# Homework Help: Length of a curve?

1. Nov 5, 2012

### monnapomona

1. The problem statement, all variables and given/known data
Find the length of the curve:

r(t) = e-2t i + e-2t*sin(t) j + e-2t*cos(t) k, 0 ≤ t ≤ 2π

2. Relevant equations

L = $\stackrel{b}{a}$ |r'(t)|

3. The attempt at a solution

I tried factoring out e-2t and got:

e-2t (i + sin(t) j + cos(t) k)

This is where I got lost. Can I take the derivative this and just leave e-2t outside of the brackets?

2. Nov 5, 2012

### haruspex

Yes, differentiate the whole thing.

3. Nov 5, 2012

### monnapomona

Okay, would this right?

-2e-2t (1 + cos(t) - sin(t))

4. Nov 5, 2012

### SammyS

Staff Emeritus
Hello monnapomona. Welcome to PF !

e-2t can be outside of the brackets, but can't be taken out of the integrand. But it may be better not to factor it out. Either way, you will need to use the product rule.

That integral needs the differential, dt:
$\displaystyle \int_{0}^{2\pi}\left|\textbf{r}'(t)\right|\,dt$​
What is $\displaystyle \left|\textbf{r}'(t)\right|\ ?$

5. Nov 5, 2012

### haruspex

What's the derivative of f(t)g(t)?

6. Nov 5, 2012

### monnapomona

I think it would be f'(t)g(t) + g'(t)f(t).

I tried not factoring out e-2t from the vector function and got this

r'(t) = e-2t - 2e-2tsin(t) +e-2tcos(t) - e-2tsin(t) - 2e-2tcos(t)

I'm lost on what to do next... :s

7. Nov 6, 2012

### SammyS

Staff Emeritus
Where are the unit vectors?

8. Nov 6, 2012

### monnapomona

Oh, whoops.

r'(t) = e-2t i - 2e-2tsin(t)+e-2tcos(t) j - e-2tsin(t) - 2e-2tcos(t) k

I'm not sure if this is okay to do but I tried factoring out the e-2t and got

r'(t) = e-2t ( i - 2sin(t) + cos(t) j - sin(t) - 2cos(t) k )

9. Nov 6, 2012

### aralbrec

It looks like the i part is wrong so you may want to doublecheck.

But I'm not sure why you stopped there :) Length of a vector is...

10. Nov 6, 2012

### monnapomona

Oh yeah! okay hopefully this is right:

r'(t) = -2e-2t i - 2e-2tsin(t)+e-2tcos(t) j - e-2tsin(t) - 2e-2tcos(t) k

i'm a little confused on how to do length...
Could I square each vector like (i, j, k) and just add what I get for it together?

11. Nov 6, 2012

### aralbrec

That looks right. I would have kept the exponential factored out to make it easier to take the magnitude.

Yes and take square root. This vector is no different than any other. It has a direction and magnitude; the only difference from what you are used to is maybe that its direction and magnitude change with t (time).

For (2d) vectors v = ai + bj, |v| = √(a2 + b2). Sometimes it is more convenient to use the dot product: v.v = a2 + b2, |v| = √(v.v)

Sometimes it's not always taught where this length integral comes from. You should notice that if r(t) is a position vector, then r'(t) is velocity, which makes speed |r'(t)|. Since distance = rate * time, the distance travelled in time dt is |r'(t)| dt. Add up the distances travelled during the entire time interval and you get the distance travelled.

12. Nov 7, 2012

### monnapomona

Okay, I took the derivative of all the components and my length formula turned out to look like this:

L = $\stackrel{2\pi}{0}$ √(4e-4t + e-4tsin2t + 4e-4tcos2t + e-4tcos2t + 4e-4tsin2t)

= e-2t√( 4 + 5(cos2t + sin2t )
= e-2t√(4+5)
= 3e-2t

Using the fundamental theorem of calc, I got the answer to be 3e-4π - 3 ? Would this be the right answer?

13. Nov 7, 2012

### SammyS

Staff Emeritus
You really need extra parentheses:
r'(t) = -2e-2t i + (- 2e-2tsin(t)+e-2tcos(t) ) j + (- e-2tsin(t) - 2e-2tcos(t) )k

Then, in you last post it appears that your integration isn't quite right.

14. Nov 7, 2012

### monnapomona

Oh wow, I forgot to integrate it... I got this as the answer:

$\int$e-2t = -1/2 e-2t

so -3/2(e^(-2t)) | $^{2π}_{0}$

= -3/2(e-4π + 1)

15. Nov 7, 2012

### SammyS

Staff Emeritus
Almost ...

It's -3/2(e-4π - 1)

16. Nov 7, 2012

### monnapomona

Oh I see where I went wrong.

Thank you for all your help!

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