Unit Tangent Vector of a curve / Arc Length

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Homework Help Overview

The discussion revolves around finding the unit tangent vector of a given curve and determining the length of a specified portion of that curve. The curve is defined parametrically, and the problem is situated within the context of calculus, specifically relating to vector functions and arc length.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the derivative of the position vector to obtain the tangent vector. There are attempts to simplify the expression for the magnitude of the tangent vector, with some participants expressing uncertainty about the simplification process and the application of trigonometric identities.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts and seeking clarification on specific steps. Some guidance has been offered regarding the use of trigonometric identities to simplify expressions, and there is an acknowledgment of the need to verify calculations.

Contextual Notes

One participant notes that they are studying independently and are not currently enrolled in a class, which may influence their approach and understanding of the material. There is also mention of a specific textbook and exercise that prompted the question.

ElijahRockers
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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\sqrt{2})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\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.
 
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ElijahRockers said:
...

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.

...
That's the tangent vector alright.

The square of its terms works out to something very simple. What did you get for it?
 
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.
 
well just remember the trig identity sin2(θ) + cos2(θ) = 1 and your radical will simplify greatly (assuming you foil correctly).
 
pergradus said:
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.
 
Oooook, I found it! Thanks for all your help!
 

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