Asked by Lela Mitevska
Starting at midnight, a 10 foot radius circular pond freezes inward from the outer edge at a rate of 4 inches per hour. How fast is the open area shrinking at when the radius is 9 feet? (Note: if I'd said "at 3am" instead of "when radius is 9 ft", this would just be a chain rule problem
Answers
Answered by
Aldrin Tolentino
y=csc^4 x - 2cot^2 x
y=x^2x+1
y=x^2x+1
Answered by
Reiny
area = πr^2
d(area)/dt = 2πr dr/dt
given: dr/dt = -4 inches/hr
find d(area)/dt when r = 108 inches
d(area)/dt = 2π(108)(-4) inches^2/hr
= appr 2714.3 inches^2/h
if you had said 3:00 am, the time would have been 3 hrs, and the radius would have shrunk by 12 inches, making r = 9, the same as the first wording.
d(area)/dt = 2πr dr/dt
given: dr/dt = -4 inches/hr
find d(area)/dt when r = 108 inches
d(area)/dt = 2π(108)(-4) inches^2/hr
= appr 2714.3 inches^2/h
if you had said 3:00 am, the time would have been 3 hrs, and the radius would have shrunk by 12 inches, making r = 9, the same as the first wording.
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