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Firefox File Edit View History Bookmarks Tools Window Help 62% QT, Fri 10:59 AM Q. Facebook WeBWork Abramson MA... x The Linear Approximation... x https /webwork2/Abramson MAT 265 Fall 2014/Section 2.7/6/?user bdalmuha&key; 8UfAOlyKHyovgloc v e CY Yahoo webwork edu asl Logged in as bdalmuha. Log Out MATHEMATICAL ASSOCIATION OF AMERICA K webwork abramson mat 265 fall 2014 section 2.7 MAIN MENU Homework Sets Section 2.7 Problem 6 Section 2.7: Problem 6 Password/Email Prev Up Next Grades (1 pt) Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second. How fast is the radius increasing after 3 minutes? Problems Note: The volume of a sphere is given by V (4/3)nr3 Problem 1 Rate of change of radius in feet per second) Problem 2 Problem 3 Problem 4 Preview Answers Submit Answers Problem 5 Problem 6 You have attempted this problem 6 times. Your overall recorded score is 0%. Problem 7 You have unlimited attempts remaining Problem 8 Display Options Email instructor View equations as ages C. Math Jax

Explanation / Answer

The volume of a sphere is:

V = (4/3)?r^3.

By implicit differentiation:

dV/dt = 4?r^2(dr/dt).

We know dV/dt and wish to find dr/dt. In order to find dr/dt, we also need to know r. After 3 minutes. Notice that 3 minutes = 180 seconds. Then, after 3 minutes, the volume of the balloon is 180(3) = 540 cubic feet. At this point:

540 = (4/3)?r^3, by substituting V = 540
==> r = (405/?)^(1/3).

Therefore:

dV/dt = 4?r^2(dr/dt)
==> dr/dt = (dV/dt)/(4?r^2)
= 3/[4?(405/?)^(2/3)]
? 0.0094 ft/s.

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