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I'm asking this on the math stack exchange because it seems that the key part of this physics problem I'm asking for help on is more related to the geometry of it than the physics of it.

I'm independently going through Physics for Scientists and Engineers, 3rd Edition, by Raymond A. Serway, Chapter 3, problem 80, which, to paraphrase the problem, goes like this:

A boy scout ties a rope to his food, throws the rope over a branch, and starts walking at a constant velocity $v_0$ and the food starts going straight up. The food starts at the level of his hands and at a height $h$ above that is the branch. The distance he is away from the vertical rope at any given time is $x$. (Here's my drawing of the situation below, the book has a fancier picture of a real human and a tree, but I've stripped it down to the essentials).

The problem goes on to say:

(a) Prove that the food's velocity $v$ when the boy's position is $x$ is equal to:

$$v = v_0 \cdot x \cdot (h^2 + x^2)^{-1/2}.$$

(b) Prove that the food's acceleration a when the boy's position is $x$ is equal to:

$$a = v_0^2 \cdot h^2 \cdot (h^2 + x^2)^{-3/2}.$$

I can solve part (a) by drawing a triangle down by the boy's hands as shown and understanding that the food's velocity wraps around the branch in the direction along $l$, and $v_0$ is then the hypotenuse, so $v$ is $v_0 \sin \theta$, which $\sin \theta$ is then equivalent to $x / (h^2 + x^2)^{1/2}$, so $v$ is $v_0 \cdot x \cdot (h^2 + x^2)^{-1/2}$, which is the answer.

I can also see that a is the time derivative of $v$. I can imagine using the chain rule to say that $a = dv/dt = dv/d \theta \cdot d \theta /dx \cdot dx/dt$. I can see that $dv/d \theta = v_0 \cos \theta $ and I can see that $dx/dt$ is the constant $v_0$ that the boy is moving at. I can look at the answer I am supposed to arrive at for part (b) and what I have so far and see that $d \theta /dx$ is supposed to be $\cos \theta$, but that's what I can't see as natuarally following.

Can someone help me visualize or analytically prove why in this setup, $d \theta /dx$ is necessarily $\cos \theta$?

enter image description here

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I would need to draw a picture, but: if on your picture you replace $v_0$ with $dx$, so that the small triangle becomes really small, you can see that the side of the small triangle orthogonal to $l$ has (approximately) length $l\,d\theta$. On the other hand, from the small triangle, it has length $cos\theta\,dx$. Therefore $d\theta/dx=(cos\theta)/l$. – user8268 Jul 5 '12 at 14:23
I see, Américo Tavares, thanks! I also see that I missed an $l$, that $d \theta / dx$ is $\cos \theta / l$ like you showed. From the perspective of a long narrow triangle with sides $l$, $\approx l$, and small angle $d \theta$, $l d \theta$ is reasoned as the arc length of a sector of length $l d \theta$. And I see that I need the $l$ to get the $-3/2$. – mring Jul 5 '12 at 14:37

2 Answers 2

up vote 8 down vote accepted

Note (from the big triangle) that $\tan\theta=\frac{x}{h}$. Differentiating both sides with respect to $x$, and using the Chain Rule, we get $$\sec^2\theta\frac{d\theta}{dx}=\frac{1}{h},$$ and therefore $$\frac{d\theta}{dx}=\frac{1}{h}\cos^2\theta.$$ Not quite the $\cos\theta$ that you expected, but kind of close.

Remark: The angle is a very useful thing from the point of view of the physical intuition. However, let $y$ be the distance of the food from the branch, and let $k$ be the total length of the rope. Then the height of the food is $h-y$, so if we know everything about $y$, we know everything about the height of the food.

Note that in the big triangle, the hypotenuse is $k-y$. It follows by the Pythagorean Theorem that $$(k-y)^2=h^2+x^2.$$ Differentiating with respect to $t$, we get $$-2(k-y)\frac{dy}{dt}=2x\frac{dx}{dt}.\tag{$1$}$$ But $k-y=(h^2+x^2)^{1/2}$ and $\frac{dx}{dt}=v_0$, so we get $$-\frac{dy}{dt}=v_0x(h^2+x^2)^{-1/2}.$$ To find the rate at which the food is rising, just change the sign.

For the acceleration, differentiating both sides of $(1)$ with respect to $t$ works very nicely. Since $x$ changes at a constant rate, the derivative of $x\frac{dx}{dt}$ with respect to $t$ is just $\left(\frac{dx}{dt}\right)^2$.

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Now $d\theta$ is so small that we can approximate the arc length to $dx\cos \theta$ $$r.d\theta=dx \cos \theta$$ So $$\frac{d\theta}{dx}=\frac{\cos \theta}{r}$$ where $r=\sqrt{x^2+h^2}$

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