Verified answer

Hi Jane!

Given curve vector r = <t^2, (2/3)t^3, t>

1.) When t = 1, r = <(1)^2, (2/3)(1)^3, 1> = <1, 2/3,1>

The osculating circle lies on the osculating plane. The osculating circle is the circle that best describes the behavior of the curve near r = <1, 2/3,1>. This circle has the same tangent, normal, and curvature as the curve does at this point.

k(t) = curvature = |r'(t) x r"(t)| / |r'(t)|^3 = 1 / (radius of circle)

r'(t) = <2t, 2t^2, 1>; |r'(t)| = SQRT[4t^2 + 4t^4 + 1]

r"(t) = <2, 4t, 0>

r'(t) x r"(t) = <- 4t, 2, 4t^2>; |r'(t) x r"(t)| = SQRT[16t^2 + 4 + 16t^4]

k(t=1) = |r'(t) x r"(t)| / |r'(t)|^3 = SQRT[16(1)^2 + 4 + 16(1)^4] / [SQRT[4(1)^2 + 4(1)^4 + 1]]^3 = 2 / 9

(radius of circle) = 1 / k = 4.5

The center of the osculating circle is 4.5 from (1, 2/3,1) and lies on the normal line to the curve at that point:

N(t) = T '(t) / |T '(t)|

T(t) = r '(t) / |r '(t)| = <2t, 2t^2, 1> / SQRT[4t^2 + 4t^4 + 1] = <2t / (2t^2 + 1), 2t^2 / (2t^2 +1), 1 / (2t^2 + 1)>

T '(t) = <(- 4t^2 + 2) / (2t^2 + 1)^2, 4t / (2t^2 + 1)^2, - 4t / (2t^2 + 1)^2>

|T '(t)| = SQRT[[(- 4t^2 + 2) / (2t^2 + 1)^2]^2 + [4t / (2t^2 + 1)^2]^2 + [- 4t / (2t^2 + 1)^2]^2]

N(t=1) = T '(1) / |T '(1)| = <-2/9, 4/9, - 4/9> / SQRT[4/81 + 16/81 + 16/81] = <-1/3, 2/3, -2/3>

So the vector containing the center and (1,2/3,1) is: <1 - t/3, 2/3 + 2t/3, 1 - 2t/3>

We want to know t such that the distance between (1,2/3,1) and the point on the normal line is 4.5.

4.5 = SQRT[t^2/9 + 4t^2/9 + 4t^2/9] = t

The center corresponds to t = 4.5, so the center is (-1/2, 11/3, -2)

2.)The normal vector to the osculating plane is the binormal vector,B.

B = T x N

T(t) = <2t / (2t^2 + 1), 2t^2 / (2t^2 +1), 1 / (2t^2 + 1)> from part 1)

N(t) = T '(t) / |T '(t)| where

T '(t) = <(- 4t^2 + 2) / (2t^2 + 1)^2, 4t / (2t^2 + 1)^2, - 4t / (2t^2 + 1)^2>

|T '(t)| = SQRT[[(- 4t^2 + 2) / (2t^2 + 1)^2]^2 + [4t / (2t^2 + 1)^2]^2 + [- 4t / (2t^2 + 1)^2]^2] = 2

N(t) = <(- 2t^2 + 1) / (2t^2 + 1)^2, 2t / (2t^2 + 1)^2, - 2t / (2t^2 + 1)^2>

B(t) = T(t) x N(t) = <- 2t / (2t^2 + 1)^2, 1 / (2t^2 + 1)^2, 2t^2 / (2t^2 + 1)^2>

B(t=1) = <-2/9, 1/9, 2/9>

So the equation of the osculating plane, which contains (1, 2/3, 1) is:

(-2/9)(x - 1) + (1/9)(y - 2/3) + (2/9)(z - 1) = 0

3.) lim k(t) as t --> infinity =

= lim [SQRT[16(t)^2 + 4 + 16(t)^4] / [SQRT[4(t)^2 + 4(t)^4 + 1]]^3] as t --> infinity

= lim [2(2t^2 + 1) / (2t^2 + 1)^3

= lim [2 / (2t^2 + 1)^2]as t --> infinity

= 0

This means the curve appears to become flatter as t --> infinity.