Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: adwight on April 17, 2005, 10:19:05 pm
-
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.
-
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
-
No diameter = height, not radius. That makes a big difference.
-
oh, I wasn't paying atention, yeah, give me a sec to recalculate some stuff
-
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
-
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.
-
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.
-
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 (http://www.google.com/search?num=100&hl=en&lr=&safe=off&q=+%28%28pi%29%28%28c%2F2%29%5E2%29*c%29%2F3+-+%28c%5E3*pi%29%2F12+%3D&btnG=Search)
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
:)
-
"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
-
I stand in awe of your mastery of the incomprehensible squiggle language commonly known as math.
-
Bobbau, I owe you dude. Thanks so much, it makes sense what I have to do now.
-
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.
-
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.
-
Someone should stick you in a lab dude, seriously.
-
Agreed, 99% of this thread went way over my head. :D
-
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:)
-
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? :p
-
Congratulations, WMCoolmon; you are the king of ambiguous nouns. May the hordes of specificity tremble in your shadow.
-
wow, i used to be able to do this. then i became a med student.
-
Just so you know, I got a perfect 9/9 on this problem, thanks to the help. Thanks so much.
-
thanks, it's nice to know I still remember this stuff.
this has reawakened my desier to get that damned moment of inertia tensor for the POF header figured out.
now this is a man's math problem :)
intigrateing along a volume defined by the volume under three points, intigrate the following six formulie:
y^2+z^2
x^2+z^2
x^2+y^2
x*y
z*x
z*y
this might not seem like that hard a problem but it isn't realy the intigrand (is that the right termonology? the formula getting integrated) that bites you in the ass, it's the volume of intigration
the first step shows this, you have to intigrate the volume under the plane defined by the normal of the polygon (N) and one of the points (P). after useing the plane forumla to get a height function pln(x,y), you have to integrate along z from z=0 to z = pln(x,y), doesn't seem to nasty untill you realise that pln(x,y)=(Nx*x-Ny*y+Nx*Px+Ny*Py+Nz*Pz)/Nz. (NOTE Px means the x component of the vector P not P*x, the variables x and y are only in that function once respectively, the rest of those are more or less constants, unfortunately they are constants who's value we do not know)
doing the first step of intigration will yeild an slightly more complicated equasion, pln(x,y)*I(x,y) so long as you are not integrateing any of the elements with a z in them, if you do integrate one of the z elements, you get an unholy beast of a horably unnatural equasion of doom! so horrable I'm not even going to try to type it out, and it still needs to be integrated two more times! along the area on the xy plane defined by the polygon's three points, and each of these intigration must be done in two parts!
I'm thinking a Jacobian transformation might posably make this problem solveable, too bad they just skimmed over that in my Calc 3 class. :doubt:
-
Jesus Christ, I don't even want to try and begin to do something like that. I barely understand this normal Calc.
-
Originally posted by 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:)
You are the only man I know of that put "helixical volume" and "fun" in a same sentence.
-
And you should be shot for doing so. You sick, sick man.
-
:lol:
-
As I am only in 7th grade Algebra, all I can tell you is to get a few equations to fit into the rate of change equation, and do the math from there. Of course, as I am only in Algebra, I could be mistaken. Also, I am too tired to read all the posts to see if this has been resolved, so forgive me if I am posting like an idiot.
-
don't worry too much Algebra is probly the hardest you'll have to deal with, exept maybe trig.
-
Are you saying Eighth Grade will be easier than what I'm doing now? Riiight...
If I had a (working) Windows OS, I would be Fredding a lot. When I get a new Windows, you should expect many missions from me. Some of which will be with you as a capship.
-
Seriously, Geometry was cake compared to my first crouse in Algebra.
-
Yay another problem that I can do half of, but I can't seem to figure out the first half.
The rate of fuel consumption, in gallons per minute, recorded during an airplane flight is given by a twice differentiable and strictly increasing function R of time t. The graph of R and a table of selected values of R(t), for the time interval 0 <= t <= 90 minutes, are shown below.
(don't have graph, but you could make one from this info positive concavity from 0 to 50, negative concavity(but still increasing) from 50 to 70, and positive concavity from 70 to 90. Looks like a demented W. )
Table
t(mins) R(t)(galls/min)
0 20
30 30
40 40
50 55
70 65
90 70
Use Data from the table to approx. for R'(45). Show computation and units of measure. (I still can't do derivatives with no equation... :()
The rate of consumption is fastest at time t = 45 minutes. What is the value of R''(45)? Explain reasoning.
There, I can do the rest, I just don't know how to find the derivative when I don't have the equation for it...
You can use a calculator. I promise this will be like the last one... I swear.
-
What does the ' and " mean?
-
' = 1st derivative
'' = 2nd derivative
-
Well, if you've got a TI Graphical Calculator, put all the values for R in L1, then do y1=nDeriv(L1). You can find nDeriv under Math, option 8. Not sure if this will work, don't think you'll get a proper graph.
Otherwise, you need to create a formula/equation for R(t). And I'll be damned if I know how.
-
The rate of consumption is fastest at time t = 45 minutes.
right
=> maximum of the rate of consumption = max(R'(t))
=> the function R'(t) has a maximum at t=45
=> R''(t) = derivation of R'(t)
=> R''(t=45) = 0