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.

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) Expressing theta as a function of x:

Yes, your expression for theta is correct. The cosine function relates the adjacent side (x) to the hypotenuse (24) of the right triangle formed by the ladder, the wall, and the ground. Therefore, theta can be expressed as theta = arccos(x/24).

b) Finding the value of x when theta is 40 degrees:
To find the value of x, you can use the equation you derived from part (a): cos(theta) = x/24. Plugging in theta = 40 degrees (or 2 Pi/9 radians) gives us:
cos(40 degrees) = x/24
x = cos(40 degrees) * 24
x ≈ 18.39 feet

Your answer of x ≈ 18.385 feet is almost accurate, just rounded to the nearest thousandth.

c) Finding how fast theta is changing when x = 6:
This part involves finding the rate of change of theta with respect to time. Since the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec, we need to find d(theta)/dt, the rate of change of theta with respect to time.

To do this, we can differentiate the equation from part (a) with respect to time (t). Since x is changing, we consider x to be a function of time x(t).

Differentiating both sides of the equation arccos(x(t)/24) = theta with respect to t, we get:

d(arccos(x(t)/24))/dt = d(theta)/dt

Differentiating the left side using chain rule, we have:

-1/sqrt(1 - (x(t)/24)^2) * (1/24) * dx(t)/dt = d(theta)/dt

When x = 6, we can plug in x(t) = 6 into the equation above to find d(theta)/dt.

-1/sqrt(1 - (6/24)^2) * (1/24) * dx(t)/dt = d(theta)/dt

Simplifying the expression gives us:

-1/sqrt(1 - (1/4)^2) * (1/24) * dx(t)/dt = d(theta)/dt

Simplifying further using the fact that sqrt(1 - (1/4)^2) = sqrt(15)/4, we get:

-4/sqrt(15) * (1/24) * dx(t)/dt = d(theta)/dt

Now, we need to find dx(t)/dt, which is the rate at which x is changing. Given that dx(t)/dt = 2 ft/sec, we have:

-4/sqrt(15) * (1/24) * 2 = d(theta)/dt

Simplifying this expression gives us:

-1/sqrt(15) * (1/6) = d(theta)/dt

Therefore, when x = 6, d(theta)/dt ≈ -1/(6√15) radians/sec.