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December 17, 2014

December 17, 2014

Posted by **Anonymous** on Friday, June 22, 2012 at 8:12am.

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)?

- Calculus -
**Reiny**, Friday, June 22, 2012 at 8:23amfor 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

etc

are you sure that you were not given the rate at which the ladder was pulled away ?

- Calculus -
**Anonymous**, Friday, June 22, 2012 at 8:46amThe first part of the problem was find a formula for g'(x) in terms of

x and g(x)?

I got -x/g(x)

- Calculus -
**Reiny**, Friday, June 22, 2012 at 9:00amAt 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.

- Calculus -
**Anonymous**, Friday, June 22, 2012 at 9:07amI 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!!

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