Question

When a hot air balloon is being inflated, the radius r meters after t minutes is given by r = 3 + 0.04t². The balloon can be assumed to be roughly spherical.

a) Work out the expressions for dr/dt and dV/dr
b) Find dV/dt
c) What is the rate of volume after 2 minutes

A) a) dr/dt = 0.08t, dV/dr = 8t, b) dV/dt = 16, c) 32
B) a) dr/dt = 0.08t, dV/dr = 8t, b) dV/dt = 8, c) 16
C) a) dr/dt = 0.04, dV/dr = 8t, b) dV/dt = 32, c) 64
D) a) dr/dt = 0.04, dV/dr = 8t, b) dV/dt = 16, c) 32

Answer 1

The expressions for dr/dt and dV/dr are dr/dt = 0.08t and dV/dr = 8πr², respectively. The expression for dV/dt is found by multiplying dV/dr by dr/dt, resulting in dV/dt = 0.64πrt. The rate of volume after 2 minutes is given by plugging t=2 into the expression for dV/dt, yielding 1.28πr.

Step-by-step explanation:

To find the expression for dr/dt, we need to take the derivative of the equation r = 3 + 0.04t² with respect to t. Taking the derivative, we get dr/dt = 0.08t. To find the expression for dV/dr, we need to take the derivative of the volume formula V = (4/3)πr³ with respect to r. Taking the derivative, we get dV/dr = 8πr².

To find dV/dt, we can use the chain rule. Since V = (4/3)πr³, we can write dV/dr = (4/3)π(3r²) = 4πr². Then, we can multiply dV/dr by dr/dt to find dV/dt. So, dV/dt = (8πr²)(0.08t) = 0.64πrt.

To find the rate of volume after 2 minutes, we can plug the value of t=2 into the expression for dV/dt. So, dV/dt = 0.64π(2)r = 1.28πr.