametric equation for the normal line to \(S\) at \((x_0,y_0,z_0)\text{.}\) Then the requirement that \((0,0,0)\) lies on that normal line gives three equations in the four unknowns \(x_0,y_0,z_0\) and \(t\text{.}\) The requirement that \((x_0,y_0,z_0)\) lies on \(S\) gives a fourth equation. Solve this system of four equations.
3.2.15. (✳).
Hint.
Two (nonzero) vectors \(\vv\) and \(\vw\) are parallel if and only if there is a \(t\) such that \(\vv=t\,\vw\text{.}\) Don’t forget that the point has to be on the hyperboloid.
3.2.16. (✳).
Hint.
(b) If \(\vv\) is tangent, at a point \(P\text{,}\) to the curve of intersection of the surfaces \(S_1\) and \(S_2\text{,}\) then \(\vv\)
- has to be tangent to \(S_1\) at \(P\text{,}\) and so must be perpendicular to the normal vector to \(S_1\) at \(P\) and
- has to be tangent to \(S_2\) at \(P\text{,}\) and so must be perpendicular to the normal vector to \(S_2\) at \(P\text{.}\)
3.2.17. (✳).
Hint.
The angle between the curve and the surface at \(P\) is \(90^\circ\) minus the angle between the curve and the normal vector to the surface at \(P\text{.}\)
3.2.18.
Hint.
At the highest and lowest points of the surface, the tangent plane is horizontal.
3.3 Surface Integrals
3.3.6 Exercises
3.3.6.1.
Hint.
\(S\) is a very simple geometric object.
3.3.6.2.
Hint.
The triangle is part of the plane \(\frac{x}{a}+\frac{y}{b} +\frac{z}{c}=1\text{.}\)
3.3.6.3.
Hint.
Flatten \(S\) out.
3.3.6.8. (✳).
Hint.
The total surface area of (b) (ii) can be determined without evaluating any integrals.
3.3.6.9.
Hint.
Rewrite the equation of the cone in the form \(y=h(x,z)\text{.}\)
3.3.6.11.
Hint.
On \(S\text{,}\) \((x,y)\) runs over the interior of \(x^2+y^2=2x\text{,}\) or equivalently, the interior of \((x-1)^2+y^2=1\text{.}\)
3.3.6.12.
Hint.
See Example 3.1.5 for a parametrization of the torus.
3.3.6.13.
Hint.
Call the part of the sphere in the first octant \(S\text{.}\) By definition, the centroid is \((\bar x,\bar y,\bar z)\) with
\begin{equation*}
\bar x = \frac{\dblInt_S x\ \dee{S}}{\dblInt_S \ \dee{S}}\qquad
\bar y = \frac{\dblInt_S y\ \dee{S}}{\dblInt_S \ \dee{S}}\qquad
\bar z = \frac{\dblInt_S z\ \dee{S}}{\dblInt_S \ \dee{S}}
\end{equation*}
The integrals will be easy if you use spherical coordinates. You can reduce the number of integrals evaluated by using symmetry.
3.3.6.14.
Hint.
Before parametrizing the cylinder, express \(x^2+y^2=2ay\) in cylindrical coordinates.
3.3.6.16.
Hint.
(a) The integral can be easily evaluated by using that the sphere has surface area \(4\pi a^2\text{.}\)
(c) Use cylindrcial coordinates for the top part of the cone.
3.3.6.20. (✳).
Hint.
Beware of signs. Note that \(0\le z\le 1\) on \(\cS\text{.}\)
3.3.6.21. (✳).
Hint.
The \(z\)-coordinate of the centre of mass is the weighted average of the \(z\)-coordinate over the cone. Since a density has not been specified, we assume that it is a constant. We may take the density to be \(1\text{,}\) so the \(z\)-coordinate of the centre of mass is \(\dblInt_S z\,\dee{S}/
\dblInt_S \dee{S}\text{.}\)
3.3.6.22. (✳).
Hint.
Use cylindrcal coordinates. Note that because of the symmetry of the cone, only the \(z\)-component of the centre of mass requires an integral to be calculated. The \(z\)-coordinate of the centre of mass is the weighted average of the \(z\)-coordinate over the cone. That is \(\bar z =\dblInt_S z\,\dee{S}/\dblInt_S\dee{S}\text{.}\)
3.3.6.24. (✳).
Hint.
Review (3.3.2).
3.3.6.25. (✳).
Hint.
Don’t be afraid to tweak spherical coordinates so as to fit the condition \(x \ge \sqrt{y^2 + z^2}\) well. To do so, first use a sketch to develop a geometric interpretation of \(\frac{\sqrt{y^2 + z^2}}{x}\text{.}\)
3.3.6.26. (✳).
Hint.
The surface \(S\) may be parametrized by observing that, for each fixed \(y\text{,}\) \(x^2 + z^2 = \sin^2y\) is a circle.
3.3.6.27. (✳).
Hint.
By symmetry, the centre of mass will lie on the \(z\)-axis. By definition, the \(z\)-coordinate of the centre of mass is the weighted average of \(z\) over \(S\text{,}\) which is
\begin{equation*}
\bar z = \frac{\dblInt_S z\,\rho(x,y,z)\ \dee{S}}
{\dblInt_S \rho(x,y,z)\ \dee{S}}
\end{equation*}
3.3.6.37.
Hint.
You can use the cylindrical coordinates \(\theta\) and \(z\) to parametrize the hyperboloid.
4 Integral Theorems
4.1 Gradient, Divergence and Curl
4.1.6 Exercises
4.1.6.2.
Hint.
Compute \(\vnabla\times \vF\) for some simple vector fields.
4.1.6.3.
Hint.
For parts(a) and (b), write out the definitions of the left and right hand sides and observe that they are equal. Part (c) can be done easily by using other, simpler, vector identities.
4.1.6.6. (✳).
Hint.
(c) can be done efficiently by using (a) and (b).
4.1.6.12.
Hint.
(a) Find the magnitude and direction of the velocity vector. Then verify that \(\vOm\times\vr\) has that magnitude and direction.
4.2 The Divergence Theorem
4.2.6 Exercises
4.2.6.3.
Hint.
(b) The integral can be trivially evaluated by exploiting oddness and the fact that \(\tripInt_V\ \dee{V}=\text{Volume}(V)\text{.}\)
4.2.6.4.
Hint.
For part (a), use spherical coordinates.
4.2.6.5.
Hint.
(a) The integral is easier in polar coordinates.
(b) Since \(x\) is odd, \(\tripInt_V x\,\dee{V}=0\text{.}\)
4.2.6.6.
Hint.
(a) The integral is easy in polar coordinates.
(b) The volume of the solid can be easily computed by decomposing the solid into thin horizontal pancakes. See Section 1.6 in the CLP-2 text.
4.2.6.7. (✳).
Hint.
The divergence theorem, of course.
4.2.6.8.
Hint.
It’s easier to use the divergence theorem. But don’t forget the base of the silo.
4.2.6.9.
Hint.
The divergence theorem, of course. The integral can be easily evaluated by using that, for any solid \(\cV\) in \(\bbbr^3\text{,}\)
\begin{gather*}
\tripInt_\cV\dee{V}=\text{Volume}(\cV)
\end{gather*}
and
\begin{gather*}
\bar x =\frac{\tripInt_\cV x\,\dee{V}}{\text{Volume}(\cV)}\qquad
\bar y =\frac{\tripInt_\cV y\,\dee{V}}{\text{Volume}(\cV)}\qquad
\bar z =\frac{\tripInt_\cV z\,\dee{V}}{\text{Volume}(\cV)}
\end{gather*}
where \((\bar x,\bar y,\bar z)\) is the centroid of \(\cV\text{.}\)
4.2.6.10. (✳).
Hint.
The complexity of \(\vF\) is a hint that the flux should not be evaluated directly.
4.2.6.11. (✳).
Hint.
The specified surface is not closed.
4.2.6.12. (✳).
Hint.
(a), (b), (c) Review warning 4.2.3.
(d) The divergence theorem can be used — with care.
(e) The equation can be made more understandable by completing a square.
4.2.6.13. (✳).
Hint.
(a) Use a suitable modification of spherical coordinate. Do not forget to specify the range of the parameters.
4.2.6.14. (✳).
Hint.
Don’t evaluate the flux directly.
4.2.6.15. (✳).
Hint.
For practice, try doing this question twice — once using the divergence theorem and once using direct evaluation.
4.2.6.16. (✳).
Hint.
The question highlights that the vector field has divergence \(0\text{.}\) Thta’s a big hint.
4.2.6.17. (✳).
Hint.
As \(\vF\) looks complicated, it is probably wise not to try and evaluate the flux integral directly.
4.2.6.18. (✳).
Hint.
As \(\vF\) looks complicated, it is probably wise not to try and evaluate the flux integral directly.
4.2.6.19. (✳).
Hint.
The vector field \(\vF\) looks complicated. Try to avoid a direct evaluation of the flux integral.
4.2.6.20. (✳).
Hint.
The divergence of \(\vF\) is a lot simpler than \(\vF\) itself. By default, we want the outward flux.
4.2.6.21. (✳).
Hint.
The vector field \(\vF\) looks very complicated. That strongly suggests that we not evaluate the integral directly.
4.2.6.22.
Hint.
The divergence of \(\vF\) is a lot simpler than \(\vF\) itself.
4.2.6.23. (✳).
Hint.
Note that \(\vF(x,y,z)\) is not defined at \((x,y,z)=(0,0,0)\text{.}\)
4.2.6.25. (✳).
Hint.
The surface S is not a closed surface.
4.2.6.29. (✳).
Hint.
The complexity of \(\vF\) is a hint that the flux should not be evaluated directly.
4.2.6.31. (✳).
Hint.
The flux can be calculated directly, but it is rather easier to calculate it using the Divergence Theorem.
4.2.6.32. (✳).
Hint.
Use that \(y\) is odd to easily evaluate some integrals.
4.2.6.34.
Hint.
(a) Use cylindrical coordinates.
(b) The volume of the \(V\) can be easily computed by decomposing \(V\) into thin horizontal washers. See Section 1.6 in the CLP-2 text.
4.2.6.35.
Hint.
Review the derivation of the heat equation in Section 4.2.1.
4.2.6.37. (✳).
Hint.
Make a judicious choice of parametrization.
4.2.6.38. (✳).
Hint.
Do not compute the integral directly.
4.2.6.39. (✳).
Hint.
Be careful about which normals to use in part (c). For practice, try to do part (c) in two different ways, with one being direct evaluation.
4.2.6.40. (✳).
Hint.
For part (b), do not evaluate the flux directly. In part (c), the flux can be related to the volume enclosed by the surface, and the centre of mass of the volume enclosed by the surface.
4.2.6.41. (✳).
Hint.
(b) We have several different methods for evaluating flux integrals. Think about what would be involved in applying each of them before settling on which one to use.
(c) Be sneaky — don’t evaluate this integral directly.
4.2.6.42. (✳).
Hint.
For parts (b) and (c), write out carefully the integral that the divergence theorem gives you.
4.2.6.43. (✳).
Hint.
Note that \(2^2+2(1^2)+3(1)^2 = 9 \lt 16\) so that \((2,1,1)\) is inside \(S\text{,}\) while \(3^2+2(2^2)+3(2)^2 = 29 \gt 16\) so that \((3,2,2)\) is outside \(S\text{.}\)
4.2.6.44. (✳).
Hint.
Review §4.2.2.
4.2.6.46. (✳).
Hint.
Consider very small \(a\)’s.
4.2.6.47. (✳).
Hint.
Carefully draw a side view of \(S\text{.}\)
4.2.6.48. (✳).
Hint.
Both the divergence theorem and a vector identity in Theorem 4.1.4 are useful.
4.2.6.49. (✳).
Hint.
\(x\) is an odd function.
4.2.6.50. (✳).
Hint.
You should be able to guess the centre of mass, \((\bar x,\bar y)\) of the disk \(D\text{.}\) Then the integrals \(\dblInt_D x\,\dee{x}\dee{y}\) and \(\dblInt_D y\,\dee{x}\dee{y}\) can be found by using \(\bar x = \frac{\dblInt_D x\,\dee{x}\dee{y}}{\dblInt_D \dee{x}\dee{y}}\) and \(\bar y = \frac{\dblInt_D y\,\dee{x}\dee{y}}{\dblInt_D \dee{x}\dee{y}}\text{.}\)
4.2.6.51. (✳).
Hint.
The divergence of \(\vF\) is a lot simpler than \(\vF\) itself. But beware of the singularity in \(\tan z\) at \(z=\tfrac{\pi}{2}\text{.}\)
4.3 Green’s Theorem
Exercises
4.3.2.
Hint.
Let \(\vr(s)=x(s)\,\hi+y(s)\,\hj\) be a counterclockwise parametrization of \(C\) by arc length. Then \(\hT(s) = \vr'(s) = x'(s)\,\hi+y'(s)\,\hj\) is the forward pointing unit tangent vector to \(C\) at \(\vr(s)\) and \(\hn(s) = \vr'(s)\times\hk=y'(s)\,\hi-x'(s)\,\hj\text{.}\) To see that \(\vr'(s)\times\hk\) really is \(\hn(s)\text{,}\) note that \(y'(s)\,\hi-x'(s)\,\hj\)
- has the same length, namely \(1\text{,}\) as \(\vr'(s)\) (recall that \(\vr(s)\) is a parametrization by arc length),
- lies in the \(xy\)-plane and
- is perpendicular to \(\vr'(s)\text{.}\) (Check that \(\vr'(s)\cdot\big[y'(s)\,\hi-x'(s)\,\hj\big]=0\text{.}\))
- Use the right hand rule to check that \(\vr'(s)\times\hk\) is \(\hn\) rather than \(-\hn\text{.}\)
4.3.3.
Hint.
Use direct evaluation!
4.3.4.
Hint.
The functions \(\frac{x}{x^2+y^2}\) and \(\frac{-y}{x^2+y^2}\) are not defined, let alone continuous or differentiable, at \(x=y=0\text{.}\)
4.3.5.
Hint.
For practice, evaluate this integral twice — once directly and once using Green’s theorem.
4.3.6.
Hint.
The \(\sin y^2\) and \(\cos y^2\) in the integrand look hard to integrate. Try Green’s theorem.
4.3.7. (✳).
Hint.
Don’t do the integral directly.
4.3.8. (✳).
Hint.
Don’t do the integral directly. Sketch the rectangle.
4.3.9. (✳).
Hint.
Do not compute the integral directly.
4.3.10. (✳).
Hint.
Don’t do the integral directly. Sketch the triangle.
4.3.11. (✳).
Hint.
The integrand for direct evaluation looks complicated — don’t evaluate this integral directly.
4.3.12. (✳).
Hint.
Direct evaluation is not the most efficient method available.
4.3.13. (✳).
Hint.
Green’s theorem must be applied to a closed curve; note that the curve \(C\) is not closed.
Consider carefully the point \((0, 0)\) in your analysis.
You may use the fact that \(\int \frac{\dee{t}}{1+t^2} = \arctan(t) + C\text{.}\)
4.3.14. (✳).
Hint.
If we were to try to evaluate this integral directly, then on the \(y=x^2-4x+3\) part of \(C\text{,}\) the integrand would contain \(x^2 e^y = x^2 e^{x^2-4x+3}\text{.}\) That looks hard to integrate, so try Green’s theorem.
4.3.15. (✳).
Hint.
Beware the point \((0,0)\text{.}\)
4.3.18. (✳).
Hint.
It is possible to evaluate this integral by three different methods, one of them being direct evaluation (though it requires some ingenuity). Try to find all three.
4.3.20. (✳).
Hint.
Write \(\oint_C\vF\cdot \dee{\vr}-A\oint_C\vG\cdot \dee{\vr}
=\oint_C(\vF-A\vG)\cdot \dee{\vr}\text{.}\)
4.3.21. (✳).
Hint.
Note that \(\vF(x,y)\) is not defined at \((x,y)=(0,0)\text{.}\)
4.3.22. (✳).
Hint.
Note that \(\vF(x,y)\) is not defined at \((x,y)=(0,0)\text{.}\)
4.3.24. (✳).
Hint.
(a) All points on the curve obey an equation that contains \(x\)’s and \(y\)’s, but no \(z\)’s.
(b) Exploit conservativeness as much as possible.
4.3.25.
Hint.
Use Green’s theorem to convert the integral over \(C\) into an integral over the region \(R\) in the \(xy\)-plane whose boundary is \(C\text{.}\) Consider the sign of the integrand of the integral over \(R\text{.}\)
4.4 Stokes’ Theorem
4.4.3 Exercises
4.4.3.1.
Hint.
One approach is to first do
Then imagine slowly deforming the sketch to the get specified \(S\)’s
4.4.3.2.
Hint.
Define the vector field \(\vF(x,y,z) = F_1(x,y)\,\hi +F_2(x,y)\,\hj\text{.}\)
4.4.3.3.
Hint.
First verify the vector identity \(\vnabla\times[\phi\vnabla\psi+\psi\vnabla\phi]=\vZero\)
4.4.3.4.
Hint.
To parametrize the curve \(x^2+y^2=1\text{,}\) \(z=y^2\text{,}\) first parametrize the circle \(x^2+y^2=1\text{.}\) That is, find \(x(t)\) and \(y(t)\) obeying \(x(t^2)+y(t)^2=1\text{.}\) Then set \(z(t) =y(t)^2\text{.}\)
4.4.3.5.
Hint.
Apply Stokes’ theorem. Note that \(\vr(t) = x(t)\,\hi+y(t)\,\hj +z(t)\,\hk\) obeys \(x(t)+y(t)+z(t)=3\text{,}\) for every \(t\text{,}\) and that \(x(t)\,\hi+y(t)\,\hj
=(1+\cos t)\,\hi+(1+\sin t)\,\hj\) runs counterclockwise around the circle of radius 1 centered on \((1,1)\text{.}\)
4.4.3.6. (✳).
Hint.
The form of the integral should be quite suggestive.
4.4.3.7. (✳).
Hint.
The form of the integral should be quite suggestive.
4.4.3.8. (✳).
Hint.
What’s the title of this section?
4.4.3.9.
Hint.
We are to evaluate a flux integral of the form \(\dblInt_S\vnabla\times\vF\cdot\hn\,\dee{S}\text{.}\) Sure looks like one side of Stokes’ theorem.
4.4.3.10.
Hint.
The vector field \(\vF\) looks too complicated for a direct evaluation of the line integral. So, try Stokes’ theorem.
4.4.3.16. (✳).
Hint.
All three vertices of part (a) lie in the plane of part (b).
4.4.3.17. (✳).
Hint.
The curve \(C\) is the boundary of a surface. To guess the surface express the \(z\) component of \(\vr(t)\) in terms of the \(x\) and \(y\) components.
4.4.3.18. (✳).
Hint.
The fact that the surface is not completely specified is a big hint.
4.4.3.19. (✳).
Hint.
We are to evaluate the line integral of a complicated vector field around a relatively complicated closed curve. (Sketch it!) That certainly suggests that we should not try to evaluate the integral directly.
4.4.3.20. (✳).
Hint.
The integral looks messy. Compute the curl of \(\vF\) to help gauge if Stokes’ theorem would be easier.
4.4.3.21. (✳).
Hint.
The form of the integrand is sugestive.
4.4.3.26. (✳).
Hint.
Let \(D\) be the disk in the plane \(x+y+z=3\) whose boundary is \(C\text{.}\) Suppose that, as \((x,y,z)\) runs over \(D\text{,}\) \((x,y)\) runs over the ellipse \(D_{xy}\text{.}\) We are told that the area of \(D\) is \(\pi R^2\text{,}\) but we are not told the area of \(D'\text{.}\) So it is easier to deal with the integral \(\dblInt_D \dee{S}\) than with the integral \(\dblInt_{D'}\dee{x}\dee{y}\text{.}\)
4.4.3.29.
Hint.
Given the form of \(\vF\text{,}\) direct evaluation looks hard.
The integral evaluations can be greatly simplified by using that the centroid \((\bar x,\bar y)\) of any region \(R\) in the \(xy\)-plane is
\begin{equation*}
\bar x =\frac{\dblInt_R x\,\dee{x}\,\dee{y}}{\text{Area}(R)}\qquad
\bar y =\frac{\dblInt_R y\,\dee{x}\,\dee{y}}{\text{Area}(R)}
\end{equation*}
4.4.3.30. (✳).
Hint.
Part (a) is a hint for part (b). Sketch the curve in part (b).
4.4.3.31. (✳).
Hint.
For practice, evaluate the flux of part (a) twice — once by direct evaluation and once using Stokes’ theorem.
4.4.3.32. (✳).
Hint.
By definition, \(D\) is connected if any two points in \(D\) can be joined by a curve that lies completely in \(D\text{.}\)
By definition, \(D\) is simply connected if any simple closed curve in \(D\) can be shrunk to a point continuously in \(D\text{.}\)
4.4.3.33. (✳).
Hint.
Review §4.1.2.
4.4.3.34. (✳).
Hint.
Considering that there are ten line segments in \(C\text{,}\) it is probably not very efficient to use direct evaluation.
4.4.3.35. (✳).
Hint.
Direct evaluation looks hard.
4.4.3.36. (✳).
Hint.
Rewrite \(\oint_C\vE\cdot\dee{\vr}\) as a surface integral.
4.4.3.37. (✳).
Hint.
What is \(x(t)^2+y(t)^2+z(t)^2=2\text{?}\) How is \(x(t)\) relatex to \(z(t)\text{?}\)
4.4.3.38. (✳).
Hint.
The intersection of the plane \(x+y+z=1\) with the sphere \(x^2+y^2+z^2=1\) is a circle. Use symmetry to guess the centre of the circle.
4.4.3.39. (✳).
Hint.
Sketch \(S\text{.}\)
4.4.3.40.
Hint.
You can avoid evaluating any integral by identifying \(S'\) as a simple geometric figure.
5 True/False and Other Short Questions
5.2 Exercises
5.2.2. (✳).
Hint.
Read (d), (e), (f), (g), (h) very carefully.
5.2.3. (✳).
Hint.
Beware that in part (f) a surface is defined to be closed if and only if it is the boundary of a solid region \(E\text{.}\) Even though that is not the usual definition, it is be used in this question.
5.2.4. (✳).
Hint.
(b) In general, for which values of \(x\) is the curvature of \(y=f(x)\) zero?
(c) First parametrize \(x^2+z^2=1\text{.}\)
(d) First determine when \(\vr(u,v)\) has \(z=0\text{.}\)
(e) What type of curve has curvature zero?
(f) What theorem relates the divergence of a vector field with flux integrals of the vector field?
(g) What is the screening test for conservativeness in two dimensions?
(h) What is the definition of “parametrized by arclength”?
(i) What theorem relates line integrals to curls?
(j) What theorem relates flux integrals to divergences?
(k) Use Stokes’ theorem.
5.2.5. (✳).
Hint.
Read all of the statements very carefully. The details are critical.
(a) Note the word anywhere.
(b) If you have not learned about simply connected domains, skip this part. If you have, read the statement very carefully.
(d) If you have not learned about Kepler’s three laws, skip this part.
(h) Read the statement very carefully. It does not specify that \(C\) is closed.
(i) Review §1.5.
5.2.8. (✳).
Hint.
Read all of the statements very carefully. The details are critical.
For part (d), note that the curve need not lie in a plane.
For part (g), note that the domain can have holes in it.
For parts (h) and (i), by definition, \(D\) is simply connected if any simply closed curve in \(D\) can be shrunk to a point continuously in \(D\text{.}\)
5.2.9. (✳).
Hint.
Read all of the statements very carefully. The details are critical.
For part (b), note that the curve need not lie in a plane.
For part (d), note that the domain can have holes in it.
For parts (i) and (j), by definition, \(D\) is simply connected if any simply closed curve in \(D\) can be shrunk to a point continuously in \(D\text{.}\)
5.2.10. (✳).
Hint.
Read all of the statements very carefully. The details are critical.
(a) The integral \(\int_C f\,\dee{s}=0\) is not of the form \(\int_C \vF \cdot \dee{\vr}\text{.}\)
(d) \(\vF\) and \(\vG\) can be any vector fields.
(e) Think about how \(\int_C f\,\dee{s}\) is defined.
(f) Look at \(\frac{\partial\vr}{\partial u}\times
\frac{\partial\vr}{\partial u}\) very closely.
(g) The integral is completely independent of \(x(u,v)\) and \(y(u,v)\text{.}\)
5.2.11. (✳).
Hint.
Read all of the statements very carefully. The details are critical.
(b) Read the statement very carefully. “simply connected” plays no role here. The vector field \(\vF\) is not required to be conservative.
(e) Recall that \(S\) is closed when it is the boundary of a solid region \(V\text{.}\)
(g) Assume that the constant \(|\vv|\) is not zero.
(j) If you have not learned about Kepler’s three laws, skip this part.
5.2.13. (✳).
Hint.
(g) Be careful. The power in the denominator is important.
(j) Beware the sign.
5.2.20. (✳).
Hint.
Review Corollary 4.3.5.
5.2.21. (✳).
Hint.
(b) \(\dblInt_{S}\vnabla\times\vF\cdot\hn\,\dee{S}\) is a flux integral over the closed surface \(S\text{.}\)
(c) Consider \(\oint_C \vF\cdot \dee{\vr}-\oint_C \vG\cdot \dee{\vr}
=\oint_C (\vF - \vG)\cdot \dee{\vr}\text{.}\)
