Calculus: Related rates
 

okay, some smarty pants explain to me how to do this problem:

Sand is being dumped on a pile in such a way that it always forms a cone whose radius equals its height. If the sand is being dumped at a rate of 10 cubic feet per minute, at what rate is the height of the pile increasing when there is 1000 cubic feet of sand in the pile?

This problem is from Ostebee Zorn Calculus Book, section 4.9 #6.

Heh, this is a very easy problem dude.

Alright, here it is:

Area of a Cone = (1/3)*(PI)*(r^2)*(h)
Since r = h, Area = (1/3)*(PI)*(r^3)
Now, we know that (1/3)*(PI)*(r^3) = 1000
This is simplified to:
(PI)*(r^3) = 3000
(r^3) = (3000/PI)
r = Cubed root of (3000/PI), which = 9.847 (approximately)

Ok, now that we have that part solved, we find out the rate. It goes in at 10 cubic feet per minute, so after 1000 cubic feet is down, 100 minutes have passed.

Therefore, when the pile is at 9.847 feet high, 100 minutes have passed, and (9.847/100) = 0.09847 feet per minute.

That is the answer you were looking for, no?

okay, the first part is right, (i've consulted some of my friends since i posted this fourm), the second part requires something to do with alot of derivatives. hm...

First, not an easy problem.
Second, 9.874/100 is change in distance divided by change in time. Think.... that's average velocity.

Here is how I would do it:

I decided to track the radius starting with your radius at 0 at the origin.
0-------->+

Okay, here's a couple of equations: (` is initial)
2a(x-x`) = (v)(v) - (v`)(v`)
x = x` + (v`)(t) + (a)(t)(t)/2

Initial velocity is 0
Initial displacement is 0 since that's where I set my axis.
Plug in numbers and you get respectively:
2a(9.847 - 0) = (v)(v)
9.847 = 0 + (a)(5000) (100seconds squared divided by 2)

Solve for a and plug into other equation:
2(.00197)(9.847) = (v)(v)

Solve for v:
v = .19694 feet/minute
---------------------------------
Check my work, that may or may not be the right method to do this problem. Hope I helped.

Wow, you are right, this is actuall kinda difficult, give me a few minutes now to concur with you...

Ok, here we go, I'll try to keep this as clear as possible:

We know from previous that V=(1/3)(PI)(h^3)
We are given dV/dt=10 meters cubed per minute.
So.... dV/dt=(PI)(h^2)(dh/dt)
Therefore, dh/dt=(10 meters cubed per minute)/(PI)(h^2)
Now we need to find out what h will equal when the volume is 1000 cubic meters.

(h^3)=(3000)/(PI)
and
h=[the cubed root of(3000)/(PI)], which =9.847
Ok, now comes the rate:
dh/dt=(10 meters cubed per minute)/(PI)(h^2), and we now know what h is, therefore, we plug it in...

dh/dt=(10 meters cubed per minute)/(PI)(9.847 squared), which = (10 meters cubed per minute)/(304.647 meters squared), which = 0.0328 meters per minute.

Now, I'm pretty sure that is correct, but double check me if you want.

the answer is.... Socks


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Had an epiphany. You do have to use derivatives, and it is quite simple if you know how.

Area of Cone: (1/3)*(pi)*(r)^3

Take the derivative with respect to time:
dAC/dt = (pi)*(r)^2*(dr/dt)

Look at what is known:
dAC/dt is 10 cu ft.
r is cubed root of (3000/pi) as previously found.

Solve for (dr/dt) which is the change in radius with respect to time.
dr/dt = .0328248340614
:) There's your answer.

Oh dear lord..... I still think the answer is 42.

wtf ur speaking in chinese, i can hardly do a geometry proof

ahhh i love calculus... thanks for the answer guys, even tho it was like a week late.

hey if heard chinese and they cannot speak it on a forum they are typing it and it is not the same.

LoL Very simple probleme no need of derivatives

1000=(1/3)*(PI)*(H^3)
H1=9.847450218

You add 10 feet cube for 1 minute

1010=(1/3)*(PI)*(H^3)
H2=9.88016624

H2-H1=.03271602179

The height increase of .03271602179 feet per minute

Again, this is wrong. Your formula finds average increase during that minute. To get the instantaneous velocity, you have to use derivatives.

Hehe, your epiphany occured 8+ hours after I posted the right answer...

Eh well I was at school... Oh well.

I tried :(
That seemed to easy to be true

Heh Heh Heh....

DIG!



Actually I just wanted to note that this is the first time Titusfied ever posted outside of the Flame/D2 forums.
I just wanted to show that people's post DRAMATICALLY improve the first time they leave the realm of those two forums.

Man, that actually does look like Greek up there, errr, at least a different language that is...

~Closed!~