Question

Assume that the radius r of a sphere is expanding at a rate of 90 cm/min. The volume of a sphere is V= 4/3 πr³ and its surface area is 4πr². Determine the rate at which the surface area is changing with respect to time at t=2 min, assuming that r=10 at t=0.

(Use symbolic notation and fractions where needed.)
dA/dt = __

Answer 1

To determine the rate at which the surface area is changing with respect to time, we differentiate the surface area with respect to radius, and multiply by the given rate of radius change. The rate of change in surface area, dS/dt, is then calculated at the specific time where the radius is known or determined by the expansion rate.

Step-by-step explanation:

The question involves using calculus to find the rate at which the surface area of a sphere changes with respect to time, given that the radius is expanding at a steady rate. Since the radius r of a sphere is expanding at a rate of 90 cm/min, and we know the surface area of a sphere is given by S = 4πr², we need to find dS/dt when r = 10 at t = 0. To find dS/dt, we can differentiate S with respect to t using the chain rule:

dS/dt = dS/dr × dr/dt

dS/dr = 8πr, and we are given dr/dt = 90 cm/min. Therefore, dS/dt = 8πr × 90. At t=2 min, since the initial radius is 10 cm and the radius is expanding at 90 cm/min, r = 10 cm + (90 cm/min)(2 min) = 10 cm + 180 cm = 190 cm. Plugging in this value of r into our derivative, we get dS/dt = 8π(190 cm) × 90 cm/min = 8π(190) × 90 cm²/min.

We are given that the radius r is expanding at a rate of 90 cm/min, so dr/dt = 90 cm/min. Substituting this value into the derivative formula gives dA/dt = 8πr(90 cm/min). At t=2 min, the given radius is 10, so plugging in these values gives dA/dt = 8π(10)(90 cm/min) = 720π cm²/min.