Homework 6 Solutions

Problem 1.

(a) We haves′(0) = ‖~x ′(0)‖ = ‖(2, 1, 2)‖ = 3 .

As for s′′(0), recall our formula for acceleration:

~x ′′ = s′′ ~T + (s′)2κ ~P .

Then s′′ = ~x ′′ · ~T =~x ′′ · ~x ′

s′=

(2, 1, 2) · (12, 9, 6)

3= 15 .

(b) We know that ~T =~x ′

s′=

1

3(2, 1, 2) . To find ~P , observe that

(s′)2κ ~P = ~x ′′ − s′′ ~T = (12, 9, 6)− 15

(1

3(2, 1, 2)

)= (2, 4,−4).

Since (s′)2κ > 0, it follows that ~P =(2, 4,−4)

‖(2, 4,−4)‖=

1

3(1, 2,−2) . Finally,

~B = ~T × ~P =1

3(2, 1, 2)× 1

3(1, 2,−2) =

1

3(−2, 2, 1)

(c) We have (s′)2κ = ‖(s′)2κ ~P‖ = ‖(2, 4,−4)‖ = 6. Since s′ = 3, it follows that κ =2

3.

(d) The circle has radius R = 1/κ = 3/2, and its center is at

~C = ~x(0) +R ~P (0) = (3, 1, 0) +3

2

(1

3(1, 2,−2)

)=

(7

2, 2,−1

)The circle lies in the plane parallel to ~T and ~P , so parametric equations for the circle are

~C + (cos t) ~T (0) + (sin t) ~P (0) =

(7

2, 2,−1

)+

cos t

2(2, 1, 2) +

sin t

2(1, 2,−2)

(e) The osculating plane is perpendicular to ~B(0), so it has an equation of the form

−2x + 2y + z = C

for some constant C. The plane goes through the point ~x(0) = (3, 1, 0), and plugging thispoint in gives C = −4. Thus the osculating plane is the plane −2x+ 2y + z = −4 .

(f) We have τ =det

([~x ′ ~x ′′ ~x ′′′])(s′)6κ2

=1

(3)6(2/3)2

∣∣∣∣∣∣2 12 111 9 −42 6 3

∣∣∣∣∣∣ = −1

2.

Problem 2.

(a) Since the curve is unit-speed, we know that ~P ′ = −κ~T + τ ~B. Then

κ = −~P ′ · ~T = −~P ′ · ~x ′ = −(4, 0,−6) · (2/7,−3/7, 6/7) = 4

(b) We have

τ ~B = ~P ′ + κ ~T = (4, 0,−6) + 4(2/7,−3/7, 6/7) =1

7(36,−12,−18).

Since τ = −6, it follows that ~B =1

7(−6, 2, 3)

(c) We have

~P = ~B × ~T =1

7(−6, 2, 3)× 1

7(2,−3, 6) =

1

7(3, 6, 2)

(d) Since the curve is unit speed, we know that ~B ′ = −τ ~P . Taking the derivative yields

~B ′′ = −τ ′ ~P − τ ~P ′ = −14(3/7, 6/7, 2/7)− (−6)(4, 0,−6) = (18,−12,−40)