A spherical balloon is released from rest and expands as it rises. After rising for t seconds, its radius is r cm, and its surface area is A cm2, where A = 4r2. The initial radius of the balloon is 16 cm. Given that the rate of increase of the radius is constant and has the value of 0.8 cm s– 1, find the rate of increase of A when t = 5.
Question
A spherical balloon is released from rest and expands as it rises. After rising for t seconds, its radius is r cm, and its surface area is A cm2, where A = 4r2. The initial radius of the balloon is 16 cm. Given that the rate of increase of the radius is constant and has the value of 0.8 cm s– 1, find the rate of increase of A when t = 5.
Solution
The problem involves the concept of related rates in calculus. Here's how to solve it:
Step 1: We know that the rate of increase of the radius, dr/dt, is constant and equal to 0.8 cm/s.
Step 2: We are asked to find the rate of increase of the surface area, dA/dt, when t = 5 s.
Step 3: We know that the surface area A of a sphere is given by A = 4πr².
Step 4: Differentiate both sides of this equation with respect to time t to get the rate of change of A with respect to time. Using the chain rule, we get dA/dt = 8πr(dr/dt).
Step 5: Substitute the given values into this equation. At t = 5 s, the radius r = 16 cm + 0.8 cm/s * 5 s = 20 cm. And we know that dr/dt = 0.8 cm/s.
Step 6: So, dA/dt = 8π * 20 cm * 0.8 cm/s = 128π cm²/s.
So, the rate of increase of the surface area of the balloon when t = 5 s is 128π cm²/s.
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