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:

b. 2

step-by-step explanation:

the slope of a line =δy/δx

δy= y₂-y₁

δx= x₂-x₁

therefore, using the points (-2,-3)   and (3,7) we can calculate gradient as follows:

m=(7--3)/(3--2) (a number minus a negative number the sign changes to a plus)

m=(7+10)/(3+2)

=10/5

=2

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl