The radius of a circle is decreasing at a constant rate of 0.1 centimeters per second. In terms of the circumference C, what is the rate of the area of the circle, in square centimeters per second?

a. -(0.2)pie C
b. -(0.1)C
c. -(0.1)C/2 pie
d. (0.1)^2C
e. (0.1)^2 pie C

C=2πr

A=πr²
dA/dr=2πr=C

dr/dt=-0.1 cm/sec

dA/dt
= dA/dr * dr/dt
= 2πr (-0.1) cm²/s
= -0.1C cm²/s

Well, to find the rate of the area of the circle, we need to differentiate the formula for the area of a circle with respect to time. So, we have:

A = πr^2

where A is the area and r is the radius.

Differentiating both sides of the equation with respect to time, we get:

dA/dt = 2πr (dr/dt)

Now, we know that the radius is decreasing at a constant rate of 0.1 centimeters per second, so dr/dt is -0.1. Plugging this into our equation, we get:

dA/dt = 2πr (-0.1)

Since we want the rate in terms of the circumference C, we can substitute C = 2πr into the equation:

dA/dt = -0.2πC

Therefore, the answer is a. -(0.2)πC.

To find the rate at which the area of the circle is changing, we need to use the formula for the area of a circle, A = πr², where A is the area and r is the radius.

First, let's differentiate both sides of the formula with respect to time (t):

dA/dt = d(πr²)/dt

To solve for dA/dt, we need to express r in terms of the circumference (C). The relationship between the radius and the circumference of a circle is given by C = 2πr.

We can solve this equation for r:

r = C/(2π)

Now, substitute this value of r into the equation for the area:

A = π(C/(2π))²

Simplifying:

A = πC²/4π²

A = C²/4π

Now, differentiate both sides of the equation with respect to time (t):

dA/dt = d(C²/4π)/dt

Using the chain rule on the right-hand side, we have:

dA/dt = (2C/4π) * (dC/dt)

Since the radius is decreasing at a constant rate of 0.1 centimeters per second, the circumference is also decreasing at the same rate. Therefore, dC/dt = -0.1 cm/s.

Substituting this value into the equation for dA/dt:

dA/dt = (2C/4π) * (-0.1)

Simplifying:

dA/dt = -0.1C/2π

Therefore, the rate of change of the area of the circle is -(0.1)C/2π.

So, the correct answer is c. -(0.1)C/2π.

To find the rate of change of the area of a circle, we need to use derivatives. The formula for the area of a circle is A = πr^2, where A represents the area and r represents the radius.

We are given that the radius of the circle is decreasing at a constant rate of 0.1 centimeters per second. Let's denote the rate of change of the radius as dr/dt, which is -0.1 cm/s since it is decreasing.

Using the formula for the area of a circle, we can differentiate both sides with respect to time (t) using implicit differentiation:

dA/dt = d/dt (πr^2)

To do this, we need to apply the chain rule. The derivative of A with respect to t (dA/dt) represents the rate of change of the area of the circle, which is what we are looking for.

dA/dt = d/dt (πr^2)

The derivative of A with respect to r (dA/dr) is 2πr, and the derivative of r with respect to t (dr/dt) is -0.1. Applying the chain rule, we have:

dA/dt = dA/dr * dr/dt

dA/dt = 2πr * (-0.1)

dA/dt = -0.2πr

Now, we need to express the rate of change of the area in terms of the circumference C. The formula for the circumference of a circle is C = 2πr. Solving for r, we get:

r = C / (2π)

Substituting this expression for r into the rate of change of the area, we have:

dA/dt = -0.2π * (C / (2π))

Simplifying, the π cancels out:

dA/dt = -0.2 * C / 2

dA/dt = -0.1 * C

Therefore, the rate of change of the area of the circle, in square centimeters per second, is given by:

b. -(0.1)C