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Let A be the area of a circle with radius r. If the radius increases at a rate of dr/dt = 3 m/s, find the rate the area is increasing, dA/dt.

User DMulligan
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Final answer:

The rate at which the area of a circle is increasing, given that the radius is increasing at a rate of 3 m/s, is dA/dt = 6πr m²/s. This rate depends on the current value of the radius. For example, at r = 4 m, the area increases at a rate of 24π m²/s.

Step-by-step explanation:

To find the rate at which the area of a circle is increasing, we first note that the area A of a circle is given by the formula ℝ A = πr², where r is the radius of the circle and π (pi) is a constant. The question provides the rate at which the radius of the circle is increasing, given as dr/dt = 3 m/s. To find the rate of change of the area, we need to differentiate the area with respect to time, which will give us dA/dt.

Applying the chain rule, we differentiate A with respect to r and then multiply by dr/dt:

dA/dt = d(πr²)/dt = 2πr (dr/dt),

Since we know dr/dt, we substitute it into the equation:

dA/dt = 2πr × 3 m/s = 6πr m²/s. Therefore, the rate at which the area is increasing depends on the current value of the radius r.

For example, if at a certain moment the radius r is 4 meters, we substitute that into the equation:

dA/dt = 6π(4 m) = 24π m²/s.

This would be the rate at which the area is increasing at that instant when the radius is 4 meters.

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