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Related Rates

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

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Related rates problem and solution: A man walks away from a light pole that casts his shadow.  How fast is the tip of his shadow moving along the ground?
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.

Two similar triangles: the hypotenuse of each goes from the light to the head of the man's shadow.

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] 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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What are your thoughts and questions?

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Nicky G

Hi there, when you find the relationship between L and x, why do you put the L-x and 1.8 on top of the cross multiplication problem? Is that like a rule or something that the smaller triangle components go on top?

Matheno

Thanks for asking, Nicky! The answer is that we didn’t have to do it that way; the only thing that matters is that when we set the two ratios equal to each other, we’re careful to *match* the two sides given the similar triangles. Specifically, we chose to set the ratio of their bases (SMALLER triangle’s base : LARGER triangle’s base) to the ratio of their heights (SMALLER triangle’s height : LARGER triangle’s height), so the “smaller” is on top for both sides of the equation. But you could have written that instead as the inversion of both sides of… Read more »

Nikhith Pagidi

Rate of increase of distance between man’s head and tip of shadow ( head )? In the above problem

Matheno

Great question! As you can see from the figures above, the distance (we’ll call “d”) between the man’s head and the shadow’s tip is
\[ d = \ell – x \]
Hence its rate of change is
\[ \dfrac{d}{dt} = \dfrac{d\ell}{dt} – \dfrac{dx}{dt}\]
You can substitute values from there to find the answer. The important thing is: does that set-up make sense to you? Please let us know!

Nikhith Pagidi

No ,I think Mr matheno you didn’t get my question
The answer you have given is correct for rate of increase of shadow of a person
I’m asking rate of increase distance from head of the man to top of shadow

Nikhith Pagidi

Mr matheno
Let man be AB ( A is on ground and B is head)
And pole of lamp be OP(O is on ground and P be tip of lamp)
AB’ be shadow (B’ is tip of head of shadow)

Now my question is that ,
Rate of increase of BB’?

Matheno

Got it. Using the notation in the left figure immediately above, you’re looking for the rate of change of the hypotenuse of the triangle with height 1.8 m (the man’s height) and base $\ell – x.$ Let’s call that hypotenuse length “h.” Then
\[ h^2 = (1.8)^2 + (\ell – x)^2 \]
You’re looking for dh/dt. We won’t work out the math for you, but if you take the derivative with respect to time (d/dt) of both sides of that last equation and solve for dh/dt you’ll find the result you’re after.

We hope that helps.

Nikhith Pagidi

Hmm… I too did the same
But getting a lengthy process
Even though thanks for replying and giving me your time…

Matheno

Thank you for your thanks, which we greatly appreciate. In case it’s helpful, here are the next few steps as we’d do them, which might make for a simpler approach. Taking the derivative with respect to time of the preceding line gives: \[ 2h \dfrac{dh}{dt} = 0 + 2(\ell – x) \cdot \left(\dfrac{d\ell}{dt} – \dfrac{dx}{dt} \right) \] You were probably given a specific value of x and also a value for $\dfrac{dx}{dt}$, and can find $\dfrac{d\ell}{dt}$ as shown above. To find h, treat it as a separate subproblem and use the pythagorean theorem as shown above: $h^2 = (1.8)^2 +… Read more »

Kim

You give an example of the person walking TOWARDS the light and asking what rate the head of the shadow is moving. The solution given is the one where the person walks AWAY from the light. Can you provide the answer to walking TOWARDS the light? I am trying to understand these problems, and this one confuses me.

Matheno

Great question! Thanks for asking. The only thing that changes is the sign of the rates: If the person walks AWAY from the the light as in our example, his position (x-value) is INCREASING, and so dx/dt is +1.5 m/s, a positive value. The rate of change of the change in position of his shadow’s head, dl/dt, is then positive as well, reflecting the fact that “l” increases as time passes. If he were to walk TOWARD the light instead, his position (x-value) would be DECREASING, and so we would have to make dx/dt a negative value, dx/dt = –1.5… Read more »

Dougie

What if the questions like this “How fast is the end of his shadow moving when he is 6 meters away from the based of the lamp?”

When would I use the 6 meters away?

Matheno

Great question — thanks for asking!

As our analysis shows, the man’s distance from the lamp (what we call x) doesn’t affect how fast the end of his shadow moves; instead, only the rate at which he walks, dx/dt = 1.5 m/s, matters. The “6 meters away” is extra information you don’t need.

Does that make sense?