# Unit Tangent Vector of a curve / Arc Length

1. Oct 6, 2011

### ElijahRockers

1. The problem statement, all variables and given/known data

Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve.

2. Relevant equations

r(t) = (-3tcost)i + (3tsint)j + (2$\sqrt{2}$)t(3/2)k

0 ≤ t ≤ ∏

3. The attempt at a solution

So I found dr/dt (I think), which is

v(t) = (3tsint - 3cost)i + (3tcost + 3sint)j + (3$\sqrt{2t}$)k

This would be the tangent vector at 't', right?

So to find the unit tangent vector, I need to divide v(t) by its length, which would be the square root of its terms squared. This looks like it's going to be very ugly, but I tried to do it anyway. I got some huge radical that went all the way across the page.

The example showed that |v| was equal something much simpler, 3(t+1). My trig is not great, and I can't see how they got it down that small. Is there some trig identity I'm overlooking?

By the way, I am not taking any classes, this is independent study. I got this question from www.interactmath.com , Thomas' Calculus, 12e - Chapter 13, Section 3, Exercise 7

Thanks.

2. Oct 6, 2011

### SammyS

Staff Emeritus
That's the tangent vector alright.

The square of its terms works out to something very simple. What did you get for it?

3. Oct 7, 2011

### ElijahRockers

Well, my work is actually gone for it, but I FOILed the i and j terms.

Like I said in the original post, I ended up with some huge radical that went all the way across the page, and I couldn't seem to simplify it. I think I remember being able to factor off a coefficient of 3, and being left with a common expression of (sint + tcost), but I wasn't sure what to do with that, or how they got the whole thing to boil down to 3(t+1).

My trig is not great.

Anyway, going out of town, be back sunday. Thanks in advance for your input.

4. Oct 7, 2011

well just remember the trig identity sin2(θ) + cos2(θ) = 1 and your radical will simplify greatly (assuming you foil correctly).

5. Oct 7, 2011

### ElijahRockers

I must have not foiled correctly then, because I was looking for that! The previous exercise made use of that, so I was looking for it, but I must've made a mistake. I will try again, thanks.

6. Oct 7, 2011

### ElijahRockers

Oooook, I found it! Thanks for all your help!