McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Radical Equations
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Exercise 8 Page 261

Practice makes perfect
a We are given a function that can approximate the maximum speed that a person can run.
S=Ď€ sqrt(9.8 l/1.6) In this function S represents the speed in meters per second and l represents the leg length of a person in meters. We will find an approximate value for the maximum running speed of a person with a leg length of 1.1 meters long. To do so, let's substitute l= 1.1 into the function and solve it for S.
S=Ď€ sqrt(9.8 l/1.6)
S=Ď€ sqrt(9.8 ( 1.1)/1.6)
â–Ľ
Solve for S
S=Ď€ sqrt(10.78/1.6)
S=Ď€ sqrt(6.7375)
S=8.154536...
S=8.2
The maximum speed that the person can run, to the nearest tenth, is 8.2 meters per second.
b We will now calculate the leg length of a person who has a running spreed of 6.7 meters per second. To do this, let's substitute S= 6.7 into the function and solve it for l.
S=Ď€ sqrt(9.8 l/1.6)
6.7=Ď€ sqrt(9.8 l/1.6)
â–Ľ
Solve for l
6.7/Ď€=sqrt(9.8 l/1.6)
(6.7/Ď€)^2=9.8 l/1.6
(6.7)^2/Ď€^2=9.8 l/1.6
44.89/Ď€^2=9.8 l/1.6
44.89/Ď€^2=9.8/1.6 l
44.89/Ď€^2=6.125 l
44.89/Ď€^2(6.125)=l
l=44.89/Ď€^2(6.125)
l=0.742580...
l=0.7
The leg length of the person, to the nearest tenth, is 0.7 meters.
c Consider the given function.

S=Ď€ sqrt(9.8 l/1.6) As the leg length increases, the numerator of the quotient goes up. Therefore, the value of the radical increases, which implies that the maximum speed also increases. Let's test our argument by examining the cases in Part A and Part B.

l π sqrt(9.8 l/1.6) S
0.7 π sqrt(9.8 ( 0.7)/1.6) 6.7
1.1 π sqrt(9.8 ( 1.1)/1.6) 8.2

As we can also see from the table, the longer the leg length, the higher the maximum speed.