Question

Doing a related rates problem to find the change in volume of a cone.

I know volume of a cone is V=(1/3)bh. What would (dV/dt) equal?

Answer 1

out of context, 0.

the full problem would be required to answer properly

Answer 2

To clarify, you know how you have to differentiate V before you can plug things in? I just want to know what that would be

Answer 3

Ok. The radius r and height h of a circular cone change at a rate of 2 cm/s. How fast is the volume of the cone increasing when r=10 and h=20?

Answer 4

\[\frac{ dV(t) }{ dt } = \frac{ 1 }{ 3 }\frac{ d }{ dt }(b(t)*h(t))\]

Answer 5

and since we use product rule:

\[\frac{ dV }{ dt } = \frac{ 1 }{ 3 }(\frac{ db }{ dt }h + b\frac{ dh}{dt })\]

Answer 6

Thanks! I'm sorry not trying to be difficult but I suck at these problems, so why/how did you get that?

Answer 7

Ohh product rule. ok gotcha!

Answer 8

db/dt and dh/dt are given

atually. hold on. this is b not r.

so it's actually:

\[\frac{ dV }{ dt } = \frac{ \pi }{ 3 }\frac{ d }{ dt }(r^2 * h)\]

Answer 9

but yes. product rule lets you plug in r, h, dr/dt and dh/dt

Answer 10

Where did pi come from? Different Equation?

Answer 11

in V = 1/3 b*h, b is the base of the cone. and the base is the area of a circle: pi*r^2

V = (1/3)bh = (1/3)pi*r^2*h

Answer 12

and we have to use r since we're given the rate of change of r and r's relation to b is not linear

Answer 13

Oh right! Ok thanks that helped a lot!

Answer 14

glad i could help. lmk if you want me to check your answer and if you have more questions :)

Answer 15

I'm sorry, I got completely stuck so I ditched the problem and just came back to it. so given that last equation, would it come out to be:

\[\frac{ dV }{ dt }= \frac{ \pi }{ 3 }2r ^{3}h\]

Answer 16

??

Answer 17

Euler signed off, could anyone explain it to me from here reading our conversation?