We are tasked with finding the rate of change of the radius ($\frac{dr}{dt}$) of a sphere when the volume is $V = 3778 \, \text{in}^3$, and the volume is decreasing at a constant rate of $\frac{dV}{dt} = -3312 \, \text{in}^3/\text{min}$. The volume of a sphere is given by:

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

Step 1: Differentiate the volume equation with respect to time ($t$):

$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

Step 2: Solve for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4 \pi r^2}$

Step 3: Find the radius when $V = 3778$:

Rearranging the volume formula to solve for $r$:

$r = \left( \frac{3V}{4 \pi} \right)^{\frac{1}{3}}$

Substituting $V = 3778 \, \text{in}^3$:

$r = \left( \frac{3 \cdot 3778}{4 \pi} \right)^{\frac{1}{3}}$

Step 4: Compute $\frac{dr}{dt}$:

Substitute $r$ and $\frac{dV}{dt} = -3312$ into the equation for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{-3312}{4 \pi r^2}$

Now, compute $r$ and then $\frac{dr}{dt}$. math

Constants

dV_dt = -3312 # Rate of change of volume in cubic inches per minute V = 3778 # Volume of the sphere in cubic inches

Step 1: Compute radius r when V = 3778

r = ((3 * V) / (4 * math.pi)) ** (1/3)

Step 2: Compute rate of change of radius dr/dt

dr_dt = dV_dt / (4 * math.pi * r**2)

r, dr_dt### Results:

1. The radius of the sphere when the volume is $3778 \, \text{in}^3$ is approximately:

   $r \approx 9.662 \, \text{inches}$

2. The rate of change of the radius ($\frac{dr}{dt}$) is approximately:

   $\frac{dr}{dt} \approx -2.823 \, \text{inches per minute}$

This means the radius is decreasing at a rate of about $2.823 \, \text{inches per minute}$ at this moment.

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Would you like a detailed breakdown of any of these steps?

Related Questions:

1. What happens to the rate of change of radius as the sphere gets smaller?
2. How is the relationship between $\frac{dV}{dt}$ and $\frac{dr}{dt}$ affected by a larger volume?
3. Can this method apply if the volume decreases at a non-constant rate?
4. How do changes in $\frac{dr}{dt}$ affect the surface area of the sphere over time?
5. What if the rate of volume change $\frac{dV}{dt}$ were positive instead of negative?

Tip:

When working with rates of change, always confirm whether they are increasing or decreasing, as this affects the sign of derivatives significantly.