Question

Medal!!
The radius of a right circular cylinder is increasing at the rate of 3 ft/sec, while the height is decreasing at the rate of 6 ft/sec. At what rate is the volume of the cylinder changing when the radius is 15 ft and the height is 10 ft? Remember to use the product rule when you find the expression for dv/dt.

30 ft3/sec
−900 ft3/sec
−450π ft3/sec
−900π ft3/sec

Answer 1

@Hero

Answer 2

@Michele_Laino

Answer 3

@ParthKohli

Answer 4

@pooja195

Answer 5

here we have to start from the formula which expresses the volume \(V\) of the cylinder:
\[\huge V = \pi {r^2}h\]
now, please write the first derivative of the volume \(V\) with respect to time, using the formula above

Answer 6

(2pi(15)*(3)(10) + (pi 15^2)(-6)

Answer 7

that's right!

Answer 8

I know how to set it up but I keep getting the wrong answer

Answer 9

I thionk it that the answer is right, since we have this first derivative:
\[\Large \frac{{dV}}{{dt}} = \pi \left\{ {2r\left( {\frac{{dr}}{{dt}}} \right)h + {r^2}\left( {\frac{{dh}}{{dt}}} \right)} \right\}\]
now, if you replace the data of the exercise, you will find your answer above

Answer 10

oops.. I think*

Answer 11

so far i got this 30(30)+ -1350

Answer 12

so -450?

Answer 13

@Michele_Laino

Answer 14

if that right can you help me with one more?