1.2: Rates of Change

Suppose \(x(t)\) gives the position, at some time \(t,\) of an object (such as Zeno's arrow) moving along a straight line. The problem we face is that of giving a determinate meaning to the idea of the velocity of the object at a specific instant of time. We first note that we face no logical difficulties in defining an average velocity over an interval of time of non-zero length. That is, if \(a<b,\) then the object travels a distance

\[\Delta x=x(b)-x(a)\]

from time \(t=a\) to time \(t=b,\) an interval of time of length \(\Delta t=b-a,\) and, consequently, the average velocity of the object over this interval of time is

\[v_{[a, b]}=\frac{x(b)-x(a)}{b-a}=\frac{\Delta x}{\Delta t} .\]

Example \(\PageIndex{1}\)

Suppose an object, such as a lead ball, is dropped from a height of 100 meters. Ignoring air resistance, the height of the ball above the earth after \(t\) seconds is given by

\[ x(t)=100-4.9 t^{2} \text { meters } , \] a result first discovered by Galileo. Hence, for example, from time \(t=0\) to time \(t=2\) we have \[ \Delta x=x(2)-x(0)=(100-(4.9)(4))-100=-19.6 \text { meters } , \] \[ \Delta t=2-0=2 \text { seconds } , \] and so \[ v_{[0,2]}=-\frac{19.6}{2}=-9.8 \text { meters / second. } \] For another example, from time \(t=1\) to time \(t=4\) we have \[ \Delta x=x(4)-x(1)=21.6-95.1=-73.5 , \] \[ \Delta t=4-1=3 \text { seconds } , \] and so \[ v_{[1,4]}=-\frac{73.5}{3}=-24.5 \text { meters / second. } \] Note that both of these average velocities are negative because we have taken the positive direction to be upward from the surface of the earth.

Exercise \(\PageIndex{1}\)

Suppose a lead ball is dropped into a well. Ignoring air resistance, the ball will have fallen a distance \(x(t)=16 t^{2}\) feet after \(t\) seconds. Find the average velocity of the ball over the intervals (a) \([0,2],\) (b) \([1,3],\) and \((\mathrm{c})\) \([1,1.5] .\)

Answer

(a) 32 feet per second, (b) 64 feet per second, (c) 40 feet per second

Letting \(\Delta t=b-a,\) we may rewrite \((1.2 .2)\) in the form

\[ v_{[a, a+\Delta t]}=\frac{x(a+\Delta t)-x(a)}{\Delta t} . \] Using \((1.2 .3),\) there are two approaches to generalizing the notion of average velocity over an interval to that of velocity at an instant. The most common approach, at least since the middle of the 19 th century, is to consider the effect on \(v_{[a, a+\Delta t]}\) as \(\Delta t\) diminishes in magnitude and defining the velocity at time \(t=a\) to be the limiting value of these average velocities. The approach we will take in this text is to consider what happens when we take \(a\) and \(b\) to be, although not equal, immeasurably close to one another.

Example \(\PageIndex{2}\)

If we have, as in the previous example,

\[x(t)=100-4.9 t^{2} \text { meters } ,\]

then from time \(t=1\) to time \(t=1+\Delta t\) we would have

\[\begin{aligned} \Delta x &=x(1+\Delta t)-x(1) \\ &=\left(100-4.9(1+\Delta t)^{2}\right)-95.1 \\ &=4.9-4.9\left(1+2 \Delta t+(\Delta t)^{2}\right) \\ &=-9.8 \Delta t-4.9(\Delta t)^{2} \text { meters. } \end{aligned}\]

Hence the average velocity over the interval \([1,1+\Delta t]\) is

\[\begin{aligned} v_{[1,1+\Delta t]} &=\frac{\Delta x}{\Delta t} \\ &=\frac{-9.8 \Delta t-4.9(\Delta t)^{2}}{\Delta t} \\ &=-9.8-4.9 \Delta t \text { meters / second. } \end{aligned}\]

Note that if, for example, \(\Delta t=3,\) then we find

\[v_{[1,4]}=-9.8-(4.9)(3)=-9.8-14.7=-24.5 \text { meters/second, } \]

in agreement with our previous calculations.

Now suppose that the starting time \(a=1\) and the ending time \(b\) are different, but the difference is so small that it cannot be measured by any real number. In this case, we call \(d t=b-a\) an infinitesimal. Similar to our computations above, we have

\[d x=x(1+d t)-x(1)=-9.8 d t-4.9(d t)^{2} \text { meters, }\]

the distance traveled by the object from time \(t=1\) to time \(t=1+d t,\) and

\[v_{[1,1+d t]}=\frac{d x}{d t}=-9.8-4.9 d t \text { meters / second, }\]

the average velocity of the object over the interval \([1,1+d t] .\) However, since \(d t\) is infinitesimal, so is \(4.9 d t .\) Hence \(v[1,1+d t]\) is immeasurably close to \(-9.8\) meters per second. Moreover, this is true no matter what the particular value of \(d t\). Hence we should take the instantaneous velocity of the object at time \(t=1\) to be

\[v(1)=-9.8 \text { meters/second. }\]

Exercise \(\PageIndex{2}\)

As in the previous exercise, suppose a lead ball has fallen \(x(t)=16 t^{2}\) feet in \(t\) seconds. Find the average velocity of the ball over the interval \([1,1+\Delta t]\) and use this result to obtain the answers to parts \((\mathrm{b})\) and \((c)\) of the previous exercise.

Answer

\(v_{[1,1+\Delta t]}=32+16 \Delta t\) feet per second

Exercise \(\PageIndex{3}\)

Find the average velocity of the ball in the previous exercise over the interval \([1,1+d t],\) where \(d t\) is infinitesimal, and use the result to find the instantaneous velocity of the ball at time \(t=1\).

Answer

\(v_{[1,1+d t]}=32+16 d t\) feet per second, \(v(1)=32\) feet per second

Example \(\PageIndex{3}\)

To find the velocity of the object of the previous examples at time \(t=3,\) we compute

\[ \begin{aligned} d x &=x(3+d t)-x(3) \\ &=\left(100-4.9(3+d t)^{2}-55.9\right.\\ &=44.1-4.9\left(9+6 d t+\left(d t^{2}\right)\right) \\ &=-29.4 d t-4.9(d t)^{2} \text { meters, } \end{aligned} \] from which we obtain \[ \frac{d x}{d t}=-29.4-4.9 d t \text { meters / second. } \] As above, we disregard the immeasurable \(-4.9 d t\) to obtain the velocity of the object at time \(t=3:\) \[ v(3)=-29.4 \mathrm{meters} / \mathrm{second.} \]

Exercise \(\PageIndex{4}\)

Find the velocity of the ball in the previous exercise at time \(t=2\).

Answer

\(v(2)=64\) feet per second

In general, if \(x(t)\) gives the position, at time \(t,\) of an object moving along a straight line, then we define the velocity of the object at a time \(t\) to be the real number which is infinitesimally close to

\[ \frac{x(t+d t)-x(t)}{d t} , \] provided there is exactly one such number for any value of the nonzero infinitesimal \(d t .\)

Example \(\PageIndex{4}\)

For our previous example, we find

\[ \begin{aligned} d x &=x(t+d t)-x(t) \\ &=\left(100-4.9(t+d t)^{2}\right)-\left(100-4.9 t^{2}\right) \\ &=-4.9\left(t+2 t d t+(d t)^{2}\right)-4.9 t^{2} \\ &=-9.8 t d t-4.9(d t)^{2} \text { meters } \\ &=(-9.8 t-4.9 d t) d t . \end{aligned} \] Hence \[ \frac{d x}{d t}=-9.8 t-4.9 d t \text { meters / second, } \] and so the velocity of the object at time \(t\) is \[ v(t)=-9.8 t \text { meters / second. } \] In particular, \[ v(1)=-9.8 \text { meters / second } \] and \[ v(3)=-9.8(3)=-29.4 \text { meters } / \text { second, } \] as previously computed.

Exercise \(\PageIndex{5}\)

Find the velocity of the ball in the previous exercise at time \(t\). Use your result to verify your previous answers for \(v(1)\) and \(v(2)\).

Answer

\(v(t)=32 t\) feet per second

Even more generally, we should recognize that velocity is but a particular example of a rate of change, namely, the rate of change of the position of an object with respect to time. In general, given any quantity \(y\) as a function of another quantity \(x,\) say \(y=f(x)\) for some function \(f,\) we may ask about the rate of change of \(y\) with respect to \(x .\) If \(x\) changes from \(x=a\) to \(x=b\) and we let

\[ \Delta x=b-a \] and \[ \Delta y=f(b)-f(a)=f(a+\Delta x)-f(x) , \] then \[ \frac{\Delta y}{\Delta x}=\frac{f(b)-f(a)}{b-a} \] is the average rate of change of \(y\) with respect to \(x ;\) if \(d x\) is a nonzero infinitesimal, then the real number which is infinitesimally close to \[ \frac{d y}{d x}=\frac{f(x+d x)-f(x)}{d x} \] is the instantaneous rate of change, or, simply, rate of change, of \(y\) with respect to \(x\) at \(x=a\) In subsequent sections we will look at this quantity in more detail, but will consider one more example before delving into technicalities.

Example \(\PageIndex{5}\)

Suppose a spherical shaped balloon is being filled with water. If \(r\) is the radius of the balloon in centimeters and \(V\) is the volume of the balloon, then

\[ V=\frac{4}{3} \pi r^{3} \text { centimeters }^{3}. \] since a cubic centimeter of water has a mass of 1 gram, the mass of the water in the balloon is \[ M=\frac{4}{3} \pi r^{3} \text { grams. } \]

To find the rate of change of the mass of the balloon with respect to the radius of the balloon, we first compute

\[ \begin{aligned} d M &=\frac{4}{3} \pi(r+d r)^{3}-\frac{4}{3} \pi r^{3} \\ &=\frac{4}{3} \pi\left(\left(r^{3}+3 r^{2} d r+3 r(d r)^{2}+(d r)^{3}\right)-r^{3}\right) \\ &=\frac{4}{3} \pi\left(3 r^{2}+3 r d r+(d r)^{2}\right) d r \text { grams. } \end{aligned} \] from which it follows that \[ \frac{d M}{d r}=\frac{4}{3} \pi\left(3 r^{2}+3 r d r+(d r)^{2}\right) \text { grams / centimeter. } \] since both \(3 r d r\) and \((d r)^{2}\) are infinitesimal, the rate of change of mass of the balloon with respect to the radius of the balloon is \[ \frac{4}{3} \pi\left(3 r^{2}\right)=4 \pi r^{2} \text { gams / centimeer. } \] For example, when the balloon has a radius of 10 centimeters, the mass of the water in the balloon is increasing at a rate of \[ 4 \pi(10)^{2}=400 \pi \text { grams } / \text { centimeter. } \] It may not be surprising that this is also the surface area of the balloon at that instant.

Exercise \(\PageIndex{6}\)

Show that if \(A\) is the area of a circle with radius \(r,\) then \(\frac{d A}{d r}=2 \pi r\).