Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic ft/ min. The diameter of the base of the cone is approximately 3x the altitude. At what rate is the height of the pile changing when the pile is 15 ft high?

Answer 1

we can do this if you tell me what the formula for the volume of a cone is

Answer 2

ok i found it it is
\[V=\frac{1}{3}\pi r^2h\]

Answer 3

yes

Answer 4

and "diameter of base is 3 times h" means
\[D=3h\] or
\[r=\frac{3h}{2}\]right? so we can write everything in terms of h which is what we want

Answer 5

\[V=\frac{\pi}{3}(\frac{3h}{2})^2h\]

Answer 6

\[V=\frac{\pi}{3}\frac{9h^3}{4}\]
\[V=\frac{3\pi h^3}{4}\] correct my algebra if it is wrong, because that is the hard part and the rest is easy

Answer 7

look ok?

Answer 8

its correct so far

Answer 9

will we b solving thsi w/ derivatives? bc taht is how i need to solve this

Answer 10

then we are almost done. take the derivative and get
\[V'=\frac{9\pi h^2}{4}h'\]

Answer 11

it that exponent a 3 or a 2 for base h

Answer 12

you are told that
\[V'=10\] it was 3, when we take the derivative it is 2

Answer 13

is v' teh same as (dv/dt)' ?

Answer 14

you want
\[h'\] so solve
\[10=\frac{9\pi h^2}{4}h'\] for
\[h'\] get
\[h'=\frac{40}{9\pi h^2}\]

Answer 15

yes maybe you wrote in class
\[\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}\] which i wrote in shorthand as
\[V'=\frac{9\pi h^2}{4}h'\]

Answer 16

ok thnks 4 cleaering taht up

Answer 17

now that you have
\[h'=\frac{40}{9\pi h^2}\] you can plug in your value for h to get the answer

Answer 18

so i substitute 15 right? i got 40/2025 pi or 8/ 405 pi

Answer 19

i guess that should be right

Answer 20

ok thnk u soooo much =)

Answer 21

yw