Question

A machine starts dumping sand at the rate of 20 m3/min, forming a pile in the shape of a cone. The height of the pile is always twice the length of the base diameter.
A. After 5 minutes, what is the height of the pile?
B. After 5 minutes, how fast is the height increasing?
C. After 5 minutes, how fast is the base radius increasing?
D. After 10 minutes, how fast is the area of the base increasing?

Answer 1

help anyone?

Answer 2

do you know the formula for the volume of a cone ?

Answer 3

@cwrw238 i dont know how to solve for the rate because it does not give a height.

Answer 4

yes

Answer 5

V=1/3pir^2h

Answer 6

this is what i have so far...
h=2d h=4r r=h/4
dV/dt=20mcu./min
20=pi/48(h^3)d/dt=pi/48(3h^2)dh/dt

Answer 7

this is where i got stuck!

Answer 8

V = pi r^2 h
-------
3

h = 2 * diameter of the base
= 4 * radius

so r = h/4

so we can write

V = ( pi *( h/4)^2 * h ) / 3

V = pi h^3
-----
48

Answer 9

you forgot to take the derivative of h^3, no?

Answer 10

i dont need it to do A

Answer 11

okay, then i have done that [art already. but this is where i got stuck!

Answer 12

do we multiply 5 by 20, then solve for h? using
V = pi h^3
-----
48

Answer 13

@cwrw238

Answer 14

yes 100 = pi* h^3
-------
48

Answer 15

i got 11.52, which doesnt make sense.

Answer 16

@cwrw238 are you there?

Answer 17

@mathstudent55 will youhelp me too?

Answer 18

hmmm does seem a high number - can you see anything wrong with my algebra?

Answer 19

no. but i need to finish this. so i'll just go with that. can we go on to B, now please?

Answer 20

ok

dV/dt = dV/dh * dh / dt

we need to find dh/dt

so dV/dt = 20

and dV/dh = 3 pi h^2
------- = pi h^2/ 16 = 11.52^2 pi / 16
48

so plug these 2 values into the above formula to get dh/dt

Answer 21

agreed?

Answer 22

agreed!

Answer 23

wait! this part is confusing: 11.52^2 pi / 16

Answer 24

why are we doing that???

Answer 25

@cwrw238

Answer 26

h= 11.52 at time 5 minutes

Answer 27

but how does that solve for dh/dt

Answer 28

we need to find dV/dh

Answer 29

i got 26.05

Answer 30

20 = [11.52^2 pi / 16] * dh/dt

dh/dt = 20* 16
-----
11.52^2 pi

Answer 31

.768
???

Answer 32

i get that 0.768 m / min yes

Answer 33

ok! C now?

Answer 34

we need to find that derivative of the radius? right?

Answer 35

you do that in a similar way to the height

dV/dt = dV/dr * dr / dt

Answer 36

find V in terms of r then differentiate

Answer 37

h = 4r
so

V = pi r^2 * 4r
--------
3

so what dv/dr ?

Answer 38

dv/dr= 8rpi/3

???

Answer 39

dV / dr = 4 pi r^2
------ right?
3

Answer 40

do you take the derivative of r^2 ??

Answer 41

no you take the derivative of (4 pi r^3) / 3

Answer 42

oh. ok. i get it. i forgot about the r^3

Answer 43

sorry i made a mistake

dV / dr = 4 pi r^2

Answer 44

its 4 * 3 pi r^(3-1)
---
3

Answer 45

i already did that, haha.

Answer 46

good for you

so we have

20 = 4pi r^2 * dr/dt

dr/dt = 20 / 4 pi r^2

so we sub in r what is the value of r?

Answer 47

the heigh over 4
so::: 11.52/4

Answer 48

@cwrw238 right?

Answer 49

yes

Answer 50

dr/dt=.012m/min

Answer 51

@cwrw238 im gonna try to do D by myself. if need help i'll ask okay?

Answer 52

thanks a bunch to you! rates of change are fun! so thank for helping!!!!

Answer 53

sorry gotta go now

Answer 54

dr/dt = 0.192 m / sec

Answer 55

in fact there was no need to do all that calculus
if dh/dt = 0.768 then dr/dt = 0.768/4 = 0.192 !!

Answer 56

as for the area the rate of increase will be pi r^2 m^2/min
that is pi * 0.192^2 = 0.116 cm^2 / min