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).