Question

Calculus AB question.

Water is flowing into a cone-shaped tank at rate of 5 cubic inches per second. If the cone has an altitude of 4 inches and a base radius of 3 inches, how fast is the water level rising when the water is 2 inches deep? Be sure to include appropriate units in your answer.

Answer 1

you need volume of a cone
\[V = \frac{1}{3}\pi r^{2} h\]
and ratio of height to radius
\[\frac{h}{r} = \frac{4}{3}\]
since they want rate h is increasing, get V in terms of only "h"
\[r = \frac{3h}{4}\]
\[V = \frac{3}{16} \pi h^{3}\]

Answer 2

from there,
- differentiate with respect to "t" (implicit)
- plug in dV/dt = 5
- solve for dh/dt
- plug in h=2

Answer 3

How did you go from 4/3 to 3h/4?

Answer 4

i solved for "r"

Answer 5

its just a proportion, i could have also said
r/h = 3/4

Answer 6

How did u go from r=3h/4 to V = 3/16 pi h^2? When you plugged r in, how did u get V = 3/16 pi h^2?

Answer 7

after you square "r" you just simplify...i figured you could simplify that without seeing every step

(3h/4)^2 = 9h^2/16 * 1/3 = 3h^2/16 * h = 3h^3/16

Answer 8

How do you do "differentiate with respect to "t" (implicit)"?