Question

A cylinder has a radius that is decreasing at a rate of 2 cm per second while its height
Is increasing at a rate of 3 cm per second .
At what rate is the volume changing when the radius is 3 cm and
the height is 2cm ?

Answer 1

can you write down the formula for the volume of a cylinder?

Answer 2

V

=



π



r

2


h

Answer 3

v=pir^2h

Answer 4

can you take the derivative (with respect to time "t") of both sides of that equation?

Answer 5

dv/dt=pi2rh*dh/dt

Answer 6

ok on dv/dt
on the right side (ignore the pi for the moment)
we have to use the product rule
\[ \frac{d}{dt} r^2 h = r^2 \frac{d}{dt} + h \frac{d}{dt} r^2\]
can you finish that ?

Answer 7

**\[ \frac{d}{dt} r^2 h = r^2 \frac{dh}{dt} + h \frac{d}{dt} r^2 \]

Answer 8

the product rule is
d( u v) = u dv + v du

Answer 9

now use the power rule on r^2 and the chain rule
d r^2 = 2 r dr

Answer 10

d is short for \(\frac{d}{dt} \)

Answer 11

dv/dt=pi*2r*dr/dh+r^2dh/dt

Answer 12

ok except pi multiplies both terms on the right side

Answer 13

\[ \frac{dv}{dt}= \pi \left( \frac{d}{dt} r^2 h \right) \]

Answer 14

last step is replace the variables with the numbers given in the problem

Answer 15

btw, you have dr/dh. that should read dr/dt
so you should get
dv/dt=pi* ( 2r*dr/dt+r^2 dh/dt)

Answer 16

dv/dt=15pi cm/s?

Answer 17

so you should get dv/dt=pi* ( 2r*h*dr/dt+r^2 dh/dt)
we should use the equation editor. Keep leaving out terms. It should read
\[ \frac{dv}{dt}= \pi \left( 2rh \dot{r} + r^2 \dot{h} \right) \]
r=3
h=2
\(\dot{r} = 2\)
\( \dot{h} = 3 \)

Answer 18

Newton used \(\dot{x} = \frac{dx}{dt} \)

Answer 19

coool

Answer 20

It's not 15 pi

Answer 21

ohhhh I left out a h

Answer 22

3 pi

Answer 23

I thought dr/dt would be -2cm/s since its decreasing at a rate of 2 cm/s

Answer 24

yes. good catch.

Answer 25

alright . thank you ;)

Answer 26

yw