A stone is dropped into a lake, creating a circular

oliviayychengwh

oliviayychengwh

Answered question

2021-12-19

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm/s. Find the rate at which the area within the circle is increasing after 1 s.

Answer & Explanation

Deufemiak7

Deufemiak7

Beginner2021-12-20Added 34 answers

Step 1 
The ripple travels outward at 60 cm/s, this is the rate of change of the radius with respect to time 
dr dt =60cms 
We can also derive an equation for the radius at time t from the problem description.
r=60t 
Step 2 
The area of a circle is A=πr2 , and we need to find the rate that this is increasing, which is dA dt . So differentiate the A equation with respect to time t. 
A=πr2 
dA dt =π2rdr dt  
We know the value of dr dt  therefore, we can change that. Furthermore We can replace r=60t so that we have an equation where all we need to do to calculate the rate of change of the area is plug in the time t.
dA dt =π2(60t)(60)=7200π 
Step 3 
So for t=1 
dA dt =7200π(1)22619cm2s

madeleinejames20

madeleinejames20

Skilled2023-06-17Added 165 answers

Step 1:
To solve the problem, let's first define some variables:
r - the radius of the circular ripple at time t in centimeters (cm).
A - the area within the circle at time t in square centimeters (cm2).
We are given that the ripple travels outward at a speed of 60 cm/s, which means that the rate of change of the radius with respect to time is constant and equal to 60 cm/s. Therefore, we can express the rate of change of the radius as:
drdt=60 cm/s.
We are asked to find the rate at which the area within the circle is increasing after 1 second, which means we need to find dAdt when t=1 second.
The area of a circle is given by the formula: A=πr2.
Differentiating both sides of this equation with respect to time t, we get:
dAdt=2πrdrdt.
Substituting the given value of drdt=60 cm/s, we have:
dAdt=2πr·60 cm2/s.
Step 2:
Now, we need to find the value of r at t=1 second. Since the ripple is traveling at a constant speed, after 1 second, the radius of the ripple will be equal to the distance traveled, which is 60 cm/s multiplied by 1 second:
r=60 cm/s·1 s=60 cm.
Substituting this value into the equation for dAdt, we get:
dAdt=2π·60cm·60 cm/s.
Simplifying, we find:
dAdt=7200π cm2/s.
Therefore, the rate at which the area within the circle is increasing after 1 second is 7200π cm2/s.
Eliza Beth13

Eliza Beth13

Skilled2023-06-17Added 130 answers

Answer:
7200πcm2/s
Explanation:
Given that the ripple travels outward at a speed of 60 cm/s, the radius of the circle will also increase at a rate of 60 cm/s.
After 1 second, the radius will be r=60cm/s×1s=60cm.
To find the rate of change of the area with respect to time, we differentiate the area formula with respect to time:
dAdt=ddt(πr2)
Using the chain rule, we can differentiate each term separately:
dAdt=2πrdrdt
Substituting the values we have, we get:
dAdt=2π×60cm×60cm/s
Simplifying, we find:
dAdt=7200πcm2/s
Therefore, after 1 second, the area within the circle is increasing at a rate of 7200πcm2/s.
Mr Solver

Mr Solver

Skilled2023-06-17Added 147 answers

The formula for the area of a circle is A=πr2. To find dAdt, we need to differentiate both sides of this equation with respect to time:
dAdt=ddt(πr2)
Using the chain rule, we can rewrite this as:
dAdt=π·ddt(r2)
To differentiate r2 with respect to time, we can apply the power rule:
dAdt=π·2r·drdt
Given that the ripple is traveling outward at a speed of 60 cm/s, we can express drdt as 60 cm/s.
Now, let's evaluate dAdt after 1 second. Since we're given the speed, we can find r after 1 second using the formula r(t)={speed}·t. Plugging in the values, we have r(1)=60{cm/s}·1{s}=60{cm}.
Substituting r=60 cm and drdt=60 cm/s into the equation for dAdt, we get:
dAdt=π·2·60{cm}·60{cm/s}
Simplifying this expression gives us the rate at which the area within the circle is increasing after 1 second.
Now, let's calculate the value:
dAdt=π·2·60{cm}·60{cm/s}=7200π{cm}2/{s}
Therefore, the rate at which the area within the circle is increasing after 1 second is 7200π{cm}2/{s}.

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