A sphere magically grows in size every second. Its radius increases at a rate of

e^t mm/s, where t is time in seconds. The initial radius of the ball is 1mm.

(1a) What is the instantaneous change of the volume of the ball at time = 2s?

We are given that the radius of the sphere grows at a rate of \( e^t \) mm/s, where \( t \) is time in seconds, and the initial radius at \( t = 0 \) is 1 mm. The goal is to find the instantaneous rate of change of the sphere's volume at \( t = 2 \) seconds.

### Step-by-Step Solution:

1. **Volume of a Sphere Formula**:
   The volume \( V \) of a sphere is given by:
   \[
   V = \frac{4}{3} \pi r^3
   \]
   where \( r \) is the radius of the sphere.

2. **Rate of Change of Volume**:
   To find the instantaneous change of the volume, we need to find the derivative of the volume \( V \) with respect to time \( t \). By the chain rule, we have:
   \[
   \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}
   \]
   First, calculate \( \frac{dV}{dr} \):
   \[
   \frac{dV}{dr} = 4 \pi r^2
   \]
   Now substitute \( \frac{dr}{dt} = e^t \) (since the rate of change of the radius is given as \( e^t \) mm/s):
   \[
   \frac{dV}{dt} = 4 \pi r^2 \cdot e^t
   \]

3. **Find \( r \) at \( t = 2 \) seconds**:
   To find \( r \) at \( t = 2 \), integrate the given rate \( \frac{dr}{dt} = e^t \) to get the radius as a function of time.
   \[
   r(t) = \int e^t \, dt = e^t + C
   \]
   Since the initial radius at \( t = 0 \) is \( r(0) = 1 \), we have:
   \[
   1 = e^0 + C \implies C = 0
   \]
   Thus, \( r(t) = e^t \).

   At \( t = 2 \):
   \[
   r(2) = e^2
   \]

4. **Find \( \frac{dV}{dt} \) at \( t = 2 \)**:
   Now substitute \( r = e^2 \) and \( t = 2 \) into the expression for \( \frac{dV}{dt} \):
   \[
   \frac{dV}{dt} = 4 \pi (e^2)^2 \cdot e^2 = 4 \pi e^6
   \]

Thus, the instantaneous change in the volume of the ball at \( t = 2 \) seconds is:
\[
\boxed{4 \pi e^6 \, \text{mm}^3/\text{s}}
\]

Would you like more details or have any questions?

Here are 5 related questions you can explore:
1. What is the volume of the sphere at \( t = 2 \) seconds?
2. How does the rate of volume change vary with time?
3. What would happen to the rate of volume change if the radius increased linearly instead of exponentially?
4. Can you compute the total volume after 5 seconds?
5. How does the surface area of the sphere change over time?

**Tip**: The chain rule is a powerful tool when dealing with composite functions, like when differentiating volume with respect to time through the radius!

A sphere magically grows in size every second. Its radius increases at a rate of e^t mm/s, where t is time in seconds. The initial radius of the ball is 1mm. What is the instantaneous change of the volume of the ball at time = 2s?

Calculate the instantaneous change in the volume of a sphere whose radius increases at an exponential rate of e^t mm/s. The problem is solved step-by-step using differentiation and the chain rule, suitable for high school calculus students in Grades 11-12.

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