Question

A spherical balloon with a 36cm diameter is being deflated. Its volume V is a function of its radius r according to

Answer 1

Question:

A spherical balloon with a 36 cm diameter is being deflated. Its volume V is a function of its radius r according to $V(r) = \frac{4}{3}\pi r^3.$

As it's deflating, it is much easier to measure its radius than its volume. Suppose the radius of the balloon is $r(t) = 18 - 18e^{-0.24t}$ cm at t seconds.

• Determine the value of V'(4) (accurate to 3 decimal places).  Write a complete sentence including the units of both 4 and V'(4) in this context.

• Determine the value of r(4) (accurate to 3 decimal places). Write a complete sentence including the units of both 4 and r(4) in this context.

Answer:

$V'(4) = 201.143$

$r(4) = 11.106$

Explanation:

Given

$V(r) = \frac{4}{3}\pi r^3.$

$r(t) = 18 - 18e^{-0.24}$

Solving (a):  V'(4)

First, we determine V('r)

$V(r) = \frac{4}{3}\pi r^3.$

Differentiate w.r.t r

$V'(r) = 3 * \frac{4}{3}\pi r^{3-1}$

$V'(r) = 3 * \frac{4}{3}\pi r^2$

$V'(r) = 4\pi r^2$

Substitute 4 for r and take $\pi = \frac{22}{7}$

$V'(4) = 4 * \frac{22}{7} * 4^2$

$V'(4) = 4 * \frac{22}{7} * 16$

$V'(4) = \frac{4 * 22* 16}{7}$

$V'(4) = \frac{1408}{7}$

$V'(4) = 201.143$

This means that the volume of the balloon when the radius is deflated to 4 seconds is 201.143 cm^3

Solving (b): r(4)

Substitute 4 for t in $r(t) = 18 - 18e^{-0.24t}$

$r(4) = 18 - 18e^{-0.24*4}$

$r(4) = 18 - 18e^{-0.96}$

$r(4) = 18 - 18*0.383$

$r(4) = 11.106$

This means that the radius of the balloon when at 4 seconds of deflation is 11.106 cm