The Pentagon is planning to build a new satellite that will be spherical. As is typical in these cases, the specifications keep changing, so that the size of the satellite keeps growing. In fact, the radius of the planned satellite is growing 0.9 foot/week. Its cost will be $1,400 per cubic foot. At the point when the plans call for a satellite 8 feet in radius, how fast is the cost growing (The volume of a solid sphere of radius r is $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ .)
A) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 3,584,000 \pi } { \text { week } }$
B) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 40,320 \pi } { \text { week } }$
C) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 230 \pi } { \text { week } }$
D) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 161,280 \pi } { \text { week } }$
E) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 322,560 \pi } { \text { week } }$

Question

The Pentagon is planning to build a new satellite that will be spherical. As is typical in these cases, the specifications keep changing, so that the size of the satellite keeps growing. In fact, the radius of the planned satellite is growing 0.9 foot/week. Its cost will be $1,400 per cubic foot. At the point when the plans call for a satellite 8 feet in radius, how fast is the cost growing  (The volume of a solid sphere of radius r is $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ .)  
A) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 3,584,000 \pi } { \text { week } }$
B) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 40,320 \pi } { \text { week } }$
C) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 230 \pi } { \text { week } }$
D) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 161,280 \pi } { \text { week } }$
E) $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { \$ 322,560 \pi } { \text { week } }$

Quizplus · Accepted Answer

The rate of change of volume with respect to time is $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. At $r = 8$ feet and $\frac{dr}{dt} = 0.9$ foot/week, $\frac{dV}{dt} = 4\pi (8)^2 (0.9) = 230.4\pi$ cubic feet/week. The cost rate is $\frac{dP}{dt} = 1400 \times \frac{dV}{dt} = 1400 \times 230.4\pi = 322,560\pi$ dollars/week.

The Pentagon Is Planning to Build a New Satellite That V=43πr3V = \frac { 4 } { 3 } \pi r ^ { 3 }V=34​πr3

The Pentagon Is Planning to Build a New Satellite That $V = \frac { 4 } { 3 } \pi r ^ { 3 }$