# Rate of change of an angle

• opus
Which seems like it would result in the same result as just using sin(y) directly.The chain rule is used for compositional operations in order to simplify the calculation. It allows you to move functions around without having to rewrite the entire equation.f

Gold Member

## Homework Statement

A pole stands 75 feet tall. An angle θ is formed when wires of various lengths of ##x## feet are attached from the ground to the top of the pole.
Find the rate of change of the angle ##\frac{dθ}{dx}## when a wire of length 90 ft is attached.

## The Attempt at a Solution

I'm not sure where to start here. So in drawing a picture of the stated problem, I started off by finding the length of the ground which was 49.75 feet. Then I took the rise (75 feet) over the run (49.75 feet), to get a slope or rate of change of 1.5 which seems to make sense because it seems that we're dealing with a right triangle, and the slope would be the hypotenuse which is linear. But then I don't understand what is being asked by a "change in the angle". The angle should stay the same no matter how long we make the wire.
Any help would be greatly appreciated!

The angle should stay the same no matter how long we make the wire.
I think you need to consider that the wire runs from the top of the pole to the ground and is taut. Therefore the angle the wire makes with the pole will change depending on how far from the pole it is attached to the ground. You need find ##\theta## as function of ##x## and get the derivative
$$\left. \frac{d\theta}{dx} \right|_{x = 90\ \mathrm{ft}}$$

• opus
I think you need to consider that the wire runs from the top of the pole to the ground and is taut. Therefore the angle the wire makes with the pole will change depending on how far from the pole it is attached to the ground. You need find ##\theta## as function of ##x## and get the derivative
$$\left. \frac{d\theta}{dx} \right|_{x = 90\ \mathrm{ft}}$$
Ohhh ok. That makes sense. I was reading it as starting from the one spot on the ground, and increasing the length of the wire until it was at the top of the pole. I'll try to work it again with this new approach.

I think I have a better idea of what's going on, but I can't seem to get the correct solution:

(i) I want to find how θ changes with respect to changes in ##x##. In this problem, as ##x## increases, θ decreases.

(ii) $$sin(θ) = \frac{opp}{hyp} = \frac{75}{x}$$

(iii) $$\frac{d}{dx}\left(sin{-1}(\frac{75}{x})\right)$$

(iv) $$Let ~y = sin^{-1}(u) ~and ~u = \frac{75}{x}$$

(v) $$\frac{dy}{du} = \frac{1}{\sqrt{1-(u)^2}}~and~\frac{du}{dx}=\frac{-75}{x^2}$$

(vi) $$\frac{dy}{dx} = \frac{-75}{x^2\sqrt{1-(\frac{75}{x})^2}}$$

(vii) $$\frac{d}{dx}\left(sin^{-1}(\frac{75}{x})\right)~|~x=90$$

(viii) $$= -0.005$$

This is wrong according to the book. Any guidance on where I went wrong?

I think I have a better idea of what's going on, but I can't seem to get the correct solution:

(i) I want to find how θ changes with respect to changes in ##x##. In this problem, as ##x## increases, θ decreases.

(ii) $$sin(θ) = \frac{opp}{hyp} = \frac{75}{x}$$

(iii) $$\frac{d}{dx}\left(sin{-1}(\frac{75}{x})\right)$$
Starting with the line above, this is overly complicated, and possibly wrong in subsequent steps. You also have "lost" one side of the equation you started with.
Differentiate both sides of this equation -- ##\sin(\theta) = \frac {75} x##.
##\frac d{dx}\left( \sin(\theta)\right) = \frac d {dx}\frac {75} x##
You need the chain rule to carry out the differentiation on the left side.

I get an answer for ##\frac {d\theta}{dx}## closer to -.02.
opus said:
(iv) $$Let ~y = sin^{-1}(u) ~and ~u = \frac{75}{x}$$

(v) $$\frac{dy}{du} = \frac{1}{\sqrt{1-(u)^2}}~and~\frac{du}{dx}=\frac{-75}{x^2}$$

(vi) $$\frac{dy}{dx} = \frac{-75}{x^2\sqrt{1-(\frac{75}{x})^2}}$$

(vii) $$\frac{d}{dx}\left(sin^{-1}(\frac{75}{x})\right)~|~x=90$$

(viii) $$= -0.005$$

This is wrong according to the book. Any guidance on where I went wrong?

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• opus
That's me. Makes things as complicated as possible Yeah I I always feel like I leave stuff out. Notation isn't my strong point with this stuff and I have troubles writing things out clearly in a step-by-step fashion.

You need the chain rule to carry out the differentiation on the left side.
I guess this would be a good time to ask this question because it's relevant to this problem (I was going to make a different thread).
Why do I need to use the chain rule for that? I know that the the chain rule is used for composition for functions, but as I've gone further, it is being used in cases that don't quite seem like compositions to me.
For example,
##\frac{d}{dx}sin(y)##
This doesn't seem like a composition to me, yet as I'm going over implicit differentiation, it says to use the chain rule which would result in ##cos(y)\frac{d}{dx}##

I'll explain a little further:

(i) ##\frac{d}{dx}(sin(y))##

(ii) Let ##y = sin(u)## and ##u=y##

(iii) ##\frac{dy}{du} = cos(u)## and ##\frac{du}{dx} = 1##

(iv) ##\frac{dy}{dx} = cosy(y) ⋅1 ##

Now this is where my work isn't the same as the book's. The way that I've done it, would result in just ##cos(y)##, and the book has it as ##cos(y)\frac{dy}{dx}## What's happening with this mismatch where I have a 1 and the book has a ##\frac{dy}{dx}##?

Once again, I think I'm just confused by the notation.

That's me. Makes things as complicated as possible Yeah I I always feel like I leave stuff out. Notation isn't my strong point with this stuff and I have troubles writing things out clearly in a step-by-step fashion.

I guess this would be a good time to ask this question because it's relevant to this problem (I was going to make a different thread).
Why do I need to use the chain rule for that? I know that the the chain rule is used for composition for functions, but as I've gone further, it is being used in cases that don't quite seem like compositions to me.
For example,
##\frac{d}{dx}sin(y)##
This doesn't seem like a composition to me, yet as I'm going over implicit differentiation, it says to use the chain rule which would result in ##cos(y)\frac{d}{dx}##
Nope: ##\frac{d}{dx} \sin y = \frac{d \sin y}{dy} \cdot \frac{dy}{dx} = \cos y \: \frac{dy}{dx}.##

• opus
Maybe this will clear up some of my confusion:
What's the difference between ##\frac{d}{dx}## and ##\frac{dy}{dx}## ?
Apologies for the bombardment of questions.

Maybe this will clear up some of my confusion:
What's the difference between ##\frac{d}{dx}## and ##\frac{dy}{dx}## ?
Apologies for the bombardment of questions.
Big difference!
The first, ##\frac{d}{dx}##, is an operator. It says that you intend to find the derivative with respect to x of something, but you haven't actually performed this operation yet.
The second, ##\frac{dy}{dx}##, is the derivative of y, with respect to x. Another notation for this is y', or sometimes ##\dot y## (the notation that Newton developed, that represents the derivative of y with respect to time, t).

• opus
You took a direct, but complicated turn after equation (ii) of post #4. Suppose you solve (ii) for ##x## as a function of ##\theta## and take the derivative (use the quotient rule). Then you have ##\frac {dx}{d\theta}## and are practically done. Remember that ##\frac {d\theta} {dx} = \frac 1 {\frac {dx}{d\theta}}##

• opus
I like the method of Mark44 and Ray Vickson. It's nice and direct. And differentiating like that is called implicit differentiation, where you differentiate both sides and use the chain rule where appropriate.

• opus
It's good to know that the direct way will give the correct result. But if some fundamental knowledge of slopes allows a simple way, IMHO, that is better.

• opus
It's good to know that the direct way will give the correct result. But if some fundamental knowledge of slopes allows a simple way, IMHO, that is better.

Your way has the advantage that it doesn't need one to know implicit differentiation. But I think it's worth learning, so I like that method. But both are good.

• opus
Ok guys I think I got it. Thanks a ton for all the helpful responses.

Here's what I've got, and I think it is correct.

(i) $$sin(θ) = \frac{75}{x}$$

(ii) $$\frac{d}{dx}\left(sin(θ)\right) = \frac{d}{dx}\left(\frac{75}{x}\right)$$
(A) $$cos(θ)\left(\frac{dθ}{dx}\right) = \frac{-75}{x^2}$$
(B) $$\frac{dθ}{dx} = \frac{-75}{x^2cos(θ)}$$
(C) At x=90, ##cos(θ) = 0.552##

(iii) $$\frac{dθ}{dx} | x=90$$
$$=\frac{-75}{90^2cos(θ)}$$
$$=\frac{-75}{90^2⋅0.552}$$
$$=-0.016 ~radians~ per~ foot$$