rate of increase of the surface area

afcwestwarrior

a spherical balloon is being inflated. find the rate of increase of the surface area (S=4 pie r squared) with respect to radius r when r is (A) 1 ft, (B) 2 ft, (C) 3ft.

Here's what i did

i found the derivative of s and i put 4 2R and then i plugged in the numbers in R
and i got 8 ft sq/ft 16 ft sq/ft and 24 ft sq/ft

except i'm not sure if im doing it right, although those are the right answers i didnt get the units along with them

afcwestwarrior

is there simple algebra involved in this calculus problem

Plastic Photon

are you not given dr/dt or dv/dt?

Schrodinger's Dog

You've got a graph there, simply find the gradient of the slope, that's the rate of increase.

(1,8)
(2,16)
(3,24)

HallsofIvy

Give more detail about what you did (I, for one, don't know what "i put 4 2R" means!). What is the formula for surface area (I see you give that)? what is its derivative?

Plastic Photon, since the problem asks for the rate of increas eof surface area with respect to r you don't need to know dr/dt.

If the surface area is in square feet and the radius in feet, then the "rate of change of surface area with respect ot radius" is in (square feet)/feet or just feet!

afcwestwarrior

ok so ur saying i dont need to find the derivative of it, ok check this mate,
the i put 4 2 R means thats the derivative of the surface area

HallsofIvy

No, "i put 4 2 R" doesn't mean anything!