**When he is $10$ feet from the base of the light, at what rate is the tip of his shadow moving?****When he is $10$ feet from the base of the light, at what rate is the length of his shadow changing?**

The purpose of this question is to find the rate of change of the length of the shadow given two different scenarios.

Proportion is primarily described using ratios and fractions. A fraction is defined as $\dfrac{a}{b}$, whereas a ratio is depicted as $a:b$, and a proportion depicts that two ratios are equal. In this case, $a$ and $b$ are two integers. The ratio and proportion are the basis for assessing different theories in science and mathematics.

The rate of change function is expressed as the ratio at which one quantity changes with respect to the other. More generally, the rate of change divides the amount of change in one object by the respective amount of change in the other. The rate of change can take a negative or a positive value. The ratio of horizontal and vertical change between two points lying on a line or a plane is called a slope, which is equal to the rise by run ratio where rise denotes the vertical difference between two points and run denotes the horizontal difference between two points.

## Expert Answer

Let $s$ be the length of the base of the light pole to the shadow, $x$ be the length of the base of the light pole to the man, then the length of the shadow will be $s-x$. Since the height of the light pole is $15\,ft$ and the height of the man is $6\,ft$, therefore using the proportion as:

$\dfrac{15}{6}=\dfrac{s}{s-x}$

$15\,s-15\,x=6\,s$

$s=\dfrac{5x}{3}$

Now, differentiating both sides with respect to time:

$\dfrac{ds}{dt}=\dfrac{5\,dx}{3\,dt}$

Now from the question $\dfrac{dx}{dt}=5\,ft/s$, so that:

$\dfrac{ds}{dt}=\dfrac{5}{3}\times 5$

$\dfrac{ds}{dt}=\dfrac{25}{3}\,ft/s$

Since the length of the shadow is $s-x$, so the rate of change of the shadow length is:

$\dfrac{ds}{dt}-\dfrac{dx}{dt}=\dfrac{25}{3}-5$

$\dfrac{ds}{dt}-\dfrac{dx}{dt}=\dfrac{10}{3}\,ft/s$

## Example

Consider a vertex down conical tank having the radius $80\,ft$ and the height $80\,ft$. Also, assume that the rate of flow of water is $100\,ft^3/min$. Work out the rate of change of radius of the water when it is $4\,ft$ deep.

### Solution

Given that:

$\dfrac{dV}{dt}=-100\,ft^3/min$, $h=4\,ft$.

Now,Â $\dfrac{r}{40}=\dfrac{h}{80}$

$h=2r$

Since $h=4\,ft$, therefore:

$r=2$

Also, $V=\dfrac{\pi}{3}r^2h$

$V=\dfrac{2\pi}{3}r^3$

$\dfrac{dV}{dt}=2\pi r^2\cdot \dfrac{dr}{dt}$

Or $\dfrac{dr}{dt}=\dfrac{-100}{2\pi(2)^2}$

$\dfrac{dr}{dt}=-\dfrac{25}{2\pi}\,ft/min$