The circumference of a circle is increasing at a rate of 2pi/5 inches per minute. When the radius is 5 inches, how fast is the area of the circle increasing in square inches per minute?
A. 1/5
B. pi/5
C. 2
D. 2pi
E. 25pi
since dC/dt = 2π/5, dr/dt = 1/5
a = πr^2, so
da/dt = 2πr dr/dt = 2π(5)(1/5) = 2π
Well, let's put on our circus hats and calculate this! The formula to find the area of a circle is A = πr², where A is the area and r is the radius. Now, we need to find the rate at which the area is changing with respect to time.
First, let's find the rate at which the radius is changing with respect to time. We know that the circumference of the circle is increasing at a rate of 2π/5 inches per minute. Since the circumference formula is C = 2πr, we can differentiate both sides with respect to time to get dC/dt = 2π(dr/dt). Notice that the dr/dt term represents the rate of change of the radius. Solving for dr/dt, we find that dr/dt = (dC/dt)/(2π).
Now, let's substitute the given value for the radius. When the radius is 5 inches, we can find the rate at which the radius is changing by plugging it into the formula: dr/dt = (2π/5)/(2π) = 1/5 inches per minute.
Finally, we can find the rate at which the area is changing with respect to time. We can use the chain rule, which states that dA/dt = dA/dr * dr/dt. To find dA/dr, we differentiate the area formula A = πr² with respect to r, which gives dA/dr = 2πr. Now, we can substitute the given value for the radius and the calculated rate at which the radius is changing: dA/dt = (2π * 5) * (1/5) = 2π square inches per minute.
So, the correct answer is option D. The area of the circle is increasing at a rate of 2π square inches per minute. Now you can juggle with this information!
To find the rate at which the area of the circle is increasing, we need to relate the rate of change of the circumference to the rate of change of the area.
The circumference of a circle is given by the formula: C = 2πr, where r is the radius of the circle.
Differentiating both sides of the equation with respect to time (t), we have: dC/dt = 2π(dr/dt).
Given that dC/dt = 2π/5, and when the radius is 5 inches, we need to find dr/dt when r = 5.
Substituting the given values into the equation, we have:
2π/5 = 2π(dr/dt)
dr/dt = (2π/5) / (2π)
dr/dt = 1/5
Therefore, the rate at which the area of the circle is increasing is 1/5 square inches per minute.
The correct answer is A. 1/5.
To find the rate at which the area of the circle is increasing, we need to use the formula for the area of a circle, which is given by A = πr^2, where A is the area and r is the radius.
Given that the circumference of the circle is increasing at a rate of 2π/5 inches per minute, we can find the rate at which the radius is increasing using the relationship between circumference and radius: C = 2πr.
Differentiating both sides of the equation with respect to time (t), we get dC/dt = 2π(dr/dt).
Since we're interested in the rate at which the area of the circle is changing, we need to find the derivative of the area with respect to time.
Differentiating both sides of the area formula (A = πr^2) with respect to time, we get dA/dt = 2πr(dr/dt).
Now, let's plug in the given values: r = 5 and dC/dt = 2π/5.
dA/dt = 2πr(dr/dt)
dA/dt = 2π(5)(2π/5) [Substituting r = 5 and dC/dt = 2π/5]
dA/dt = 2π^2
Therefore, the area of the circle is increasing at a rate of 2π^2 square inches per minute.
The correct answer choice is D. 2π.