#  ## Related Rates

### 15 What are your thoughts and questions?

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Marissa

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

Matheno

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

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 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

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

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)

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

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

We hope that helps!

Kim

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

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??

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?