Question

a ladder which is 15ft leans against a wall. suppose that when the bottom of the ladder is x feet away from the wall,the bottom is being pushed towards the wall at a rate of 1/2x feet per second. how fast is the top of the ladder rising at the moment the bottom is 5 feet from the wall?


so i did pythagerous and i figured out the other side which was 14.14, since that is the y side or the height of the wall at a point where the ladder is 5 feet from the wall

from there i ddid the derivative of
x^2 + y^ 2 = c^2
which was
2x(dx/dt) + 2y(dy/dt) = 0 ( this is 0 cuz the hypothenouse is constant)

and then i subbed in the numbers and solved for dy/dt
the numbers i subbed in were
a = 5
y = 14.14
dx/dt= 0.5

and im getting -.0.176 as my answer but the correct answer for some reason is 0.884 , can someoen tell me what im doing wrong?

thanks a lot

Answers

Damon
x/dt = - 1/2 it is getting smaller.
5 (-.5) + 14.14 (dy/dt) = 0
dy/dt = .176
so I agree with you except for the sign.
just wondering
okay so other than the sign are you saying what im doing is correct? and the answer is most likely wrong?
Damon
Yes.

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