Unit Tangent Vector of a curve / Arc Length

[ElijahRockers]

Homework Statement

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

Homework Equations

r(t) = (-3tcost)i + (3tsint)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) = (3tsint - 3cost)i + (3tcost + 3sint)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.

 

[SammyS]

...

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

v(t) = (3tsint - 3cost)i + (3tcost + 3sint)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.

...

That's the tangent vector alright.

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

 

[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.

 

[pergradus]

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

 

[ElijahRockers]

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

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.

 

[ElijahRockers]

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