If the radius of a circle changes at the rate of −2/π m/sec, at what rate does…

Concept: When two quantities are tied together by an equation, differentiating that equation with respect to time ties their rates together as well. A circle’s area and radius satisfy A = πr2, so by the chain rule dA/dt = (dA/dr)·(dr/dt) = 2πr·(dr/dt). The factor 2πr is the circumference at that instant, and dA/dt inherits the sign of dr/dt — a radius that is shrinking forces a shrinking area.

Application: here the radial rate is dr/dt = −2/π m/sec and the instant of interest is r = 10 m.

1. Write the relation between the quantities: A = πr2.

2. Differentiate both sides with respect to time t using the chain rule: dA/dt = 2πr·(dr/dt).

3. Substitute the radius at the required instant, r = 10 m: dA/dt = 2π(10)·(dr/dt) = 20π·(dr/dt).

4. Substitute the given radial rate, dr/dt = −2/π m/sec: dA/dt = 20π × (−2/π).

5. Cancel π: dA/dt = 20 × (−2) = −40.

6. Attach the units, (m) × (m/sec) = m2/sec: dA/dt = −40 m2/sec.

Cross-check: both the dimension and the size agree. The factor 2πr at r = 10 m is the circumference 20π m, and a length in metres multiplied by a radial speed in m/sec gives m2/sec, which is exactly the dimension of an area rate. For the size, 20π ≈ 62·8 and the radial speed has magnitude 2/π ≈ 0·64, whose product is about 40; the radius is decreasing, so the area rate carries a minus sign.

Hence the area of the circle is changing at −40 m2/sec — that is, shrinking by 40 m2 every second.