 	A sphere magically grows in size every second. Its radius increases at a rate of 
e^t mm/s, where t is time elapsed in seconds. The initial radius of the ball is 10mm. 
How fast is the surface area of the ball increasing at t = 2s?


To find how fast the surface area of the sphere is increasing at \( t = 2 \) seconds, we can use the following steps:

### 1. **Surface Area of a Sphere:**
The surface area \( A \) of a sphere with radius \( r \) is given by:
\[
A = 4 \pi r^2
\]

### 2. **Rate of Change of the Radius:**
We are given that the radius of the sphere changes with time according to the rate:
\[
\frac{dr}{dt} = e^t \text{ mm/s}
\]
At \( t = 2 \), this becomes:
\[
\frac{dr}{dt} \bigg|_{t=2} = e^2 \text{ mm/s}
\]

### 3. **Differentiate the Surface Area:**
We need to find how fast the surface area is changing. Using the chain rule, we differentiate the surface area with respect to time:
\[
\frac{dA}{dt} = \frac{d}{dt} \left( 4 \pi r^2 \right) = 8 \pi r \frac{dr}{dt}
\]

### 4. **Substitute the Values:**
At \( t = 2 \), we need to find the radius first. The radius increases according to the equation:
\[
r(t) = 10 + \int_0^t e^t \, dt = 10 + (e^t - 1)
\]
At \( t = 2 \):
\[
r(2) = 10 + (e^2 - 1) = 9 + e^2
\]

Now, substitute \( r(2) \) and \( \frac{dr}{dt} \) into the expression for \( \frac{dA}{dt} \):
\[
\frac{dA}{dt} = 8 \pi (9 + e^2) \cdot e^2
\]

### 5. **Final Calculation:**
Now, let's compute the final value:

\[
\frac{dA}{dt} = 8 \pi (9 + e^2) \cdot e^2
\]
Let me calculate this value.

The surface area of the sphere is increasing at a rate of approximately \( 3043.57 \, \text{mm}^2/\text{s} \) at \( t = 2 \) seconds.

Would you like more details or have any further questions?

Here are 5 related questions to expand your understanding:
1. How would the result change if the initial radius were different?
2. What is the volume of the sphere at \( t = 2 \) seconds?
3. How fast is the volume of the sphere increasing at \( t = 2 \)?
4. What is the surface area of the sphere at \( t = 2 \)?
5. How does the rate of change of the surface area compare to the rate of change of the radius over time?

**Tip:** The chain rule is a powerful tool when dealing with rates of change in related quantities like area and volume.

A sphere magically grows in size every second. Its radius increases at a rate of e^t mm/s, where t is time elapsed in seconds. The initial radius of the ball is 10mm. How fast is the surface area of the ball increasing at t = 2s?

Explore how fast the surface area of a sphere is increasing at t = 2 seconds when the radius grows exponentially at a rate of e^t mm/s. This problem uses key calculus concepts like differentiation, chain rule, and geometry of spheres. Suitable for advanced high school or early college students.

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