Question

A 13 ft ladder is leaning against a house when its base begins to slide away. By the time the base is 3 ft from the house, the base is moving at the rate of 2 ft/sec. Answer the following:

At what speed is the top of the ladder sliding down the wall at that time?

Answer 1

Keep getting -12/(2sqrt160) but that's wrong..

Answer 2

Do I differentiate x^2+y^2=13^2 or do I differentiate 3^2+y^2=13^2?

Answer 3

Refer to the attachment from Mathematica 9.

Answer 4

to start
H^2 + W^2 = 13^2

then, after slipping in x (to left) and y (down wall) directions:
(H-y)^2 + (W+x)^2 = 13^2

differentiate this 2nd equation wrt time (t):
2(H-y)dy/dt + 2(W+x)dx/dt = 0

dy/dt = -dx/dt {(W+x)/(H-y) }

plugging in some of the numbers gives
dy/dt = -(2) { 3/ (H-y) } = -6 / (H-y)

so what is (H - y)?

well from the original equation
(H-y)^2 + (W+x)^2 = 13^2

H - y = sqrt ( 13^2 - 3 ^ 2) = sqrt (160) = sqrt (16 x 10) = 4 sqrt(10)

thus dy/dt = -6sqrt(10)/4 = etc etc

Answer 5

slight correction:
-2(H-y)dy/dt + 2(W+x)dx/dt = 0

NB the minus sign at the start ==> then cancels the minus sign at the end. dy/dt is positive.....

Answer 6

***Keep getting -12/(2sqrt160) but that's wrong.. ***

numerically, that seems to be the correct answer
take the derivative with respect to time of
x^2 + y^2 = 169

Perhaps they want the answer "rationalized" and written as
\[ - \frac{3 \sqrt{10}}{20} \]

Answer 7

well my answer, 6/4sqrt(10) = 3/2sqrt(10) = 3sqrt(10)/2(10) = 3sqrt(10)/20

Answer 8

also, they ask for the speed, so they may expect a positive value. (the sign will depend on where we put the coordinate axes, but in general, speed is a positive quantity)

Answer 9

Awesome yep I just had to remove the -.... That's annoying hahah I redid the problem like 3 times and couldnt find my error!

Answer 10

Thanks everyone