In SERIOUS need of help!

[cv2yanks13]

I have a practice problem that is confusing me.. .and I have a test coming up in a few days... I need to know the answer to this problem to do well on the test!

A ladder 41 feet long that was leaning against a vertical wall begins to slip. Its top slides down the wall while its bottom moves along the level ground at a constant speed of 4 ft/s. How fast is the top of the ladder moving when it is 9 ft above the ground?

Clearly show all steps in the process of solving the problem and use correct derivative notation.

I've done similar problems regarding rate of change... but this was is confusing me for some reason. Can anyone help me out? I really need to know the answer to this problem so I can

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ Hypotenuse \ = \ 41 \ ft. \ and \ \frac{dx}{dt} \ = \ 4 \ ft./s, \ find \ \frac{dy}{dt} \ when \ y \ = \ 9.\)

\(\displaystyle h^{2} \ = \ x^{2}+y^{2}, \ \implies \ 0 \ = \ 2x\frac{dx}{dt}+2y\frac{dy}{dt}, \ x\frac{dx}{dt} \ = \ -y\frac{dy}{dt}\)

\(\displaystyle Now, \ when \ y \ = \ 9, \ \sqrt{41^{2}-9^{2}} \ = \ x \ = \ 40 \ ft.\)

\(\displaystyle Hence, \ (40)(4) \ = \ -9\frac{dy}{dt}, \ therefore \ \frac{dy}{dt} \ = \ -\frac{160}{9} \ ft./s\)