## 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

[Scroll down for text (non-video) version of the 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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(Archived comments from before we started our Forum are below.)

Shan, who is 2 meters tall, is approaching a post that holds a lamp 6 meters above the ground. If he is walking at a speed of 1:5 m/s, how fast is the end of his shadow moving (with respect to the lamp post) when he is 6 meters away from the base of the lamp post?

Can you help me on how to solve this?

Thanks for asking, Emily.

The solution to this problem is the same as the solution above, with only two changes: (1) the man’s height is now 2 m instead of 1.8 m, and (2) the sign of dx/dt is negative, dx/dt = -1.5 m/s, since he is moving toward instead of away from the post. You can read more about that sign-change in our reply to Kim in the comments below.

If you make those two substitutions in the solution above, you should arrive at the answer you’re after.

We hope that helps!

how did you get .70l=x, from l-.30l=x

Thanks for asking, Marissa! Factor the $\ell$ out and you’ll see:

$$ \ell – 0.30 \ell = (1 – 0.30) \ell = 0.70 \ell $$

We hope that helps!

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?

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 that equation (putting the larger values on top for BOTH sides), and the math would come out the same in the end.

So no, there’s no rule that the smaller components go on top; it’s just what we happened to do here.

Does that answer your question? We hope so,and thanks again for asking!

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

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!

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

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’?

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.

Hmm… I too did the same

But getting a lengthy process

Even though thanks for replying and giving me your time…

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

xand also a value for $\dfrac{dx}{dt}$, and can find $\dfrac{d\ell}{dt}$ as shown above. To findh, treat it as a separate subproblem and use the pythagorean theorem as shown above: $h^2 = (1.8)^2 + (\ell -x)^2$. That should give you all the values you need to substitute in and find your final answer.We hope that helps!

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.

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 m/s. You’d just put that negative sign in there “by hand” yourself, to capture the fact that x decreases as time passes. Then following exactly the same analysis (all the way up to the end when you substitute values), you’d get a negative value for dl/dt, reflecting the fact that “l” decreases as time passes. That is, if he walks toward the light at the same rate, dl/dt = –2.1 m/s.

[That said, a warning: If you are using an online homework system and it asks for “the rate at which the shadow’s head changes position when he walks toward the light,” they’ve *probably* already taken the negative sign into account, and so you’d have to enter a positive value to be marked correct. Unfortunately they can be (annoyingly) inconsistent about that, so you might have to try both the positive and negative values to see which it accepts.]

Does that make sense? What further questions do you have??

Thanks again for asking!