CLP-1 Differential Calculus

Suppose that you are moving along the \(x\)–axis and that at time \(t\) your position is given by

We're going to try and get a good picture of what your motion is like. We can learn quite a bit just by looking at the sign of the velocity \(v(t)=x'(t)\) at each time \(t\text{.}\)

• If \(x'(t) \gt 0\text{,}\) then at that instant \(x\) is increasing, i.e. you are moving to the right.
• If \(x'(t)=0\text{,}\) then at that instant you are not moving at all.
• If \(x'(t) \lt 0\text{,}\) then at that instant \(x\) is decreasing, i.e. you are moving to the left.

From the given formula for \(x(t)\) it is straight forward to work out the velocity

This is zero only when \(t=-1\) and when \(t=+1\text{;}\) at no other value ¹This is because the equation \(ab=0\) is only satisfied for real numbers \(a\) and \(b\) when either \(a=0\) or \(b=0\) or both \(a=b=0\text{.}\) Hence if a polynomial is the product of two (or more) factors, then it is only zero when at least one of those factors is zero. There are more complicated mathematical environments in which you have what are called “zero divisors” but they are beyond the scope of this course. of \(t\) can this polynomial be equal zero. Consequently in any time interval that does not include either \(t=-1\) or \(t=+1\text{,}\) \(v(t)\) takes only a single sign ²This is because if \(v(t_a) \lt 0\) and \(v(t_b) \gt 0\) then, by the intermediate value theorem, the continuous function \(v(t)=x'(t)\) must take the value \(0\) for some \(t\) between \(t_a\) and \(t_b\text{.}\) . So

• For all \(t \lt -1\text{,}\) both \((t+1)\) and \((t-1)\) are negative (sub in, for example, \(t=-10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)
• For all \(-1 \lt t \lt 1\text{,}\) the factor \((t+1) \gt 0\) and the factor \((t-1) \lt 0\) (sub in, for example, \(t=0\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \lt 0\text{.}\)
• For all \(t \gt 1\text{,}\) both \((t+1)\) and \((t-1)\) are positive (sub in, for example, \(t=+10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)

The figure below gives a summary of the sign information we have about \(t-1\text{,}\) \(t+1\) and \(x'(t)\text{.}\)

It is now easy to put together a mental image of your trajectory.

• For \(t\) large and negative (i.e. far in the past), \(x(t)\) is large and negative and \(v(t)\) is large and positive. For example ³Notice here we are using the fact that when \(t\) is very large \(t^3\) is much bigger than \(t^2\) and \(t^1\text{.}\) So we can approximate the value of the polynomial \(x(t)\) by the largest term — in this case \(t^3\text{.}\) We can do similarly with \(v(t)\) — the largest term is \(3t^2\text{.}\) , when \(t=-10^6\text{,}\) \(x(t)\approx t^3=- 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.
• For \(t \lt -1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are moving to the right.
• At \(t=-1\text{,}\) \(v(-1)=0\) and you have come to a halt at position \(x(-1)=(-1)^3-3(-1)+2=4\text{.}\)
• For \(-1 \lt t \lt 1\text{,}\) \(v(t)=x'(t) \lt 0\) so that \(x(t)\) is decreasing and you are moving to the left.
• At \(t=+1\text{,}\) \(v(1)=0\) and you have again come to a halt, but now at position \(x(1)=1^3-3+2=0\text{.}\)
• For \(t \gt 1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are again moving to the right.
• For \(t\) large and positive (i.e. in the far future), \(x(t)\) is large and positive and \(v(t)\) is large and positive. For example ⁴We are making a similar rough approximation here. , when \(t=10^6\text{,}\) \(x(t)\approx t^3= 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.

Here is a sketch of the graphs of \(x(t)\) and \(v(t)\text{.}\) The heavy lines in the graphs indicate when you are moving to the right — that is where \(v(t)=x'(t)\) is positive.

And here is a schematic picture of the whole trajectory.

In this example we are going to figure out how far a body falling from rest will fall in a given time period.

• We should start by defining some variables and their units. Denote

  • time in seconds by \(t\text{,}\)
  • mass in kilograms by \(m\text{,}\)
  • distance fallen (in metres) at time \(t\) by \(s(t)\text{,}\) velocity (in m/sec) by \(v(t)=s'(t)\) and acceleration (in m/sec\(^2\)) by \(a(t)=v'(t)=s''(t)\text{.}\)

  It makes sense to choose a coordinate system so that the body starts to fall at \(t=0\text{.}\)

• We will use Newton's second law of motion
  \begin{gather*} \text{the force applied to the body at time } t = m \cdot a(t). \end{gather*}
  together with the assumption that the only force acting on the body is gravity (in particular, no air resistance). Note that near the surface of the Earth,
  \begin{align*} \text{the force due to gravity acting on a body of mass } m &= m \cdot g. \end{align*}
  The constant \(g\text{,}\) called the acceleration of gravity ⁵It is also called the standard acceleration due to gravity or standard gravity. For those of you who prefer imperial units (or US customary units), it is about \(32\) ft/sec\(^2\text{,}\) \(77165\) cubits/minute\(^2\text{,}\) or \(631353\) furlongs/hour\(^2\text{.}\) , is about 9.8m/sec\(^2\text{.}\)
• Since the body is falling from rest, we know that its initial velocity is zero. That is
  \begin{align*} v(0) &= 0. \end{align*}
  Newton's second law then implies that
  \begin{align*} m\cdot a(t) &= \text{force due to gravity}\\ m \cdot v'(t) &= m \cdot g & \text{ cancel the } m\\ v'(t) &=g \end{align*}

• In order to find the velocity, we need to find a function of \(t\) whose derivative is constant. We are simply going to guess such a function and then we will verify that our guess has all of the desired properties. It's easy to guess a function whose derivative is the constant \(g\text{.}\) Certainly \(gt\) has the correct derivative. So does

  \begin{gather*} v(t) = gt + c \end{gather*}

  for any constant \(c\text{.}\) One can then verify ⁶While it is clear that this satisfies the equation we want, it is less clear that it is the only function that works. To see this, assume that there are two functions \(f(t)\) and \(h(t)\) which both satisfy \(v'(t)=g\text{.}\) Then \(f'(t)=h'(t) = g\) and so \(f'(t)-h'(t) = 0\text{.}\) Equivalently \(\diff{}{t}\left( f(t)-h(t) \right) = 0.\) The only function whose derivative is zero everywhere is the constant function (see Section 2.13 and Theorem 2.13.12). Thus \(f(t)-h(t) = \text{constant}\text{.}\) So all the functions that satisfy \(v'(t)=g\) must be of the form \(gt + \text{constant}\text{.}\) that \(v'(t)=g\text{.}\) Using the fact that \(v(0)=0\) we must then have \(c=0\) and so

  \begin{align*} v(t) &= gt. \end{align*}
• Since velocity is the derivative of position, we know that
  \begin{align*} s'(t) &= v(t) = g \cdot t. \end{align*}
  To find \(s(t)\) we are again going to guess and check. It's not hard to see that we can use
  \begin{align*} s(t) &= \frac{g}{2} t^2 + c \end{align*}
  where again \(c\) is some constant. Again we can verify that this works simply by differentiating ⁷To show that any solution of \(s'(t)=gv\) must be of this form we can use the same reasoning we used to get \(v(t) = gt + \text{constant}\text{.}\). Since we have defined \(s(t)\) to be the distance fallen, it follows that \(s(0)=0\) which in turn tells us that \(c = 0\text{.}\) Hence
  \begin{align*} s(t) &= \frac{g}{2} t^2 & \text{but $g=9.8$, so}\\ &= 4.9 t^2, \end{align*}
  which is exactly the \(s(t)\) used way back in Section 1.2.

A car's brakes can decelerate the car at 64000\(\textrm{km/hr}^2\text{.}\) How fast can the car be driven if it must be able to stop within a distance of 50m?

Solution Before getting started, notice that there is a small “trick” in this problem — several quantities are stated but their units are different. The acceleration is stated in kilometres per hour\(^2\text{,}\) but the distance is stated in metres. Whenever we come across a “real world” problem ⁸Well — “realer world” would perhaps be a betterer term. we should be careful of the units used.

• We should first define some variables and their units. Denote

  • time (in hours) by \(t\text{,}\)
  • the position of the car (in kilometres) at time \(t\) by \(x(t)\text{,}\) and
  • the velocity (in kilometres per hour) by is \(v(t)\text{.}\)

  We can also choose a coordinate system such that \(x(0)=0\) and the car starts braking at time \(t=0\text{.}\)

• Now let us rewrite the information in the problem in terms of these variables.

  • We are told that, at maximum braking, the acceleration \(v'(t)=x''(t)\) of the car is \(-64000\text{.}\)
  • We need to determine the maximum initial velocity \(v(0)\) so that the stopping distance is at most \(50m = 0.05km\) (being careful with our units). Let us call the stopping distance \(x_{stop}\) which is really \(x(t_{stop})\) where \(t_{stop}\) is the stopping time.
• In order to determine \(x_{stop}\) we first need to determine \(t_{stop}\text{,}\) which we will do by assuming maximum braking from a, yet to be determined, initial velocity of \(v(0)=q\) m/sec.

• Assuming that the car undergoes a constant acceleration at this maximum braking power, we have

  \begin{align*} v'(t) &= -64000 \end{align*}

  This equation is very similar to the ones we had to solve in Example 3.1.2 just above.

  As we did there ⁹Now is a good time to go back and have a read of that example. , we are going to just guess \(v(t)\text{.}\) First, we just guess one function whose derivative is \(-64000\text{,}\) namely \(-64000 t\text{.}\) Next we observe that, since the derivative of a constant is zero, any function of the form

  \begin{gather*} v(t) = -64000\,t + c \end{gather*}

  with constant \(c\text{,}\) has the correct derivative. Finally, the requirement that the initial velocity \(v(0)=q\)" forces \(c=q\text{,}\) so

  \begin{gather*} v(t) = q - 64000\,t \end{gather*}
• From this we can easily determine the stopping time \(t_{stop}\text{,}\) when the initial velocity is \(q\text{,}\) since this is just when \(v(t) = 0\text{:}\)
  \begin{align*} 0 = v(t_{stop}) &= q-64000 \cdot t_{stop} & \text{ and so}\\ t_{stop} &= \frac{q}{64000}. \end{align*}
• Armed with the stopping time, how do we get at the stopping distance? We need to find the formula satisfied by \(x(t)\text{.}\) Again (as per Example 3.1.2) we make use of the fact that
  \begin{align*} x'(t) &= v(t) = q - 64000t. \end{align*}
  So we need to guess a function \(x(t)\) so that \(x'(t) = q-64000 t\text{.}\) It is not hard to see that
  \begin{align*} x(t) &= qt - 32000t^2 + \text{constant} \end{align*}
  works. Since we know that \(x(0)=0\text{,}\) this constant is just zero and
  \begin{align*} x(t) &= qt - 32000 t^2. \end{align*}
• We are now ready to compute the stopping distance (in terms of the, still yet to be determined, initial velocity \(q\)):
  \begin{align*} x_{stop} &= x(t_{stop}) = q t_{stop} - 32000 t_{stop}^2\\ &= \frac{q^2}{64000} - \frac{32000 q^2}{64000^2}\\ &= \frac{q^2}{64000} \left(1 - \frac{1}{2} \right)\\ &= \frac{q^2}{2 \times 64000} \end{align*}
  Notice that the stopping distance is a quadratic function of the initial velocity — if you go twice as fast, you need four times the distance to stop.
• But we are told that the stopping distance must be less than \(50m = 0.05km\text{.}\) This means that
  \begin{align*} x_{stop} = \frac{q^2}{2 \times 64000} &\leq \frac{5}{100}\\ q^2 &\leq \frac{2 \times 64000 \times 5}{100} = \frac{64000 \times 10}{100} = 6400 \end{align*}
  Thus we must have \(q \leq 80\text{.}\) Hence the initial velocity can be no greater than \(80km/h\text{.}\)