Asked by Bubba
The sun is shining and a spherical snowball of volume 210 ft3 is melting at a rate of 14 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 3 hours?
Answers
Answered by
bobpursley
Volume= 4/3 PI r^3
dV/dt=4PI r^2 dR/dt
you are given dV/dt as 14ft^2/hr
you are looking for dr/dt
so what is the radius at t=3?
210-14*3=4/3 PI r^3, solve for r.
put that r into the dV/dt equation, and solve for dR/dt
dV/dt=4PI r^2 dR/dt
you are given dV/dt as 14ft^2/hr
you are looking for dr/dt
so what is the radius at t=3?
210-14*3=4/3 PI r^3, solve for r.
put that r into the dV/dt equation, and solve for dR/dt
Answered by
Paula
I followed these steps, but the answer is wrong. Is there a typo here?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.