Exercise 2 Page 421

Practice makes perfect

a Let's first create a visual of the circular waves created every second as George jumps off the tire swing.

As time increases, the radius of the circle increases at a rate of 16 ft/sec. r=16t To find the radius at any specific second, take the rate and multiple it by that time, as seen by the equation above.

b The area of any circle can be found with just the radius. Recall the formula for the area of a circle is A= π r^2. This equation represents how the area of the circle depends on the radius.

c To get a composite function that represents the relationship between area and time, we must bring our two equations together. Both equations have the radius as a variable. We can use this to relate them.

A= π r^2

Substitute

r= 16t

A= π ( 16t)^2

Simplify power

A= 256 π t^2

d Finally, let's substitute A=22,000 into the equation from Part C, and solve for t. We will use the time in the equation from Part A to find the radius, and double it to get the diameter.

A= 256 π t^2

Substitute

A= 22,000

22,000 = 256 π t^2

Div

Divide by 256 π

22,000/256 π = t^2

ReduceFrac

a/b=.a /16./.b /16.

1,375/16 π = t^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(1,375/16 π) = t

RearrangeEqn

Rearrange equation

t = sqrt(1,375/16 π)

Simplify radical and terms

t = 5.230177 ...

If the area of the pond is 22,000 ft^2 then it will take sqrt(1,37516 π) or approximately 5.23 seconds for the circle to cover the pond. Now let's plug this into the equation from Part A to get the radius of the pond.

r=16t

Substitute

t= 5.230177

r=16 ( 5.230177)

Multiply

r=83.682832 ...

The radius of the pond is 83.682832 ft, multiplying it by two will give the diameter. 83.682832 * 2 = 167.365664

The diameter of the pond is 167.365664 ft.

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Exercise 2 Page 421

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