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!