Question

Calculas 1help

A spherical balloon is to be deflated so that its radius decreases at a constant rate of . At what rate must air be removed when the radius is

Answer 1

when the radius is ?

Answer 2

you want to find the rate of change of the volume:

Answer 3

dV/dt

Answer 4

use the chain rule:
dV/dt = dV/dr * dr/dt

Answer 5

constant rate of 15cm/min and radius 9cm can you show me the steps im confused

Answer 6

dr/dt = 15 cm per minute ... given

Answer 7

ok check

Answer 8

and what't the formula for Volume of sphere?

Answer 9

V=pi/3r^2h

Answer 10

so the r is plugged in?

Answer 11

SPHERE...
\[V(r)=\frac 43 \pi r^3\]
correct?

Answer 12

oh yeahhh

Answer 13

so dV/dr = ?

(you need to get the formula for dV/dr before plugging r)

Answer 14

so I derive the equation with respect to r instead of t?

Answer 15

yes

Answer 16

is dv/dr=3r^2??

Answer 17

that times 4pi/3 ...

Answer 18

\[dV/dr=4\pi r^2\]

Answer 19

wait isn't the derivative of 4pi/3=0???

Answer 20

the derivative of x^3 is: 3x^2

the derivative of 2x^3 is 6x^2

the derivative of ax^3 = 3ax^2

Answer 21

yes I know that but how come I didn't derive 4pi/3??

Answer 22

*... for any constant a. 4pi/3 is just a constant.

Answer 23

so constants I leave alone??
and I plugged into the equation and I have dv/dr=4pi(81) is that right so far?

Answer 24

yes just like when taking the derivative of 2x^3 ... the 2 doesn't just disappear

Answer 25

how does the pi disapper I get 4860pi cm^3/min but my book answer has no pi

Answer 26

right,

dV/dr= 324 pi cm^3 per cm
dr/dt = 15 cm per min

dV/dt= dV/dr * dr/dt
= 324pi*15
= 4860pi cm^3/min

Answer 27

pi is just a number 3.14159...

Answer 28

\[\Large 4860\pi \approx 15268.1403\]

Answer 29

oh yeah it does have pi thank you!