Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common screen. x=t, y=e^-t, z=2t-t^2; (0, 1, 0)

Step-by-step explanation:

Given

(0, 1, 0)

The vector equation is given as:

Substitute values for x, y and z

Differentiate:

The parametric value that corresponds to (0, 1, 0) is:

Substitute 0 for t in r'(t)

The tangent line passes through (0, 1, 0) and the tangent line is parallel to r'(0)

It should be noted that:

The equation of a line through position vector a and parallel to vector v is given as:

Such that:

and

The equation becomes:

By comparison:

and

The parametric equations for the tangent line are: