Gas is escaping from a spherical balloon at 45 ext m3/ extmin. How fast is the radius decreasing when r = 5…

Question

Gas is escaping from a spherical balloon at 45 ext m3/ extmin. How fast is the radius decreasing when r = 5 ext m?

ScanSolve AI Tutor · Accepted Answer

Here's the solution to the next rate problem:

2. Gas is escaping from a spherical balloon at $45 \text{ m}^3/\text{min}$ . How fast is the radius decreasing when $r = 5 \text{ m}$ ?

Given:
Rate of change of volume, $\frac{dV}{dt} = -45 \text{ m}^3/\text{min}$ (negative because the volume is decreasing)
Radius, $r = 5 \text{ m}$

The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$ .

Differentiate the volume formula with respect to time $t$ .
$V = \frac{4}{3}\pi r^3$
Using the chain rule, we differentiate both sides with respect to $t$ :
$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$ $\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt}$ $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Substitute the given values into the derivative and solve for $\frac{dr}{dt}$ .
$-45 \text{ m}^3/\text{min} = 4\pi (5 \text{ m})^2 \frac{dr}{dt}$ $-45 = 4\pi (25) \frac{dr}{dt}$ $-45 = 100\pi \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{-45}{100\pi}$ $\frac{dr}{dt} = \frac{-9}{20\pi} \text{ m/min}$
The negative sign indicates that the radius is decreasing.

The rate at which the radius is decreasing is \boxed{\textbf{\frac{9}{20\pi} m/min}}.

Gas is escaping from a spherical balloon at 45 ext m3/ extmin. How fast is the radius decreasing when r = 5 ext m?