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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 (a) 1 s, (b) 3 s, and (c) 5 s.  What can you conclude?

 

If anyone who knows how to solve for the answer and can write down the steps, I would really appreciate it.  Thanks.

 May 22, 2018
 #1
avatar+8852 
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The circle travels outward at a speed of 60 cm/s, which means the radius of the circle is increasing at a rate of 60 cm per second.

 

radius after  0  sec  =  0

radius after  1  sec  =  60

radius after  2  sec  =  60 + 60  =  60(2)

radius after  3  sec  =  60 + 60 + 60  =  60(3)

radius after  s  sec  =  60s

 

Or we can say...

 

radius  =  60s    , where   s  is the number of seconds after the stone hit the water.

 

Let   s  =  the number of seconds after the stone hit the water   and   a  =  the area of the circle

 

We know that the equation for the area of a circle is...

 

a  =  pi (radius)2

                            Substitute  60s  in for radius.

a  =  pi( 60s )2

 

a  =  3600 pi s2

 

This equation tells us the area, a , at any  s . But we want to know the rate of change in  a  per change in  s  at any  s . So take  d / ds  of both sides of the equation.

 

da / ds  =  d / ds [ 3600 pi s2 ]

 

da / ds  =  ( 3600 )( pi )( d/ds s2 )

 

da / ds  =  ( 3600 )( pi )( 2s )

 

da / ds  =  7200 pi s

 

To find the rate at which the area is increasing after 1 second, plug in  1 for  s  and solve for  da / ds .

 

(a)   when  s  =  1 ,          da / ds   =   7200 pi (1)   =   7200 pi     (sq cm per second)

 

(b)   when  s  =  3 ,          da / ds   =   7200 pi (2)   =   14400 pi

 

I'll let you do part  (c)  .

 

Notice that the bigger the number of seconds, the bigger the rate at which the area is increasing.

 May 22, 2018

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