[adwight]

Another problem that I yet again have some idea, but not much of how to do.  Stupid calculus, it will be the death of me.  This should be easy, but these are the types of problems I do not understand.


A container has the shape of an open right circular cone.  The height of the container is 10 cm, and the diameter of the opening is 10cm.  Water in the container is evaporating so that its depth h is changing at the constant rate of -3/10 cm/hr.

volume = 1/3(pi)r^2h

a) find the volume v of water in the container when h = 5 cm, indicate units of measure.  

Found this to be 125(pi)/3 cm^3, tell me if im right.

b)(this is the gay part...)Find the rate of change of the volume of water in the container, with respect to time, when h = 5 cm.  Indicate units of measure.


c)Show that the rate of change of the volume of water in the container due to evaporation is directly proportional to the exposed surface area of the water.  What is the constant of proportionality?


Any help will be GREATLY appreciated.

I have an idea how to do part b, but I dont know if its right.  Do I just derive the function after I plug h in, or before I plug h in, in that case I would need a differential equation because of the 2 variables?  If deriving is the answer, after plugging h in, would that be the solution, just the derivative.

[Bobboau]

ok, let me make sure I understand this corectly first

has the shape of an open right circular cone. The height of the container is 10 cm, and the diameter of the opening is 10cm.

so, r == h?
right?

if that's true then your volume formula becomes a lot simpler,

(h^3*pi)/3

acording to this formula your volume calculation is right.
ok so the rate of change of h is -3/10 cm/hr, hmm so your hight formula is -3/10t, so the volume formula with respect to time would be (-9*pi*t^3)/1000, and it's dirivitive would be (-27*pi*t^2)/1000, so 125(pi)/3 = (-9*pi*t^3)/1000, will give us the time when it is at the desiered hight -> t=50/3. so...  the rate of change of volume with respect to time when h=5 would be (-15*pi)/2

[adwight]

No diameter = height, not radius.  That makes a big difference.

[Bobboau]

oh, I wasn't paying atention, yeah, give me a sec to recalculate some stuff

[Bobboau]

ok volume(h) = (h^3*pi)/12

so volume(5) = (125*pi)/12

hight formula is -3/10t+10

v(t) = (-pi*(3*t-100)^3)/12000

v'(t) = (-3*pi*(3*t-100)^2)/4000

v(t)=v(h=5) -> hmm it's still t=50/3

so v'(t=50/3) = (-15*pi)/8

[Bobboau]

alright and surface area is pi*r^2

s(h)=pi*(h/2)^2 = (h^2*pi)/4

s(t) = (pi*(3*t-100)^2)/400

s(t)*c=v'(t) -> c=-3/10
...hmm, it's been a while sence I've done this sort of thing, and that dosen't seem right.

[adwight]

Your problem with the calculating is still the volume.  You said h^3 is the volume, which is incorrect, because the height is 10, while the radius is 5...  This problem confuses me because I don't know what to do after I find the volume.

[Bobboau]

volume(h,r)=((pi)r^2h)/3

r = h/2

volume(h,h/2) = ((pi)((h/2)^2)*h)/3 = (h^3*pi)/12

if you don't beleve me look at this

acording to google ((pi)((h/2)^2)*h)/3 - (h^3*pi)/12 == 0, therefore ((pi)((h/2)^2)*h)/3 = (h^3*pi)/12

[Bobboau]

"Find the rate of change of the volume of water in the container, with respect to time, when h = 5 cm. Indicate units of measure."

you need dv/dt(t) at the time that h=5

wich is 50/3 (now that I think about it I was figuring that stupidly, I should have just thought h=-3/10t+10, setting h to 5 you get the time)

so first you need to get the volume function in terms of h. simply take the v(h) function and plug in -3/10*t+10. now you have v(t)

wich I got as (-pi*(3*t-100)^3)/12000=v(t)

then just diferentiate that to get dv/dt

 (-3pi*(3*t-100)^2)/4000=v'(t)

finaly you plug into v'(t) the time t=50/3

and you get (-15pi)/8, the rate of change of the volume at time 50/3, the time when -3/10*t+10 = 5

[Rictor]

I stand in awe of your mastery of the incomprehensible squiggle language commonly known as math.

[adwight]

Bobbau, I owe you dude.  Thanks so much, it makes sense what I have to do now.

[adwight]

Originally posted by Bobboau
alright and surface area is pi*r^2

s(h)=pi*(h/2)^2 = (h^2*pi)/4

s(t) = (pi*(3*t-100)^2)/400

s(t)*c=v'(t) -> c=-3/10
...hmm, it's been a while sence I've done this sort of thing, and that dosen't seem right.

I still don't get your figuring for the surface area, but oh well.

[Bobboau]

well I wasn't completely sure on that one either, I wanted to make sure I got the first one right before delving too far into the next one.

"Show that the rate of change of the volume of water in the container due to evaporation is directly proportional to the exposed surface area of the water. What is the constant of proportionality?"

so (the surface area at a given time)*a_constant=rate_of_change_of_volume with respect to time at a given time
ie
s(t) *c = v'(t)

we have v'(t) frome the previous question, so we need s(t).

s is surface area, and s(r)=pi*r^2, IIRC.

r= h/2, and h=-3/10t+10
so s((-3/10t+10)/2)= pi*((-3/10t+10)/2)^2, wich if you simplify out I beleive turns out to be (pi*(3*t-100)^2)/400

so s(t)= (pi*(3*t-100)^2)/400
and v'(t)=(-3pi*(3*t-100)^2)/4000

so: s(t)*c=v'(t)
((pi*(3*t-100)^2)/400)*c=(-3pi*(3*t-100)^2)/4000

solve for c, the 't's should eventualy cancel out leaveing you with a constant of proportionality I beleve to be -3/10.

but someone should check my math over, especaly the algebra.

[BlackDove]

Someone should stick you in a lab dude, seriously.

[Fury]

Agreed, 99% of this thread went way over my head.

[Bobboau]

bah, this isn't hardly even calculus 1, give me some 4th dimentional gradients integrated along a hellixical volume, that'd be a bit of fun there:)

[WMCoolmon]

Does that mean you can make the purply thing in that one movie that blew up after it made that guy go insane and kill those people?

[Ford Prefect]

Congratulations, WMCoolmon; you are the king of ambiguous nouns. May the hordes of specificity tremble in your shadow.

[icespeed]

wow, i used to be able to do this. then i became a med student.

[adwight]

Just so you know, I got a perfect 9/9 on this problem, thanks to the help.  Thanks so much.