Related Rates - Speed of his Shadow

Question

I took an entire semester of calculus, and now, after suggestion by my professor, I'm applying for a calculus tutoring position. However, I've never been asked how to find a related rate using angles, and yet that's one of the questions for the interview. Here it is:

A street light is mounted at the top of a $~15~$-ft-tall pole. A $~6~$ ft man walks away from the pole with a speed of $~5~$ ft/s along a straight path. How fast is the tip of his shadow moving when he is $~40~$ ft from the pole?

I was able to find the length of shadow, call that $~b~$, but I wasn't able to find $~b~$ prime or $~c~$ prime. A prime is of course $~0~$.

Thanks in advance for any help!

Answer 1

Assuming that your image is two similar triangles inscribed on each other,

http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates_Files/image007.gif

the height of the lamp $h_1=15ft$, the height of the person, $h_2=6ft$. The distance between the lamp and the person $d$, and the length of the shadow, $x$.

Since the person is moving constantly at $5ft/sec$, then $d=5t$ where $t$ is the time travelled.

Assuming the angle of the shadow to the lamp as $\theta$, then we can get 2 equations by pythagorean theorem:

$$tan\theta=h_1 / (d + x)$$ $$tan\theta=h_2/x$$

equating both equations, we get:

$$h_1/(d+x) = h_2/x$$

After combining similar terms, you get:

$$(h_1-h_2)x = h_2d$$

Substituting their values, we get:

$$(15-6)x = 6(5t)$$

or

$$x = (10/3)t$$

getting the derivative, we get $dx$ as:

$$dx/dt = (10/3)dt$$

which means that for every change in time, there's $3.33 ft/sec$ increase in the shadow's length.

Answer 2

Consider this picture:

The distance from the pole and the man is a linear function of time $x(t)=5t\ \text{ft}$ because the velocity with which he is moving away from the pole is constant: $\frac{dx}{dt}=5\ \text{ft/s}$. Technically, this isn't relevant to how we're going to solve this problem, but it's just to give you a better idea of what's going on.

The length of his shadow, $y(t)$, is also a function of time because it changes with time as he walks away from the pole. What the problem is asking you to find is the velocity (derivative) of $y(t)$ when he is $40$ feet away from the pole.

Use similar triangles: $$ \frac{x+y}{15}=\frac{y}{6}\implies y=\frac{2}{3}x. $$

Now, use implicit differentiation to find the derivative of $y(t)$: $$\frac{dy}{dt}=\frac{2}{3}\frac{dx}{dt}.$$

We know what $\frac{dx}{dt}$ is: $$\frac{dy}{dt}=\frac{2}{3}\cdot 5=\frac{10}{3}\ \text{ft/s}.$$

This result tells us that because the derivative of $y(t)$ is a constant function of time, the tip of his shadow is moving with the same velocity at any given point in time during his trip, including when he is $40$ feet away from the pole. So, the answer to the problem could be worded something like this: the velocity with which the tip of his shadow is moving when he is 40 feet away from the pole is $\frac{10}{3}\ \text{ft/s}$.