- #1

ElijahRockers

Gold Member

- 270

- 10

## Homework Statement

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

## Homework Equations

**r**(t) = (-3t

**cos**t)

**i**+ (3t

**sin**t)

**j**+ (2[itex]\sqrt{2}[/itex])t

^{(3/2)}

**k**

0 ≤ t ≤ ∏

## The Attempt at a Solution

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

**v**(t) = (3t

**sin**t - 3

**cos**t)

**i**+ (3t

**cos**t + 3

**sin**t)

**j**+ (3[itex]\sqrt{2t}[/itex])

**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.