+1904 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.
+9466 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.
