# Finding a curve's unit tangent vector

• Ascendant78
In summary, the conversation is about finding an easier way to solve a problem involving derivatives and trigonometric identities. The participants discuss the use of the identity cos^2(t)+sin^2(t)=1 and the tediousness of the problem. One participant ultimately figures out the correct solution using a combination of inspection and factoring.
Ascendant78

## Homework Equations

I know the equations. See question below.

## The Attempt at a Solution

I am just wondering with this problem, how is it that they go from that derivative to the magnitude at the bottom of that image? I know the formula, but what I mean is they have no steps for something that to me seems extremely tedious, so I'm assuming they are applying some type of identity to save time and trouble, but I'm not seeing it? If someone can show me an easier way than squaring each of those, splitting it all up, factoring, then trying to make it so you can take the square root of it all, I'd really appreciate it.

Ascendant78 said:

## Homework Equations

I know the equations. See question below.

## The Attempt at a Solution

I am just wondering with this problem, how is it that they go from that derivative to the magnitude at the bottom of that image? I know the formula, but what I mean is they have no steps for something that to me seems extremely tedious, so I'm assuming they are applying some type of identity to save time and trouble, but I'm not seeing it? If someone can show me an easier way than squaring each of those, splitting it all up, factoring, then trying to make it so you can take the square root of it all, I'd really appreciate it.

Just how tedious is it? Have you tried it? I haven't worked it but it is surely workable and I would guess it uses the identity ##\sin^2t + \cos^2 t = 1##. Dive right in.

LCKurtz said:
Just how tedious is it? Have you tried it? I haven't worked it but it is surely workable and I would guess it uses the identity ##\sin^2t + \cos^2 t = 1##. Dive right in.

Well, there are 9 problems on the assigned homework just like this one (on top of 12 others on other topics), so cumulatively, if I don't have a way to simplify things like this, it will be very tedious. Besides, I didn't want to waste time if there was a more efficient way I was missing out on.

Anyway, I just took the time to factor it out and it's telling me my answer was wrong. I really, truly hate MyMathLab with a passion.

**Update** - I did figure out where I went wrong and got the right answer, but I still feel it was extremely tedious and that if they are skipping all the work, they are most likely applying some type of identity, but maybe not. They don't typically skip that much work in an example though.

Did you get ##4\sqrt 2(1+t)##? That's what I get. But in any case, it was quicker to just work it than type it into this forum.

1 person
LCKurtz said:
Did you get ##4\sqrt 2(1+t)##? That's what I get. But in any case, it was quicker to just work it than type it into this forum.

Oh, my specific problem was a bit different. The one I posted was just an example. Same exact setup, different coefficients and a couple different signs. I do appreciate you working it out though.

Ascendant78 said:
If someone can show me an easier way than squaring each of those, splitting it all up, factoring, then trying to make it so you can take the square root of it all, I'd really appreciate it.
As LCKurtz wrote, it's a matter of using cos2+sin2=1, but also of spotting that you can use it even before you write anything down.
By inspection, the leading terms of the i.i and j.j contributions sum to 16. Similarly the trailing terms sum to 16t2. Next, the middle terms (+/- 32t sin(t)cos(t)) cancel. It remains to add the k.k term and factorise.

1 person

## 1. What is a unit tangent vector?

A unit tangent vector is a vector that is tangent to a curve at a specific point and has a magnitude of 1. It represents the direction and rate of change of the curve at that point.

## 2. How is the unit tangent vector calculated?

The unit tangent vector is calculated by taking the derivative of the curve function and then dividing it by its magnitude. This can also be done by finding the slope of the tangent line at the point and then normalizing it to a unit vector.

## 3. Why is the unit tangent vector important in calculus?

The unit tangent vector is important in calculus because it helps us understand the behavior of a curve at a specific point. It also allows us to calculate the curvature of a curve and find the direction of motion of an object along the curve.

## 4. Can the unit tangent vector change at different points on a curve?

Yes, the unit tangent vector can change at different points on a curve. This is because the slope of the tangent line and the magnitude of the derivative can vary at different points, resulting in a different unit tangent vector at each point.

## 5. How can the unit tangent vector be used in real-world applications?

The unit tangent vector can be used in real-world applications such as motion analysis, robotics, and computer graphics. It can help determine the direction and speed of an object's motion along a curved path, and can also be used to create smooth and realistic animations in computer graphics.

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