Okay, I'm in Calc BC, and I have a hw problem I can't seem to get.

We JUST learned about the existance of the method of integration, so I don't think you use that here...
If somebody could just... translate this problem from words into math, I could do it. I am just not getting what it's trying to say in equation/proportion form...

dV/dT = k/r
dV = (k/r)dT (edit: something seems wrong with that).

Not sure, and too lazy to do it. But then you will be able to set the two V equations equal to each other. It's been a while since I've done those problems, but I think that should get you started.

Edit: I'm rusty, so whatever. I'll try and do it on paper.

Yeah umm any way you can solve this with Calc A knowledge? I just started BC, and I know absolutely nothing about finding integrals other than from geometric shapes...

You could try deriving the volume equation to come up with the rate of change of radius.

dV/dt = 4 pi r^2 dr/dt

dr/dt should also be 1/15 because of the slope between the two points given, but this maybe is put in later...

Then... dV/dt = 4 pi r^2 dr/dt = (k/r)?

That's as far as I can go, lol.

Yeah, then dr/dt = k/(4 pi r^3).

4 pi r^3 dr = kdt Integrate this.

pi r^4 = kt + C

With the two time values (t=0, r=1 and t=15, r=2) you can solve for k to get the answer to part A.

Again, I'm rusty. I don't think what I'm doing is right (something is seriously wrong, like my dr/dt).

Post has been edited 1 time(s), last time on Sep 12 2008, 1:43 am by Fwop_.

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dr/dt should also be 1/15 because of the slope between the two points given, but this maybe is put in later...

No, because r(t) is not linear. It varies with respect to itself.


Fwop got it while I was solving this (he is quite correct), but


General given: dV/dt = k/r

r(0) = 1, r(15) = 2.

V = 4πr³/3

---

By the Chain Rule, dV/dt = (dV/dr)(dr/dt).

dV/dt = 4πr²(dr/dt)

4πr²(dr/dt) = k/r

4πr³dr = k dt

∫4πr³dr = ∫k dt

πr⁴ = kt + c

r(t) = (kt/π + c)^1/4

r(0) = c^1/4 = 1; c = 1.

r(t) = (kt/π + 1)^1/4

r(15) = (15k/π + 1)^1/4 = 2; k = π.

r(t) = (t + 1)^1/4

etc.

It's awfully advanced for the first three weeks of a Calc BC course, but once you firmly master the Chain Rule in this situation, it's pretty simple.

Post has been edited 1 time(s), last time on Sep 12 2008, 1:52 am by DT_Battlekruser.

Don't we assume that dV/dt = k/r?

This is the kind of stuff we did for AB, there were even a couple of these problems on the AP test.

Yes, my mistake (I forgot the constant of proportionality and remembered it when that stupid pi was still in the denominator so i didn't correct everything).

Was it in AB? It's been wayy too long What do you do in BC?

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I'm taking AP Calc this year and right now we're just learning implicit differentiation. Finished third day of school today.

The trick is not the differentiation, but implementing the Chain Rule based on the givens to see that dV/dt = (dV/dr)(dr/dt). The rest is just by rote. And I could be off on when you learn stuff, but you don't learn the nice big Chain Rule until multivariable, so I tend to think of it as an "advanced concept".

k stands for a constant?

Lol BC is just AB on steroids. We've been doing the exact same thing as AB so far. In BC we are supposed to get to integration by parts and La Hospital's Rule, and maybe some more things, I don't know yet.

Edit: Yeah, k is as DTBK said, the constant of proportionality. You do that in, what, Algebra 1? more in 2.

Yes, if a is proportional to b, then that's saying a = kb, where k is some constant.

Quote from Fwop_

Lol BC is just AB on steroids. We've been doing the exact same thing as AB so far. In BC we are supposed to get to integration by parts and La Hospital's Rule, and maybe some more things, I don't know yet.

I thought L'Hospital's rule was AB.. have fun memorizing the trig decomposition algorithms when you integrate by parts You can always derive them, too.

I think it's in the book, but I don't think we ever "officially" learned it. I know I've used it though =p.

We did implicit differentiation in AB. Our teacher was also awesome.

edit: We haven't gotten into anything new since school started, there are a couple kids that skipped AB so we have to slow down for them.

OMG thank you guys for everything! You guys rox!

It's preposterous scenario problems like these that give calc a bad name. Who the heck fills a sphere at a rate inversely proportional to the radius?

"Wow this water balloon is getting awfully big. I better fill it up more slowly." Urgh, I guess that's somewhat reasonable.

Oh it happens quite often. Especially in the field of engineering. Helps with theory and practical applications for the future