- #1
Cilabitaon
- 69
- 0
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
[tex]\dot{y} = \frac{dy}{dt}[/tex],
[tex]\dot{x} = \frac{dx}{dt}[/tex],
[tex]y' = \frac{dy}{dx}[/tex]
[tex]y' = \dot{y} \times \dot{x}^{-1}[/tex] (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;
[tex]x = 12 , \dot{x} = 2[/tex]
Also;
[tex]y = \sqrt{13^{2} - x^{2}}[/tex]
The next part I did only because I had to use implicit differentiation at some point in this.
[tex]y^{2} = 13^{2} - x^{2}[/tex]
[tex]2yy' = -2x[/tex]
[tex] y' = \frac{-2x}{2y}[/tex]
Now rearrange the chain rule equation to get;
[tex]\dot{y} = y' \times \dot{x}[/tex]
[tex]\dot{y} = \frac{-4x}{2y}[/tex]
[tex]\dot{y} = \frac{-2x}{y}[/tex]
To which I then sub in my values for when x = 12
[tex]\dot{y} = \frac{-2(12)}{({13^{2} - 12^{2})^{\frac{1}{2}}}[/tex]
[tex]\dot{y} = -4.8ms^{-1}[/tex]
I think the first part is right, but I am having trouble with part (b)
I assume it has something to do with
[tex]\tan\Theta = \frac{y}{x}[/tex]
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;
[tex]\frac{d}{dx}(\tan\Theta) = -\frac{y}{x^2}[/tex]
Does anybody know if this is the right way to go?
Last edited: