## Introduction

We are given a curve r(t) = (x(t), y(t), z(t)). We want to find the unit tangent vector t(t) and the unit normal vector n(t) at the point on the curve at time t. We will also find the binormal vector b(t).

## Theoretical Background

In this section we will provide the necessary theoretical background for our work. We begin with a review of some basic concepts in differential geometry which will be used throughout the paper. We then recall the notions of tangent space and curvature.

### Curves

In mathematics, a curve is, generally speaking, an object similar to a line in that it is smooth and can be parametrized by a single parameter, but unlike a line, a curve is not necessarily straight. Intuitively, one can think of a curve as the path traced out by a point moving along some continuous function f(t). For example, the graph of the function f(x) = x2 is a parabola, and therefore its trace is also a parabola. More precisely, a curve can be defined as the image of some continuous function f: I → Rn (where I is some interval in the real numbers), or equivalently, as the set of points (f(t), t) ∈ Rn+1.

### Tangent and Normal

At any given point on a curve, the tangent line is the line that best approximates the curve at that point. The normal line is the line perpendicular to the tangent line at the point of tangency. If you are given a point on the curve and the equation of the curve, you can find the equation of the tangent line using calculus.

## Methodology

We will use the Euler method to find the tt nt at and an at the given time t for the curve rt. The Euler method is a simple numerical method for approximating the solution to a first-order ordinary differential equation (ODE).

### find tt

To find tt, we use the following equation:

tt = ((2*nt)/rt) – ((1)/(rt^2))

where:

rt is the rate of change of the curve at time t

nt is the number of t’s that have elapsed since the beginning of the curve

### find nt

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## Results and Discussion

The position of the particle is given by r(t)=t^2i+4t^3j−6t^4k. At time t=1s,

find tt nt at and an at the given time t for the curve rt:

The position of the particle is given by r(t)=t^2i+4t^3j−6t^4k. At time t=1s, we have

r(1)=(1)^2i+4(1)^3j−6(1)^4k

r(1)=(1)i+4j−6k

Therefore, the tangent to the curve at r(1) is given by

r'(1)=2i+12j−24k

and the normal to the curve at r(1) is given by n(1)=12i+4j−24k

## Conclusion

In conclusion, to find tt nt at and an at the given time t for the curve rt, one must first determine the function that describes the curve. Once this function is determined, one can then take the derivative of the function to find tt nt at and an at the given time t.