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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 $\left( \dfrac{1.8\, \text{m}}{6.0\, \text{m}}\right)$:
\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

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

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

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

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1 year ago

That’s a wonderful explanation, but I’m having a bit of a problem understanding the 3d step. As an eastern European – we use the f'(x) notation more often, so I blatantly just don’t understand the example :D.
Could u give a solution based on v(t)=s'(t) and a(t)=v'(t)?
I also don’t really get the “in respect to time” part…

Last edited 1 year ago by Justas
Reply to  Justas
1 year ago

Thank you for your question! We now use our Forum for such questions and answers since it offers a LOT more functionality than the ‘comments’ here.

Please see our reply there, which we hope will help:

Emily Dulpina
3 years ago

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?

5 years ago

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

Nicky G
5 years ago

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?

Nikhith Pagidi
5 years ago

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

Nikhith Pagidi
Reply to  Matheno
5 years ago

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
Reply to  Matheno
5 years ago

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’?

Nikhith Pagidi
Reply to  Matheno
5 years ago

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

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