posted by Anonymous .
Suppose a ladder of length 36 feet rests against a wall, and we pull the base of the ladder horizontally away from the wall. For a given distance x that we pull the ladder horizontally, let g(x) be the height of the ladder on the wall, under the assumption that the other end of the ladder always contacts the wall. We know from Pythagorean theorem that for every x:
x2 + [g(x)]2 = 36^2
At what rate is the height changing when
x = 8?
When the ladder is 6 feet from the ground (g(x) = 6)?
for easier typing, I will let y = g(x)
then x^2 + y^2 = 1296
2x dx/dt + 2y dy/dt = 0
dy/dx = -x dx/dt / y
when x=8, y = √1232 or appr 35.1
dy/dx = (-8/√1232)(dx/dt)
for the second part, let y = 6, then x = √1260 or appr 35.5
are you sure that you were not given the rate at which the ladder was pulled away ?
The first part of the problem was find a formula for g'(x) in terms of
x and g(x)?
I got -x/g(x)
At what point are you in the study of related rates?
Are you not differentiating with respect to "time" ?
If not, then you would simply have
x^2 + y^2 = 1296
2x + 2y dy/dx = 0
y dy/dx = -x
dy/dx = -x/y or -x/g(x), so that agrees with your answer.
I figured it out, for each part I just plug in the given x and the g(x) answer you gave me to the formula -x/g(x) and simplify.
Thanks for your help!!