A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to r(t)=20√t+2, find the area of the ripple as a function of time. Find the area of the ripple at t=2

Enter the exact answer. A=? in^2

The area \( A(t) \) of a circle is given by the formula:

\[
A(t) = \pi r(t)^2
\]

where \( r(t) \) is the radius of the circle as a function of time. Given the radius function:

\[
r(t) = 20\sqrt{t} + 2
\]

we can substitute this into the formula for the area.

### Step 1: Substituting the radius function into the area formula
\[
A(t) = \pi \left( 20\sqrt{t} + 2 \right)^2
\]

### Step 2: Expanding the expression
Now, we expand the expression \( \left( 20\sqrt{t} + 2 \right)^2 \).

Using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), we get:

\[
(20\sqrt{t} + 2)^2 = (20\sqrt{t})^2 + 2(20\sqrt{t})(2) + 2^2
\]

First, calculate each term:

- \( (20\sqrt{t})^2 = 400t \)
- \( 2(20\sqrt{t})(2) = 80\sqrt{t} \)
- \( 2^2 = 4 \)

So, the expanded form is:

\[
(20\sqrt{t} + 2)^2 = 400t + 80\sqrt{t} + 4
\]

### Step 3: Substituting this back into the area formula
\[
A(t) = \pi (400t + 80\sqrt{t} + 4)
\]

Thus, the area of the ripple as a function of time is:

\[
A(t) = \pi (400t + 80\sqrt{t} + 4)
\]

### Step 4: Finding the area at \( t = 2 \)
Now, we substitute \( t = 2 \) into the area formula:

\[
A(2) = \pi (400(2) + 80\sqrt{2} + 4)
\]
\[
A(2) = \pi (800 + 80\sqrt{2} + 4)
\]
\[
A(2) = \pi (804 + 80\sqrt{2})
\]

Thus, the exact area at \( t = 2 \) is:

\[
A(2) = \pi (804 + 80\sqrt{2}) \text{ square inches}
\]

### Conclusion
The exact area of the ripple at \( t = 2 \) is \( \pi (804 + 80\sqrt{2}) \, \text{in}^2 \).

Would you like further clarification on this problem or any additional details? 

Here are five related questions for practice:
1. How would you compute the derivative of \( A(t) \) to find the rate of change of the area?
2. What is the radius of the ripple at \( t = 2 \)?
3. How does the area of the ripple change as time increases?
4. Can you express the rate of change of the radius as a function of time?
5. If the radius function changed to \( r(t) = 10\sqrt{t} + 5 \), how would the area formula be affected?

**Tip**: Expanding squared terms like \( (a + b)^2 \) helps in simplifying area and volume calculations involving growing shapes!

A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to r(t)=20√t+2, find the area of the ripple as a function of time. Find the area of the ripple at t=2. Enter the exact answer. A=? in^2

Learn how to calculate the area of a circular ripple formed by a raindrop using the radius function r(t) = 20√t + 2. This detailed solution involves geometric formulas and algebraic expansions. The area at t=2 is provided as an exact answer.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation