Question

A balloon is inflated at a rate of 200cm³s⁻¹. After T seconds, when the balloon has a radius r cm and volume V cm³, the following formulae apply.

V=200t and V=4/3(πr³)

a) Write down dV/dt and dV/dr

b) By using the chain rule, dV/dt = dV/dr * dr/dt, work out an expression for dr/dt. Find the volume and radius when t=1 and so find the rate at which the radius is changing when t=1.

A) dV/dt = 200, dr/dt = 3/4π
B) dV/dt = 400, dr/dt = 3/2π
C) dV/dt = 200, dr/dt = 3/2π
D) dV/dt = 400, dr/dt = 3/4π

Answer 1

dV/dt = 200, dr/dt = 3/4π

Step-by-step explanation:

Given the equations V=200t and V=4/3(πr³), we need to find dV/dt and dV/dr. Taking the derivative of V=200t with respect to t gives us dV/dt = 200. Taking the derivative of V=4/3(πr³) with respect to r gives us dV/dr = 4πr². Using the chain rule, we have dV/dt = dV/dr * dr/dt. Solving for dr/dt, we find dr/dt = dV/dt / dV/dr = 200 / (4πr²) = 50/(πr²). When t=1, we can substitute t=1 into V=200t to find V=200(1) = 200 cm³. Similarly, we can substitute t=1 into V=4/3(πr³) to find 200 = 4/3(πr³). Solving for r, we get r = (3/4π)^(1/3).