# Finding T(t) and N(t) for the Curve r(t) = 4t^2 i + 6t j

In this article, we will explore the process of finding the tangent vector `T(t)`

and the normal vector `N(t)`

for the curve `r(t) = 4t^2 i + 6t j`

, which is a vector-valued function.

## Vector-Valued Functions

A vector-valued function is a function that maps a real number (or a set of real numbers) to a vector in a vector space. In this case, the vector-valued function `r(t)`

maps a real number `t`

to a vector in the two-dimensional Cartesian coordinate system.

The vector-valued function `r(t)`

is defined as:

`r(t) = 4t^2 i + 6t j`

where `i`

and `j`

are the standard unit vectors in the x and y directions, respectively.

## Finding the Tangent Vector T(t)

The tangent vector `T(t)`

at a point on the curve `r(t)`

is the derivative of the vector-valued function `r(t)`

with respect to `t`

. To find `T(t)`

, we need to differentiate the vector-valued function `r(t)`

component-wise.

The derivative of `r(t)`

is:

`T(t) = r'(t) = 8t i + 6 j`

Therefore, the tangent vector `T(t)`

at any point on the curve `r(t)`

is `8t i + 6 j`

.

## Finding the Normal Vector N(t)

The normal vector `N(t)`

at a point on the curve `r(t)`

is a vector that is perpendicular to the tangent vector `T(t)`

at that point. To find `N(t)`

, we can use the following formula:

`N(t) = (-T_y(t), T_x(t))`

where `T_x(t)`

and `T_y(t)`

are the x and y components of the tangent vector `T(t)`

, respectively.

Substituting the values of `T(t)`

, we get:

`N(t) = (-6, -8t)`

Therefore, the normal vector `N(t)`

at any point on the curve `r(t)`

is `(-6, -8t)`

.

## Conclusion

In this article, we have learned how to find the tangent vector `T(t)`

and the normal vector `N(t)`

for the vector-valued function `r(t) = 4t^2 i + 6t j`

. The tangent vector `T(t)`

is the derivative of `r(t)`

, which is `8t i + 6 j`

, and the normal vector `N(t)`

is perpendicular to `T(t)`

, which is `(-6, -8t)`

.