Question

HELPP MEDAL!!

Answer 1

@geerky42 please help!

Answer 2

is it 50pi?

Answer 3

By fixed radius, does that mean radius doesn't change at all?

Answer 4

In that case, you just treat radius as constant.

Answer 5

yes

Answer 6

Well, \(V_{cone} = \dfrac{1}{3}\pi r^2h\), right?

So taking derivative of volume, you have \(\dfrac{dV_{cone}}{dt}=\dfrac{1}{3}\pi r^2\dfrac{dh}{dt}\). I don't take derivative of radius because it is fixed.

And you were given \(r = 10\) and \(\dfrac{dh}{dt}=.5,~\)

So just plug in values and you have your answer.

Answer 7

50pi? @geerky42

Answer 8

oh wait

Answer 9

\(\dfrac{dV}{dt}=\dfrac{1}{3}\pi(10)^2(\dfrac{1}{2}) = \dfrac{100\pi}{6}=\boxed{\dfrac{50\pi}{3}}\)

Answer 10

@geerky42

Answer 11

Oh... I used volume of cone, not cylinder...

Silly me

Answer 12

ohhh

Answer 13

Well, after taking derivative of volume of cylinder, you have \(\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}\), right?

Just plug in given values: \(\dfrac{dV}{dt} = \pi (100) (1/2) = \boxed{50\pi}\)

Answer 14

Does that make sense?

Answer 15

yes! thank you