Question

related rates help. see picture.

Answer 1

I can't figure out how to find the radius

Answer 2

So first off draw a picture and literally list out all the values you're given or need to solve for along with derivatives like dV/dt. From there do you know any formulas that relate volume, height, and diameter or radius of a cone? Put those down too and we'll work through this together.

Answer 3

cone V=1/3 pi r^3h

Answer 4

So what is height in terms of the diameter? What is the diameter in terms of the radius?

Answer 5

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

Answer 6

the height is 3/8 of the diameter

Answer 7

Exactly, so write an equation for both cases and plug those in so that you have volume as a function of only the radius, not the height nor diameter.

Answer 8

so i plug in 3d/8 to the h?

Answer 9

Yup, and then after that you want it in terms of radius, so just use d=2r and plug that in. Then you can take the derivative with respect to time.

Answer 10

when i plugged in 2r for d, i got 1/2 pi r^2 dr/dt. Is that right?

Answer 11

No, you need to plug it into your original formula before you take the derivative. You took the derivative before you had the right formula, so you weren't taking into consideration the fact that h was dependent upon r.

Answer 12

so plug 2r for d in V=1/3 pi r^2 h

v= 1/3 pi r^2 (2r)?

Answer 13

You plugged in 2r for h, not d.

Answer 14

\[V=\frac{ 1 }{ 3 } \pi r^2h\]\[h=\frac{ 3 }{ 8 }d\]\[d=2r\] Plugging these all together gives:\[V=\frac{ 1 }{ 4 } \pi r^3\]\[\frac{ dV }{ dt }=\frac{ 3 }{ 4 } \pi r^2\frac{ dr }{ dt }\]
Now what do we plug in? Remember the initial conditions dV/dt=10, and h=4. But we have r, so we simply use the equations from above to solve for r... 4/3h=r

Then we can solve for dr/dt.