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I stumbled across a related rates problem which involved using implicit differentiation:

A rock is initially dropped at height h a horizontal distance d from a street lamp that's H tall. The lamp casts a shadow on the ground along the line between the top of the lamp and the rock (it is a horizontal distance s from the rock). As the rock falls by h', the shadow also changes position by s'. Find s' (the distance d should drop out in the equation).

From the similar triangles formed in the situation, I have assumed that the ratio $$\frac{H}{d+s} = \frac{h}{s}$$ holds true at any time, t, during the rock's descent. Next, I am assuming that the equation is valid for any form of the equation (ie solving for any variable or moving things around yields same overall result).

Now, I must take the implicit derivative of the equation. The base of my question is moreso rooted in the act/results of doing this. I included the context in case it turns out that it presents a certain case I have overlooked.

My question arises from these cases: 1) I can use the form of the equation $$Hs = hd + hs$$ by cross multiplying. Taking implicit differentiation with respect to t then yields $$Hs' = h'd + h's + hs'$$ Then I can solve for s': $$s' = \frac{h' (d + s)}{H-h}$$

2) I can solve for s first and obtain $$s = \frac{hd}{H-h}$$ giving the result $$s' = \frac{h'd (H-h) - hd (-h')}{(H-h)^2} = \frac{h'dH}{(H-h)^2}$$ I have a feeling (have not tried to prove it yet) that 1) and 2) yield the same s' but in different forms. But the next case is the meat of my confusion.

3) I can take the inverse of each side of the equation to get $$\frac{d+s}{H} = \frac{s}{h}$$ This gives a constant term d/H which drops when we implicitly take the derivative: $$\frac{s'}{H} = \frac{s'h - sh'}{h^2}$$ which will yield a result of s' that is independent of d, which is what the original question wants. But 1) and 2) depend on d, so there seems to be a contradiction.

Are all of these equivalent or am I missing something? I assume they should all be equivalent or else implicit differentiation would not be valid. Does implicit differentiation yield different results based on the form of the equation in question?

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  • $\begingroup$ I would suggest doing a little algebra to see whether your three expressions are equivalent. $\endgroup$ – Gerry Myerson Mar 8 at 0:24
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First to reassure you: all your answers are correct!

You can use your original equation to eliminate $d$. Thus $$d+s=\frac{Hs}{h}\ ,$$ and substituting into your first answer gives $$s'=\frac{h'Hs}{h(H-h)}\ .$$ Meanwhile, solving your third answer for $s'$, you get $$s'=\frac{sh'H}{hH-h^2}\ ,$$ which is the same.

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