Question

Can somone help me with this word problem: Sand is falling of a converyor belt onto a conical pile at the rate of 15ft^3/min. The diameter of the base of the cone is approxiamtly twice the altitude. At what rate is the height of the pile changing when it is 10 feet high?

Answer 1

You need to to relate the height to the diameter to the volume.
The volume of a cone is:
(1/3)pi r^2 h
But you know that the DIAMETER is twice the HEIGHT. Or, the radius is 4 times as much.
So:
V=(1/3)pi (4h)^2*h
V=(1/3)pi(16h^3)
Differentiate it with respect to time.
\[\frac{dV}{dt}=\frac{16 \pi}{3}3h^2\frac{dh}{dt}\]
dV/dt=15ft^3/min
h=10
Solve for dh/dt.
\[15=16\pi(10)(\frac{dh}{dt}) \rightarrow \frac{dh}{dt}=\frac{15}{160 \pi}\]

Answer 2

That what you got satellite?

Answer 3

\[V=\frac{1}{3}\pi r^2 h =\frac{1}{3} \pi (4h)^2h=\frac{1}{3}\pi 16h^3\]\]

Answer 4

yeah looks good to me.

Answer 5

I confirm.

Answer 6

I have the answer as being 0.048 ft/min

Answer 7

\[V'=16 \pi h^2h'=15\]

Answer 8

replace h by ten, solve for h' get just what you have. i am slow.

Answer 9

oh hold on

Answer 10

should be
\[h'=\frac{15}{16 \pi ^2h}\] yes?

Answer 11

so if
\[h=10\] get
\[h'=\frac{15}{1600 \pi}\]

Answer 12

Sorry but malevolence was wrong.. I was wrong too..

Answer 13

sorry i meant
\[h'=\frac{15}{16 \pi h^2}\]

Answer 14

replace h by ten get
\[h'=\frac{15}{1600 \pi}\]

Answer 15

i was able to get rid of one variable r. By saying that in this case d(diameter)=2h(height). And i know diameter=2 times radius, so 2r=2h which is r = h. SO the equation is: v=1/3*pi*h^3

Answer 16

Okay guys look:
V = (1/3)pie*r^2*h
diameter is twice the height, therefore:
d = 2h
2r = 2h
r = h. So plug that in:
V = (1/3)pie*h^2*h

Answer 17

V = (pi/3 ) * h^3
dV/dt = pi * h^2 (dh/dt)

Answer 18

Now the question i have is what does thhat equation look like when i differntiate it?

Answer 19

15 = pi * 100 dh/dt
dh/dt = 15/(100pi)

Answer 20

yeah looks like we all blew it in two ways.
\[r=h\] and so
\[V=\frac{1}{3} \pi h^3\]

Answer 21

Also i know dv/dt=15*ft^3/min

Answer 22

I just posted the solution, dont u guys see it?

Answer 23

He's right. Mine is wrong, I miswrote it :(

Answer 24

\[V'=\pi h^2h'=15\] or
\[h'=\frac{15}{\pi h^2}\]

Answer 25

V = (1/3)pie*r^2*h
diameter is twice the height, therefore:
d = 2h
2r = 2h
r = h. So plug that in:
V = (1/3)pie*h^2*h
V = (pi/3 ) * h^3
dV/dt = (3pi/3) * h^2 (dh/dt)
dV/dt = pi * h^2 (dh/dt)
dV/dt is 15, so
15 = pi * 10^2 dh/dt
dh/dt = 15/(100pi)

Answer 26

but how do i get rid of the cube part in 15ft^3/min

Answer 27

what cube?

Answer 28

that just means the units are cube units

Answer 29

you mean the units? ft^3/min? ignore them...

Answer 30

volume comes in the cube

Answer 31

how come the answer does not have a cube in it?

Answer 32

You have ft^3 for volume and you're dividing it by h^2 or ft^2
so ft^3/ft^2=ft

Answer 33

the answer is the change in height.. change in height is in feet per minute.. height is not volume..

Answer 34

malevolence nailed it. the units work themselves out

Answer 35

Height is like length - feet
Area - ft^2
Volume - ft^3

Answer 36

can I get a medal too?

Answer 37

exactly so if you take volume divided by area you get height.

Answer 38

Medal awarded to bahroom

Answer 39

lol tnx guys xD