# Find the unit tangent vector tt rt 5 cos t 5 sin t 4 p 5 2 5 2 4

## Introduction

In mathematics, a tangent vector to a curve at a point is a vector that is tangent to the curve at that point. In other words, it is a vector that “points in the direction” of the curve at that point. More precisely, given a smooth curve c: I -> R^n (that is, a smoothly varying function from some interval I into Euclidean space), a point p in the interior of I, and a unit tangent vector t to c at p, we can write c(p + h) = c(p) + ht + O(h^2) for h small.

## What is the unit tangent vector tt rt?

The unit tangent vector tt rt at a point on a curve is a vector that points in the direction of the curve at that point and has a magnitude of 1. The unit tangent vector is often denoted by the letter tt.

For example, consider the curve defined by the parametric equations x=5cost, y=5sint, z=4. At the point p=(5,2,4), the unit tangent vector is t=(cos(2), sin(2), 0).

To find the unit tangent vector at any point on a curve, we first need to find the tangent vector. The tangent vector at a point on a curve is a vector that points in the direction of the curve at that point. To find the tangent vector, we take the derivative of each component of the parametric equations with respect to t (i.e., we take partial derivatives). For example, if x=5cost and y=5sint, then dx/dt=−5sint and dy/dt=5cost. Therefore, the tangent vector at p=(5,2,4) is t=(−5sin(2), 5cos(2), 0). Finally, to get the unit tangent vector we simply need to normalize this Tangential Vector (divide by its magnitude).

## How to find the unit tangent vector tt rt

The unit tangent vector tt rt is a vector that points in the direction of the tangent of a curve at a given point. To find the unit tangent vector, you must first find the tangent vector, which is the derivative of the curve at the given point. Once you have the tangent vector, you can normalize it to get the unit tangent vector.

### Using the definition of the unit tangent vector

The definition of the unit tangent vector is simply a vector that points in the direction of the tangent line at a certain point on a curve. In other words, if you have a curve defined by a function (r(t)), the unit tangent vector is:

[t=\frac{r'(t)}{||r'(t)||}]

Where (r'(t)) is the derivative of (r(t)). In our example, we have:

[r(t)=\langle 5\cos t,5\sin t,4 \rangle]

Which means that:

[r'(t)=\langle -5\sin t,5\cos t,0 \rangle]

Plugging this into our equation for the unit tangent vector gives us:

[t=\frac{\langle -5\sin t,5\cos t,0 \rangle}{||\langle -5sin t,5cos t,0 \rangle||}]

All we need to do now is simplify and calculate the norm. We get:

[t=\frac{1}{||\langle 5sin t,5cos t \rangle||}\langle -5sin t,5cos t \rangle={\frac {1}{{\sqrt {\left({5}^{2}{sin}^{2}t+{5}^{2}{cos}^{2}t}\right)}}}\left\<-5sint ,5cost \right>}]

This gives us the final answer:

[t={\frac {1}{{\sqrt {25{sin}^{2}t+25{cos}^{2}ttf }}}}\left\<-sint ,cost ,0tfright>}]

### Using the formula for the unit tangent vector

The unit tangent vector of a curve at a point is a vector that points in the direction of the tangent line to the curve at that point. It is a vector of length 1, and so it is also called a unit vector.

We can find the unit tangent vector using the formula:

tt rt 5 cos t 5 sin t 4 p 5 2 5 2 4

where t is the angle measured from the positive x-axis, r(t) is the position vector of the point on the curve at angle t, and p is the position vector of the point on the curve where we want to find the unit tangent vector.

## Examples

To find the unit tangent vector, we need to find the derivative of r(t), which is r'(t).

r(t) = <5cos(t), 5sin(t), 4>
r'(t) = <-5sin(t), 5cos(t), 0>

Now we need to find the magnitude of r'(t).

||r'(t)|| = sqrt((-5sin(t))^2 + (5cos(t))^2 + (0)^2)
= sqrt((25sin^2(t)) + (25cos^2(t)))
= sqrt((25)(sin^2(t) + cos^2(t)))
= sqrt((25)(1))
= 5

Now we can divide each component of r'(t) by the magnitude to get the unit tangent vector.

tt = <(-5sin(t))/5, (5cos

## Conclusion

In conclusion, to find the unit tangent vector of the function r(t)=5cos(t)+5sin(t)+4 we simply need to take the derivative of r(t) and divide it by the norm of the derivative. In this case, the derivative is r'(t)=-5sin(t)+5cos(t) and the norm is sqrt((-5sin(t)+5cos(t))^2). Therefore, the unit tangent vector is (-5sin(t)+5cos(t))/sqrt((-5sin(t)+5cos(t))^2).