At what rate (with respect to time) is the angle θ between the ground and the ladder changing, if the top of the ladder is sliding down the wall at the rate of r inches per second, at the moment that the top of the ladder is h feet from the ground? (You're looking here for an equation in terms of h and θ.)

Question

anonymous · Answer

I know that dtheta/dt=dtheta/dh * dh/dt

anonymous · Answer

and that arcsin(h/25)=theta

anonymous · Answer

and the derivative of arcsin is equal to 1/sqrt(1-x^2)

anonymous · Answer

Pleas help...I just don't know how to put it altogether.

nincompoop · Answer

is this a general question without dimensions?

anonymous · Answer

yes

nincompoop · Answer

can you draw what the problem is asking?

anonymous · Answer

I got : r/sqrt(1/(h/25)^2)
but the answer choices say Otherwise :(
I have the answer choices and picture attached

nincompoop · Answer

|dw:1398228246079:dw|

nincompoop · Answer

you have dimensions!

anonymous · Answer

yes, just the height of the ladder aka 
hypotenuse of the triangle

anonymous · Answer

...any ideas about the next step?

nincompoop · Answer

yeah

anonymous · Answer

I got r/sqrt(1/(h/25)^2) but I know I tripped cause its not in the answer choice.  can you tell me where I missed up.

nincompoop · Answer

how did you get that is what I want to know

anonymous · Answer

dtheta/dt=dtheta/dh * dh/dt

anonymous · Answer

and that arcsin(h/25)=theta

nincompoop · Answer

:(

anonymous · Answer

I also know by the formula of the derivative of aarcsin that it is equal to 1/sqrt(1-x^2) and so I plugged in h/25 for x and multiplied everything by r

nincompoop · Answer

can you set it up using tangent?

nincompoop · Answer

tanθ = h/√(625-h^2)

nincompoop · Answer

secθ = 25/√(625-h^2)

nincompoop · Answer

do you know where to take from here?

anonymous · Answer

no, I got confuse. WHy is it in terms of secant instead of sine?

nincompoop · Answer

why sine?

anonymous · Answer

cause sintheta= (h/25)-->arcsin(h/25)=theta

nincompoop · Answer

how far does it get you?

anonymous · Answer

then I took the derivative of arcsin(h/25)=theta to get 1/sqrt(1/(h/25)^2)

anonymous · Answer

I multiplied that with r to get r/sqrt(1/(h/25)^2)

anonymous · Answer

and this was my final answer, but it is  not on the answers. I attached them if you want to check them out.

nincompoop · Answer

you need to justify why you're doing every step

nincompoop · Answer

look at this 
tanθ = h/√(625-h^2)

anonymous · Answer

Your final answer, was it on the answer choices I gave?

nincompoop · Answer

it is not finished

nincompoop · Answer

dθ/dt = 1/√(625-h^2) dh/dt

anonymous · Answer

what do I do after that? can you give me the next step?

anonymous · Answer

A solution using Mathematica 9 is attached.