your a) and b) are correct
c) write your opening equation as
x = 24cosØ
differentiate with respect to t, (t in seconds)
dx/dt = -24sinØ dØ/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 sinØ = √540/24
so from #1
dØ/dt = (dx/dt) / (-24sinØ)
= 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.
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.
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