Calculus Related Rates Problem:
Lamp post casts a shadow of a man walking.

A 1.8-meter tall man walks away from a 6.0-meter lamp post at the rate of 1.5 m/s. The light at the top of the post casts a shadow in front of the man. How fast is the “head” of his shadow moving along the ground?

Calculus Solution

To solve this problem, we will use our standard 4-step Related Rates Problem Solving Strategy.
1. Draw a picture of the physical situation.
See the figure. We’re calling the distance between the post and the “head” of the man’s shadow $\ell$, and the distance between the man and the post x.

2. Write an equation that relates the quantities of interest.
We are given that the man is walking away from the post at the rate $\dfrac{dx}{dt} = 1.5$ m/s. We are looking for the rate at which the “head” of the man’s shadow moves, which is $\dfrac{d \ell}{dt}$. We thus need to somehow relate $\ell$ to x, so we can then develop the relationship between their time-derivatives.

There’s a subtlety to this problem that typically goes unaddressed: We’re focusing on $\ell$ and $\dfrac{d \ell}{dt}$ here because $\ell$ is the distance from the shadow’s tip to the stationary post. We’re not examining the shadow’s length itself (labeled $\ell – x$ in the left figure below) because that length is relative to the man’s feet, which are also moving. So we’d find a different answer if we calculated the rate at which that gray shadow is changing. This problem asks us to find the rate the shadow’s head as it moves along the (stationary) ground, so it’s best to make our measurements from a point that isn’t also moving—namely, from the post. Hence we focus on $\ell$ and aim to compute $\dfrac{d \ell}{dt}$.

B. To develop your equation, you will probably use . . . similar triangles.

In the figure above we’ve separated out the two triangles. Notice that the angles are identical in the two triangles, and hence they are similar. The ratio of their respective components are thus equal as well. Hence the ratio of their bases $\left(\dfrac{\ell – x}{\ell} \right)$ is equal to the ratio of their heights $\left( \dfrac{1.8\, \text{m}}{6.0\, \text{m}}\right)$:
\begin{align*}
\dfrac{\ell – x}{\ell} &= \frac{1.8 \, \text{m}}{6.0 \, \text{m}} \\[12px] &= 0.30 \\[12px] \ell – x &= 0.30 \ell \\[12px] \ell – 0.30 \ell &= x \\[12px] (1 – 0.30) \ell &= x \\[12px] 0.70 \ell &= x
\end{align*}

3. Take the derivative with respect to time of both sides of your equation.
\begin{align*}
\dfrac{d}{dt}(0.70 \ell) &= \dfrac{d}{dt}(x) \\[12px] 0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt}
\end{align*}

4. Solve for the quantity you’re after.
We’re looking for $\dfrac{d \ell}{dt}$:

\begin{align*}
0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt} \\[12px] \dfrac{d \ell}{dt} &= \frac{1}{0.70} \dfrac{dx}{dt} \\[12px] &= \frac{1}{0.70} \left( 1.5 \, \tfrac{\text{m}}{\text{s}}\right) \\[12px] &= 2.1\, \tfrac{\text{m}}{\text{s}} \quad \cmark
\end{align*}

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