# 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!
http://show.imagehosting.us/show/762659/0/nouser_762/T0_-1_762659.jpg

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mr_coffee said:
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! http://show.imagehosting.us/show//0023a0u/0023a0u
The link is broken. Maybe try posting it again?

Alex

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sorry i fixed it, wow u got that message fast :)

You are thinking of s as $\int_c^xf(t)\,dt$ when in fact, it's not. What you need to do is apply the fund. theorem of calculus to this:

$$\frac{ds}{dx}=\frac{d}{dx}\int_c^x\left|\mathbf{r}'(t)\right|\,dt$$

Alex

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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?

mr_coffee said:
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?
Remember which variable you are differentiating with respect to. |r'(c)| and C are just constants, so those go to zero when you take the derivative. Now, what about the x? Well, if you look at the other part of the page that says g'(x)=f(x), you can apply that here. Since s is really s(x), you can treat it like g(x). The same goes for f(x) and |r'(x)|. Do you see what happens?

Alex

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?

mr_coffee said:
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?
Yes, and in the above post I mean to say d/dx not d/dt.

Alex

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?

mr_coffee said:
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?
Correct. Just note here that you are using x and not t in your final answer.

Alex

ohh good call, thanks again!! :)

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!

Pyrrhus
Homework Helper
mr_coffee, Let's say you have

$$L = \int_{a}^{x} f(t) dt$$

L will be a function of x

$$L = F(x) - F(a)$$

so

$$\frac{dL}{dx} = \frac{dF(x)}{dx}$$

now consider

$$L = \int_{a}^{b} f(t) dt$$

L is a number

$$L = F(b) - F(a)$$

so

$$\frac{dL}{dx} = 0$$

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

$$s(t) = \int_{a}^{t} |\vec{r}'(u)| du$$

where u is the dummy variable so

$$\frac{ds(t)}{dt} = |\vec{r}'(t)|$$

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?

Pyrrhus
Homework Helper
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.

k thanks for all the help!