# Applying the fundamental theorem of calc to calc III, confused!

1. Oct 5, 2005

### mr_coffee

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

Last edited by a moderator: Apr 21, 2017
2. Oct 5, 2005

### amcavoy

The link is broken. Maybe try posting it again?

Alex

Last edited by a moderator: Apr 21, 2017
3. Oct 5, 2005

### mr_coffee

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

4. Oct 5, 2005

### amcavoy

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

Last edited: Oct 5, 2005
5. Oct 5, 2005

### mr_coffee

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?

6. Oct 5, 2005

### amcavoy

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

7. Oct 5, 2005

### mr_coffee

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?

8. Oct 5, 2005

### amcavoy

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

Alex

9. Oct 5, 2005

### mr_coffee

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?

10. Oct 5, 2005

### amcavoy

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

Alex

11. Oct 5, 2005

### mr_coffee

ohh good call, thanks again!! :)

12. Oct 5, 2005

### mr_coffee

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!

13. Oct 5, 2005

### Pyrrhus

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

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

14. Oct 5, 2005

### mr_coffee

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?

15. Oct 6, 2005

### Pyrrhus

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.

16. Oct 6, 2005

### mr_coffee

k thanks for all the help!