# Related Rates - Speed of his Shadow

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!

• Hi Zach. Your proposed solution mentions letters, like a,b, and c, but you never explain what they mean. It is best that you make a sketch of the problem and attach it to the question. Also, please write your equations using MathJax. Finally, not really my place to judge, but maybe it is too early for you to do tutoring, if you took calculus for one semester and still struggle with some exercises? Jul 22, 2019 at 7:38
• I appreciate the criticism. I’m thinking the same thing to be honest, but I haven’t looked at calculus since the end of the semester in June, so maybe I’m just a little rusty? I will let them decide 😄 still appreciate it, though. Jul 22, 2019 at 17:07
• “Related rates using angles” made me expect a problem that requires an angle to occur in at least one formula either as the variable to differentiate over or as part of something that is to be differentiated. As shown below, you don’t need either of those things. But if you included the steps of your own calculations in the question, we might better see what the issue is. Jul 22, 2019 at 17:35
• Somehow this reminds me of one time in high school when I was sitting with some other students outside the office of the chair of another school’s math department. A man was applying for a teaching job there and the chair gave him an exercise that required differentiating a ratio of two functions. So the applicant asked us students how to do it. You’re getting better treatment here than we gave him! Jul 22, 2019 at 17:41

## 2 Answers

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 $$\frac{10}{3}\ \text{ft/s}$$.

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

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.