A4 A pebble is dropped into still water
and sets off ripples forming a circular
wave. The radius of the wave after
seconds is meters. In the first few
seconds, the radius is defined by the
function .
(i) Write down an expression for the
area covered by the ripples after
(ii) What is the rate of change of this
area with respect to time?
(iii) How fast is this area changing after
2 seconds?

Ask by Schultz Bernard. in Singapore

Mar 28,2025

Question

A4 A pebble is dropped into still water
and sets off ripples forming a circular
wave. The radius of the wave after
seconds is meters. In the first few
seconds, the radius is defined by the
function .
(i) Write down an expression for the
area covered by the ripples after
(ii) What is the rate of change of this
area with respect to time?
(iii) How fast is this area changing after
2 seconds?

Ask by Schultz Bernard. in Singapore

Mar 28,2025

UpStudy AI · Accepted Answer

Given that the radius of the wave after $$ t $$ seconds is defined by the function $$ r = 3t - 0.5t^2 $$, we can find the area covered by the ripples after $$ t $$ seconds by using the formula for the area of a circle, which is $$ A = \pi r^2 $$.

(i) To find the expression for the area covered by the ripples after $$ t $$ seconds, we substitute the given function for the radius into the formula for the area of a circle:
$$ A = \pi (3t - 0.5t^2)^2 $$

(ii) To find the rate of change of this area with respect to time, we need to find the derivative of the area function with respect to time $$ t $$. This will give us the rate at which the area is changing with respect to time.

(iii) To find how fast the area is changing after 2 seconds, we substitute $$ t = 2 $$ into the derivative of the area function and calculate the result.

Let's calculate the area function and its derivative.
Simplify the expression by following steps:
- step0: Solution:
  $$\pi \left(3t-0.5t^{2}\right)^{2}$$
- step1: Convert the expressions:
  $$\pi \left(3t-\frac{1}{2}t^{2}\right)^{2}$$
- step2: Multiply the expression:
  $$\frac{\pi }{2^{2}}\times \left(6t-t^{2}\right)^{2}$$
- step3: Evaluate the power:
  $$\frac{\pi }{4}\left(36t^{2}-12t^{3}+t^{4}\right)$$
- step4: Apply the distributive property:
  $$\frac{\pi }{4}\times 36t^{2}-\frac{\pi }{4}\times 12t^{3}+\frac{\pi }{4}t^{4}$$
- step5: Multiply the numbers:
  $$9\pi t^{2}-3\pi t^{3}+\frac{\pi }{4}t^{4}$$
The expression for the area covered by the ripples after $$ t $$ seconds is $$ 9\pi t^2 - 3\pi t^3 + \frac{\pi}{4}t^4 $$.

(ii) To find the rate of change of this area with respect to time, we need to find the derivative of the area function with respect to time $$ t $$. This will give us the rate at which the area is changing with respect to time.

Let's calculate the derivative of the area function.
Evaluate the derivative by following steps:
- step0: Evaluate the derivative:
  $$\frac{d}{dt}\left(9\pi t^{2}-3\pi t^{3}+\frac{\pi }{4}t^{4}\right)$$
- step1: Use differentiation rules:
  $$\frac{d}{dt}\left(9\pi t^{2}\right)-\frac{d}{dt}\left(3\pi t^{3}\right)+\frac{d}{dt}\left(\frac{\pi }{4}t^{4}\right)$$
- step2: Calculate:
  $$18\pi t-9\pi t^{2}+\pi t^{3}$$
The derivative of the area function with respect to time $$ t $$ is $$ 18\pi t - 9\pi t^2 + \pi t^3 $$.

(iii) To find how fast the area is changing after 2 seconds, we substitute $$ t = 2 $$ into the derivative of the area function and calculate the result.
Calculate the value by following steps:
- step0: Calculate:
  $$18\pi \times 2-9\pi \times 2^{2}+\pi \times 2^{3}$$
- step1: Multiply the terms:
  $$36\pi -9\pi \times 2^{2}+\pi \times 2^{3}$$
- step2: Multiply the terms:
  $$36\pi -36\pi +\pi \times 2^{3}$$
- step3: Multiply the terms:
  $$36\pi -36\pi +8\pi $$
- step4: Simplify:
  $$8\pi $$
The area is changing at a rate of $$ 8\pi $$ square meters per second after 2 seconds.
(i) The area covered by the ripples after $$ t $$ seconds is $$ A = \pi (3t - 0.5t^2)^2 $$. (ii) The rate of change of the area with respect to time is $$ \frac{dA}{dt} = 18\pi t - 9\pi t^2 + \pi t^3 $$. (iii) After 2 seconds, the area is changing at a rate of $$ 8\pi $$ square meters per second.

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