SOLUTION: The spread of a contaminant on the surface of a lake is increasing in a circular pattern. The radius of the contaminant can be modeled by f(t)=6 square root t ( just 6 squart of t

Question 1055332: The spread of a contaminant on the surface of a lake is increasing in a circular pattern. The radius of the contaminant can be modeled by f(t)=6 square root t ( just 6 squart of t no parenthesis) where f(t) is in meters and t is hours since contamination. Reminder: The area of a circle is pie r^2. (piesign r to the power of 2)
A. Find a function (A of F)(t) that gives the area of the circular leak in terms of the time t since the spread began. Simplify this function as much as possible.
B.) Find the size of the contaminated area after 20 hours. Include units and round to the nearest whole number.
c.) When will the size of the contaminated area be 850 square meters? Include units and round to one decimal place.
I'm in desperate need of help with this problem. I am really lost.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
f(t)=6 sqrt(t). That is the radius
The area is pi*r^2=pi*36t. 6 sqrt(t) squared is 6 squared* sqrt(t) squared,or 36 t.
A(t)=36t*pi m^2.
after 20 hours, the area is 720 pi m^2=2262 m^2. Check with the radius, which is 6sqrt(20)=26.83 m
The area is 26.83^2*pi=2261.47, close enough with rounding.
850 m^2=36*pi*t
divide 850 by 36*pi=7.5 hours.
--------------------
Check
radius after 7.5 hours is 6*sqrt(7.5)=16.43 m
The area is pi*(16.43^2) m^2=848.05 m^2, close enough with rounding.
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