Hello, Ideasman!

The first one is very straight-forward. .Where is your difficulty?

1) Find the unit tangent vector to the curve at the following points:

. . r(t) .= .< 3·cos(t), 2·sin(t) >, .t = 0, -π/2, π/2

The tangent vector is: .r'(t) .= .< -3·sin(t), 2·cos(t) >

At t = 0: .r'(0) .= .< -3·sin(0), 2·cos(0) > .= .< 0, 2 >

. . The unit vector is: .< 0, 1 >

At t = -π/2: .r'(-π/2) .= .< -3·sin(-π/2), 2·cos(-π/2) > .= .< 3, 0 >

. . The unit vector is: .< 1, 0 >

At t = π/2 .r'(π/2) .= .< -3·sin(π/2), 2·cos(π/2) > .= .< -3, 0 >

. . The unit vector is: .< -1, 0 >

2.) Sketch this curve (the one above) with the addition of the following vectors:

. . r(0), T(0), r(π/2), and T(π/2)

To identify this curve, eliminate the parameter.

. . x .= .3·cos(t) . → . x/3 .= .cos(t) .[1]

. . y .= .3·sin(t) . → . y/2 .= .sin(t) .[2]

Square [1]: .x²/9 .= .cos²(t)

Square [2]: .y²/4 .= .sin²(t)

Add: .x²/9 + y²/4 .= .sin²(t) + cos²(t)

. . . . . . . . . . . . . . . . . . . x² . .y²

And we have the ellipse: . --- + --- .= .1

. . . . . . . . . . . . . . . . . . . 9 . . .4

Iassumeyou can find those four vectors.