13.3 Rates of Change

Suppose in the given figure, I want to find the shortest path from a to a point p on the segment $[\mbox{{\bf b}} \mathbf{{\bf c}}]$ and back to d. Any such path will be uniquely defined by giving any one of the six numbers:

$$\begin{array}{ll} x=\mbox{ dist} (\mbox{{\bf b}}, \mbox{{\bf ... ...p}}\mathbf{{\bf c}}, & B\leq\beta\leq {\pi\over 2}. \end{array}$$

Here $A,B$ are as shown in the figure below:

$${3\over {12}}={{\sin (A)}\over {\cos (A)}}=\tan (A) \mbox{ and }{6\over {12}}={{\sin (B)}\over {\cos (B)}}=\tan (B).$$

For a given point p, any of the six numbers is a function of any of the others. For example, we have $l$ is a function of $x$

$$l=\sqrt{x^2+9},$$

and $l$ is a function of $m$, since for $x\in [0,12]$ and $y\in [0,12]$ we have

$$\begin{eqnarray*} m^2=(12-x)^2+36 &\mbox{$\Longrightarrow$}& 12-x=\sqrt{m^2-36}\... ...box{$\Longrightarrow$}& l=\sqrt{\Big(12-\sqrt{m^2-36}\Big)^2+9}. \end{eqnarray*}$$

Also $l$ is a function of $\alpha$, since by similar triangles $${{{\sin(\alpha})\over 1}={3\over l}}$$ and hence

$$l={3\over {\sin (\alpha)}}=3\csc (\alpha ).$$

We have

$${{dl}\over {dx}}={1\over 2} {{2x}\over {\sqrt{x^2+9}}}={x\over {\sqrt{x^2+9}}}$$

and

$${{dl}\over {d\alpha}}=-3\csc (\alpha)\cot (\alpha).$$

I refer to $${ {{dl}\over {dx}}}$$ as the rate of change of $l$ with respect to $x$ and to $${ {{dl}\over {d\alpha}}}$$ as the rate of change of $l$ with respect to $\alpha$. Note that the `` $l$"'s in `` $${ {{dl}\over {dx}}}$$'' and `` $${ {{dl}\over {d\alpha}}}$$'' represent different functions. In the first case $${ l(x)=\sqrt{x^2+9}}$$ and in the second case $l(\alpha)=3\csc \alpha$. Here $${ {{dl}\over {dx}}}$$ is positive, indicating that $l$ increases when $x$ increases, and $${ {{dl}\over {d\alpha}}}$$ is negative, indicating that $l$ decreases when $\alpha$ increases.

I want to find the path for which $l+m$ is shortest; i.e., I want to find the minimum value of $l+m$. I can think of $l$ and $m$ as being functions of $x$, and then the minimum value will occur when $${ {d\over {dx}}(l+m)=0}$$; i.e.,

$${{dl}\over {dx}}+{{dm}\over {dx}}=0.$$ (13.16)

Now $l^2=x^2+9$, so $${ 2l\cdot {{dl}\over {dx}}=2\cdot x}$$; i.e.,

$${{dl}\over {dx}}={x\over l}=\cos\alpha,$$

and $m^2=(12-x)^2+6^2$, so $${ 2m {{dm}\over {dx}}=2(12-x)(-1)}$$, i.e.,

$${{dm}\over {dx}}=-{{(12-x)}\over m}=-{y\over m}=-\cos\beta.$$

Equation (13.16) thus says that for the minimum path $\cos\alpha-\cos\beta=0$; i.e., $\cos\alpha=\cos\beta$, and hence $\alpha=\beta$. Thus the minimizing path satisfies the reflection condition, angle of incidence equals angle of reflection. Hence the minimizing triangle will make $\triangle \mbox{{\bf b}}\mbox{{\bf p}}\mbox{{\bf a}}$ and $\triangle \mathbf{{\bf c}}\mbox{{\bf p}}\mbox{{\bf d}}$ similar, and will satisfy

$${6\over y}={3\over x} \mbox{ and } x+y=12,$$

so

$$6x=3y=3(12-x)=36-3x$$

or

$$9x=36 \mbox{ so } x=4 \mbox{ and } y=8.$$

This example was done pretty much as Leibniz would have done it. You should compare the solution given here to your solution of exercise 13.15A.

The problem in the last example was solved by Heron (date uncertain, sometime between 250 BC and 150 AD) as follows[26, page 353]. Imagine the line $[{\bf b}{\bf c}]$ to be a mirror. Let ${\bf a}'$ and ${\bf d}'$ denote the images of ${\bf a}$ and ${\bf d}$ in the mirror,

i.e. $[{\bf aa}']$ and $[{\bf dd}']$ are perpendicular to $[{\bf bc}]$ and dist $({\bf a},{\bf b})$ = dist $({\bf a}',{\bf b})$, dist $({\bf d},{\bf c})$ = dist $({\bf d}',{\bf c})$. Consider any path ${\bf apd}$ going from a to a point p on the mirror, and then to d. Then triangle$({\bf pcd})$ and triangle$({\bf pcd}')$ are congruent, and hence

$$\mbox{dist}({\bf p},{\bf d}) = \mbox{dist}({\bf p},{\bf d}').$$

and hence the paths apd and ${\bf apd}'$ have equal lengths. Now the shortest path ${\bf apd}'$ is a straight line, which makes the angles $\alpha$ and $\beta$ are vertical angles, which are equal. Hence the shortest path makes the angle of incidence equal to the angle of reflection, as we found above by calculus.

Remark: We can think of velocity as being rate of change of position with respect to time.

13.17   Exercise. Consider a conical tank in the shape of a right circular cone with altitude $10'$ and diameter $10'$ as shown in the figure.

Water flows into the tank at a constant rate of 10 cubic ft./minute. Let $h$ denote the height of the water in the tank at a given time $t$. Find the rate of change of $h$ with respect to $t$. What is this rate when the height of the water is $5'$? What can you say about $${ {{dh}\over {dt}}}$$ when $h$ is nearly zero?

13.18   Exercise. A particle p moves on the rim of a wheel of radius 1 that rotates about the origin at constant angular speed $\omega$, so that at time $t$ it is at the point $\Big( \cos (\omega t),\sin(\omega t)\Big)$. A light at the origin causes p to cast a shadow at the point $(2,y)$ on a wall two feet from the center of the wheel.

Find the rate of change of $y$ with respect to time. You should ignore the speed of light, i.e. ignore the time it takes light to travel from the origin to the wall.