Related Rates - Differential Equations

Cilabitaon
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Homework Statement


A ladder, of length 13m, is leaning against a wall. The base of the ladder begins to slide, frictionless, away from the wall; by the time the base is horizontally 12m away from the wall it is moving at a rate of 2ms-1

a)How fast is the top of the ladder sliding down the wall?

b)At what rate is the angle subtended from the ground by the ladder changing by?

Homework Equations


Assuming that
\dot{y} = \frac{dy}{dt},
\dot{x} = \frac{dx}{dt},
y' = \frac{dy}{dx}

y' = \dot{y} \times \dot{x}^{-1} (chain rule)
Implicit Differentiation(basic)
Pythagoras' Theorem

The Attempt at a Solution


First I drew a right angle triangle with the ladder as the hypotenuse and labelled the sides y(the wall) and x(the ground).

Then I get that, when;
x = 12 , \dot{x} = 2

Also;
y = \sqrt{13^{2} - x^{2}}

The next part I did only because I had to use implicit differentiation at some point in this.

y^{2} = 13^{2} - x^{2}

2yy' = -2x

y' = \frac{-2x}{2y}

Now rearrange the chain rule equation to get;

\dot{y} = y' \times \dot{x}

\dot{y} = \frac{-4x}{2y}

\dot{y} = \frac{-2x}{y}

To which I then sub in my values for when x = 12

\dot{y} = \frac{-2(12)}{({13^{2} - 12^{2})^{\frac{1}{2}}}

\dot{y} = -4.8ms^{-1}

I think the first part is right, but I am having trouble with part (b)

I assume it has something to do with
\tan\Theta = \frac{y}{x}

But I can't differentiate that(actually, I can, I just can't seem to get any sensible kind of answer).

All I'm thinking right now is;
\frac{d}{dx}(\tan\Theta) = -\frac{y}{x^2}

Does anybody know if this is the right way to go?
 
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Since you're looking for dθ/dt, try differentiating with respect to t instead of x.
 
vela said:
Since you're looking for dθ/dt, try differentiating with respect to t instead of x.

I would, but the point is I have to use the chain rule as many times as I can. The points scored in the exercise are 5 for a correct answer and 2 for either the use of implicit differentiation or the chain rule. This is basic, I know, but we fail if we get less than 16 points.
 
You get to use the chain rule and implicit differentiation if you differentiate with respect to t.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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