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Applying the fundamental theorem of calc to calc III, confused! |
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| Oct5-05, 03:33 PM | #1 |
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Applying the fundamental theorem of calc to calc III, confused!
Hello everyone, i have the following problem i'm confused about! Can anyone guide me to what i'm suppose to do? I tried the following but it was wrong, he then told me I can just apply the theorem to |r'(u)| insteed of the stuff under the square root which would be more difficult. This is what i have, any ideas? Thanks!
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| Oct5-05, 03:34 PM | #2 |
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Alex |
| Oct5-05, 03:35 PM | #3 |
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sorry i fixed it, wow u got that message fast :)
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| Oct5-05, 04:03 PM | #4 |
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Applying the fundamental theorem of calc to calc III, confused!
You are thinking of s as [itex]\int_c^xf(t)\,dt[/itex] when in fact, it's not. What you need to do is apply the fund. theorem of calculus to this:
[tex]\frac{ds}{dx}=\frac{d}{dx}\int_c^x\left|\mathbf{r}'(t)\right|\,dt[/tex] Tell me if this answers your question ![]() Alex |
| Oct5-05, 05:48 PM | #5 |
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thanks for the responce alex, i'm confused how to apply this therom to that vector...
can i write... |r'(x)| - |r'(c)| + c is that what he wants? |
| Oct5-05, 06:47 PM | #6 |
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Alex |
| Oct5-05, 06:56 PM | #7 |
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so since i was given r, does that mean all i have to do is take the derivative of r and that is the answer?
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| Oct5-05, 08:38 PM | #8 |
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Alex |
| Oct5-05, 09:19 PM | #9 |
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awesome thank you so much alex! so if r(t) = f(t)i + g(t)j + h(t)k;
then r'(t) = f'(t)i + g'(t)j + h'(t)k, and that should be it then? |
| Oct5-05, 10:05 PM | #10 |
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Alex |
| Oct5-05, 10:34 PM | #11 |
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ohh good call, thanks again!! :)
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| Oct5-05, 11:22 PM | #12 |
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Alex, i had another question, why do you use x in the final answer if it wants ds/dt? and u also said, "and in the above post I mean to say d/dx not d/dt." The only time i see that, u do say d/dx, not d/dt. Thanks!
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| Oct5-05, 11:51 PM | #13 |
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Recognitions:
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mr_coffee, Let's say you have
[tex] L = \int_{a}^{x} f(t) dt [/tex] L will be a function of x [tex] L = F(x) - F(a) [/tex] so [tex] \frac{dL}{dx} = \frac{dF(x)}{dx} [/tex] now consider [tex] L = \int_{a}^{b} f(t) dt [/tex] L is a number [tex] L = F(b) - F(a) [/tex] so [tex] \frac{dL}{dx} = 0 [/tex] Basicly t in both cases is called the dummy variable, and x is the "real" variable, you're working with. In your case is the time variable therefore [tex] s(t) = \int_{a}^{t} |\vec{r}'(u)| du [/tex] where u is the dummy variable so [tex] \frac{ds(t)}{dt} = |\vec{r}'(t)| [/tex] |
| Oct5-05, 11:58 PM | #14 |
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thanks for the explanation! So if u is the dummy variable, and x is the "real" should i make the answer this: ds/dx = |r(x)|; r(x) = f(x) i + g(x) j + h(x) k; r'(x) = f'(x) i + g'(x) j + h'(x) k?
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| Oct6-05, 12:04 AM | #15 |
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Recognitions:
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t,x,u, etc.. are simply parameters of your curve C, given by your vectorial function r, it doesn't matter, but to be consistent with your book, i'd use t.
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| Oct6-05, 12:06 AM | #16 |
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k thanks for all the help!
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