Posted by **CMM** on Wednesday, March 23, 2011 at 12:39am.

A 24 foot ladder is leaning against a building. Let x be the distance between the bottom of the ladder and the building and let theta be the angle between the ladder and the ground.

a)Express theta as a function of x.

cos theta = x/24

theta = arccos(x/24) correct/incorrect?

b)When the angle is 40 degrees (2 Pi/9 radians) how big is x?

40 degrees = arccos(x/24)

cos(40 deg.)= cos(arccos)(x/24)

x= 18.385 feet coorect/incorrect?

c)Suppose the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. When x=6, how fast is theta(in radians/sec.) changing?

I really don't know about this last part... help/hints would be greatly appreciated! Seems sort of like a physics problem.

- Calculus -
**Reiny**, Wednesday, March 23, 2011 at 8:35am
your a) and b) are correct

c) write your opening equation as

x = 24cosŲ

differentiate with respect to t, (t in seconds)

dx/dt = -24sinŲ dŲ/dt (#1)

we are given dx/dt = 2 when x = 6

when x=6 we can get the height h of the triangle by Pythagoras h^2 + 6 ^2 = 24^2

h = √540 and sinŲ = √540/24

so from #1

dŲ/dt = (dx/dt) / (-24sinŲ)

= 2/(-24(√540/24)

= -2/√540 radians/sec or appr. -0.086 radians/sec

You titled your subject as "Calculus".

This last part belongs to a part of Calculus called,

"Rates of Change" or "Related Rates"

Several chapters of your text should be devoted to it, depending on the depth of the course.

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