#  ## Related Rates

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Emily Dulpina
1 year 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?

Matheno
Editor
1 year ago

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!

Marissa
3 years ago

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

Matheno
Editor
3 years ago

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!

Nicky G
3 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?

Matheno
Editor
3 years ago

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.

Nikhith Pagidi
4 years ago

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

Matheno
Editor
4 years ago

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
4 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

Nikhith Pagidi
4 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)

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

Matheno
Editor
4 years ago

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
4 years ago

Hmm… I too did the same
But getting a lengthy process

Matheno
Editor
4 years ago

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: $h^2 = (1.8)^2 + (\ell -x)^2$. That should give you all the values you need to substitute in and find your final answer.

We hope that helps!

Kim
4 years ago

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
Editor