+0  
 
0
237
1
avatar+1868 

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.

gibsonj338  May 22, 2018
 #1
avatar+7339 
+2

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.

hectictar  May 22, 2018

30 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.