Posted by **Amy** on Saturday, February 12, 2011 at 9:24am.

A 15ft ladder leans against a building and its base starts to slide down the wall. The base is moving at a rate of 5ft/sec when the base is 9ft from the wall. How fast is the top of the ladder moving at this moment? What is the rate of change of the angle formed by the ladder and the ground?

I know that this forms a right triangle with a hypotenuse of 15 an two legs that measure 9 (for x) and 12 (for y).

My first step was:

x dx/dt + y dy/dt = z dz/dt

9(5) + 12 dy/dt = 15(0)

dy/dt = 3.75 ft/sec

Then I attempted to answer the second part of the question, but I wasn't sure how to do this.

What I did was:

A=1/2bh or 1/2xy

dA/dt = 1/2 x dy/dt + 1/2 y dx/dt

dA/dt = 1/2(9)(3.75) + 1/2(-12)(5)

dA/dt = -105/8 ft^2/sec

-------> I'm not sure if this is the right process or if I'm supposed to be looking for an angle? Any help is appreciated, thanks

- Calculus - not sure if my work is correct? -
**Reiny**, Saturday, February 12, 2011 at 9:40am
Your first part is correct, except your

dy/dt should be -3.75 (showing that is slides "down")

for your second part , since we are involving an angle, we have to introduce an equation containing an angle.

Trig will do this.

So let the base angle be Ø

cosØ = x/15

15cosØ = x

-15sinØ dØ/dt = dx/dt

for our given case, sinØ = 12/15 = 4/5

dØ/dt = -5/(15(4/5)) = -5/12 radians per

The negative says that the angle is decreasing, which makes sense as the ladder is falling.

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