please help me out this calculus questions!?

Question

1. Let f(x)= sin^-1(x), then there exists a number c in the interval [-1,1] that satisfies the conclusion of the mean value theorem. Find C



2.A ferris wheel, 50 ft in diameter, revolves at the rate of 10 radians per minute. how fast is a passeger rising (vertically) when the passenger is 15ft higher (vertical distance) than the center of the Ferris wheel and is rising?


these two questions , help me plzzzzzzzzz

Answer 1

1)

Let f(x) = sin^(-1)(x) in the interval [-1, 1]

Remember that by the mean value theorem, for an interval [a, b], there exists a value c such that

f'(c) = [f(b) - f(a)] / [b - a]

In order to find c, we first require the derivative.

f'(x) = 1/sqrt(1 - x^2)

Since we wanted to calculate f'(c), we let x = c. Then

f'(c) = 1/sqrt(1 - c^2)

We want this to equal [f(b) - f(a)] / [b - a]. In our case, a = -1 and b = 1, so

[f(1) - f(-1)] / [1 - (-1)]

Therefore, since f(x) = sin^(-1)(x), then we require the value

[sin^(-1)(1) - sin^(-1)(-1)] / [1 + 1]

Sine is equal to one at precisely one point: pi/2. It is equal to -1 at -pi/2 (since sine inverse is defined on the interval -pi/2 to pi/2). So we plug these values in like so.

[pi/2 - (-pi/2)] / 2
[pi/2 + pi/2] / 2
(2pi/2) / 2
pi/2

Now that we have this value, we WANT to equate this to f'(c).

1/sqrt(1 - c^2) = pi/2

Now we solve for c. Taking the reciprocal of both sides, we have

sqrt(1 - c^2) = 2/pi

Squaring both sides,

1 - c^2 = 4/(pi)^2

Moving the -c^2 to the right hand side, and the 4/(pi)^2 to the left,

1 - 4/(pi)^2 = c^2

Making a single fraction on the left hand side,

(pi^2 - 4)/(pi)^2 = c^2

Therefore, we're going to get two values for c.

c = {sqrt(pi^2 - 4)/pi , -sqrt(pi^2 - 4)/pi}

Answer 2

Great work, Puggy!