When the area in square units of an expanding circle is increasing twice as fast as its radius in linear units, the radius is

(a) 1/4pi
(b) 1/4
(c) 1/pi
(d) 1
(e) pi

since a = pi r^2

da = 2 pi r dr
2 dr = 2 pi r dr
1 = pi r
r = 1/pi

Let's solve this problem step-by-step:

Step 1: Let's assume the radius of the expanding circle is "r" units.

Step 2: The area of a circle is given by the formula A = πr^2.

Step 3: According to the problem, the area of the circle is increasing twice as fast as its radius. This means that the rate of change of the area with respect to time (dA/dt) is equal to 2 times the rate of change of the radius with respect to time (dr/dt).

Step 4: We can find the rate of change of the area with respect to the radius by differentiating the area formula with respect to the radius: dA/dr = 2πr.

Step 5: Setting the rate of change of the area with respect to the radius equal to the given rate of change, we have: 2πr = dr/dt.

Step 6: Now we can solve this first-order differential equation. Separating the variables, we have: (1/2π) dr/r = dt.

Step 7: Integrating both sides, we get: (1/2π) ln|r| = t + C, where C is the constant of integration.

Step 8: Exponentiating both sides, we have: e^[(1/2π) ln|r|] = e^(t + C).

Step 9: Simplifying, we get: r^(1/2π) = ke^t, where k is the constant e^C.

Step 10: Solving for r, we have: r = k^(2π) e^(2πt).

Step 11: Since "k" is an arbitrary constant, we can rename it as "k" and write: r = ke^(2πt).

Step 12: Now, we need to find the value of "k" that satisfies the given condition. At time t = 0, the radius is equal to 1. Therefore, we have: 1 = ke^(2π * 0).

Step 13: Simplifying, we find: 1 = k.

Step 14: Thus, the equation for the radius becomes: r = e^(2πt).

Step 15: Now we can find the value of the radius when the area is increasing twice as fast as its radius. We can differentiate the equation for the area with respect to time to find the rate of change of the area: dA/dt = 2πr dr/dt.

Step 16: Substituting the equation for the radius, we have: dA/dt = 2π(e^(2πt))(2πe^(2πt)) = 4π^2 e^(4πt).

Step 17: According to the problem, the area is increasing twice as fast as the radius. Therefore, we have: 4π^2 e^(4πt) = 2e^(2πt).

Step 18: Simplifying, we get: 2π^2 e^(2πt) = 1.

Step 19: Dividing both sides by 2π^2 e^(2πt), we have: e^(2πt) = (1/2π^2).

Step 20: Taking the natural logarithm of both sides, we get: 2πt = ln(1/2π^2).

Step 21: Simplifying, we find: t = (1/2π) ln(1/2π^2).

Step 22: Therefore, the value of the radius at this time is: r = e^(2πt) = e^(2π[(1/2π) ln(1/2π^2)]) = e^ln(1/2π^2) = 1/2π^2.

Step 23: Comparing this value with the given options, we see that the correct answer is (b) 1/4.

So, the radius of the expanding circle is 1/4.

To solve this problem, we need to use the formula for the area of a circle and relate it to the change in radius.

The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.

We are given that the area is increasing twice as fast as the radius, so we have the relationship: dA/dr = 2.

To find the correct choice for the radius, we need to differentiate the area formula with respect to the radius and set it equal to 2.

Let's differentiate the area formula with respect to r:

dA/dr = d(πr^2)/dr = 2πr

Now we can set this expression equal to 2 and solve for r:

2πr = 2
r = 1/π

Therefore, the correct choice for the radius is (c) 1/π.