(2) Oil spilled from a ruptured tanker spreads in a circle. (Three separate, yet
ID: 2866554 • Letter: #
Question
(2) Oil spilled from a ruptured tanker spreads in a circle. (Three separate, yet related, problems.) (i) If the area of the circle increases at a constant rate of 4 miles squared per hour, how fast is the radius of the spill increasing when the area is 15 miles squared? (ii) If the radius is increasing at a rate of approximately .2913 miles per hour, how fast is the area of the circle increasing when the area is 15 miles squared? (iii) How large is the circle (in terms of area) if the area of the circle is increasing at 4 miles squared per hour and simultaneously the radius is increasing at approximately .2913 miles per hour?Explanation / Answer
2i)
A = 15
15 = pi * r^2
r = sqrt(15/pi)
A = pi * r^2
dA/dt = 2pi * r * (dr/dt)
4 = 2pi * sqrt(15/pi) * dr/dt
4 = 13.729368492956534 * dr/dt
dr/dt = 4 / 13.729368492956534
dr/dt = 0.291346 miles per hr ----> ANSWER for i
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ii)
A = pi * r^2
dA/dt = pi * 2r * dr/dt
dA/dt = 2pi * r * dr/dt
dA/dt = 2pi * r * 0.2913
Now, when A = 15, r = sqrt(15/pi) miles, which we plug in in place of r....
dA/dt = 2pi * sqrt(15/pi) * 0.2913
dA/dt = 3.9993650419982384
So, dA/dt = 4 miles squared per hr
So, area is increasing at 4 squared miles per hr ----> ANSWER for ii
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iii)
A = pi * r^2
dA/dt = pi * 2r * dr/dt
4 = 2pi * r * 0.2913
r = 4 / (2pi * 0.2913)
r = 2.1854437774376291
So, A = pi * r^2 becomes :
A = pi * 2.1854437774376291^2
A = 15.0047633191735609146
So, the circle is 15 squares miles in area -----> ANSWER for iii
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