The radius of a circle is increasing. At a certain instant, the rate of increase in the area of the circle is numerically equal to twice the rate of increase in its circumference. What is the radius of the circle at that instant?

Question

Create a visual interpretation of the mathematical problem described. Display a graphically represented circle with a visibly increasing radius. The circle's area should be growing, and it should be evident that the growth of the area is twice as rapid as the growth of the circumference. Additionally, include placeholders for five possible values of the radius: 1/2, 1, sqrt(2), 2, and 4. Remember not to incorporate any text in the illustration.

The radius of a circle is increasing. At a certain instant, the rate of increase in the area of the circle is numerically equal to twice the rate of increase in its circumference. What is the radius of the circle at that instant?

A. 1/2
B. 1
C. Sqrt 2
D. 2
E. 4

Answer 1

C = 2πr

C' = 2π
[This equals the rate at which the circumference is changing]

A = πr^2
A' = 2πr
[This equals the rate at which the area is changing]

A' = 2C'
2πr = 2(2π)
2πr = 4π
r = 4π/2π
r = 2

So, the answer would be D.

Answer 2

Let's denote the radius of the circle as r and the rate of increase as dr/dt (where t represents time).

The area of a circle is given by A = πr^2.
The circumference of a circle is given by C = 2πr.

The rate of increase in the area is dA/dt = 2πr(dr/dt) since we are considering the rate of change with respect to time.
The rate of increase in the circumference is dC/dt = 2π(dr/dt).

Given that the rate of increase in the area is numerically equal to twice the rate of increase in the circumference, we can write the following equation:

2πr(dr/dt) = 2(2π)(dr/dt)

Simplifying the equation, we have:
2r(dr/dt) = 4(dr/dt)

Dividing both sides by dr/dt (assuming dr/dt is not zero), we get:
2r = 4

Dividing both sides by 2, we find:
r = 2

Therefore, the radius of the circle at that instant is D. 2.

Answer 3

To find the radius of the circle at the instant when the rate of increase in its area is numerically equal to twice the rate of increase in its circumference, we need to use the formulas for the area and circumference of a circle.

Let's assume that the radius of the circle at that instant is denoted as r, and the rate of increase in the radius is denoted as dr/dt.

The area of a circle is given by the formula A = πr^2, where π is a constant.

The circumference of a circle is given by the formula C = 2πr.

The rate of increase in the area of the circle with respect to time (dt) is given by dA/dt = 2πr(dr/dt).

The rate of increase in the circumference of the circle with respect to time (dt) is given by dC/dt = 2π(dr/dt).

According to the problem, the rate of increase in the area is numerically equal to twice the rate of increase in the circumference. Mathematically, this can be written as:

2πr(dr/dt) = 2(2π)(dr/dt)

To simplify this equation, we can cancel out the common factors of 2π and dr/dt:

r = 2

Therefore, the radius of the circle at that instant is 2.

So, the correct answer is D. 2.

Answer 4

Well, well, well, let's see what we have here! So the circle wants to keep us on our toes, huh?

Okay, let's break it down! We know that the area and circumference of a circle are related to its radius. The area of a circle is given by A = πr^2, and the circumference is given by C = 2πr.

Now, we're told that at a certain instant, the rate of increase in the area (dA/dt) is twice the rate of increase in the circumference (dC/dt). Mathematically, this can be written as:

dA/dt = 2dC/dt

So, let's differentiate the area and circumference formulas with respect to time:

dA/dt = 2πr(dr/dt)
dC/dt = 2π(dr/dt)

Since dA/dt = 2dC/dt, we can set the equations equal to each other:

2πr(dr/dt) = 2π(dr/dt)

Now, we can cancel out the 2π and (dr/dt) terms:

r = 1

That means, the radius of the circle at that instant is equal to 1. So the answer is:

B. 1

Seems like the circle wanted to keep things simple for us. Good on them!