Calculus of vector valued functions

Curvature - suggested problems - solutions

P1: For r(t) =< t, t, 1 + t2 >

(a) Find the curvature function κ(t) (use the formula κ(t) =||r′(t)× r′′(t)||

||r′(t)||3 ).

r′(t) = < 1, 1, 2t >

||r′(t)|| =√

2 + 4t2

r′′(t) = < 0, 0, 2 >

r′(t)× r′′(t) =

∣

∣

∣

∣

∣

∣

i j k

1 1 2t0 0 2

∣

∣

∣

∣

∣

∣

=< 2,−2, 0 >

||r′(t) × r′′(t)|| =√

8

κ(t) =||r′(t) × r′′(t)||

||r′(t)||3

=

√8

(√

4t2 + 2)3

=

√23

(√

2(2t2 + 1))3

=

√23

√23(

√2t2 + 1)3

=1

(√

2t2 + 1)3

(b) What is the curvature at r(1)?

When t = 1,

κ(1) =1

(√

2(1)2 + 1)3=

1

3√

3≈ .192

(c) What is the radius of curvature at r(1)?

ρ =1

κ= 3

√3 ≈ 5.196

(d) Plot the curve in MVT and sketch the tangent circle at r(1).

Attached at end.

P2: Find the curvature and radius of curvature of r(t) =<√

2 t, et, e−t > at the point (0, 1, 1).Graph the curve and sketch the tangent circle at that point.

Note first <√

2 t, et, e−t >=< 0, 1, 1 > solves to t = 0.

r′(t) = <√

2, et,−e−t >

r′′(t) = < 0, et, e−t >

One trick to make these more manageable is to realize that that’s it for the Calculus(differentiation). After that, it’s all vector algebra. If you are looking for curvature ata point (instead of a curvature function), your best bet is to go and plug the t value inright now, and save yourself a lot of algebra...

r′(0) = <√

2, e0,−e−0 >=<√

2, 1,−1 >

r′′(0) = < 0, e0, e−0 >=< 0, 1, 1 >

||r′(0)|| =√

2 + 1 + 1 = 2

r′(0) × r′′(0) =

∣

∣

∣

∣

∣

∣

i j k√2 1 −1

0 1 1

∣

∣

∣

∣

∣

∣

=< 2,−√

2,√

2 >

||r′(0) × r′′(0)|| =√

4 + 2 + 2 =√

8

κ(0) =||r′(0) × r′′(0)||

||r′(0)||3

=

√8

23=

√8

8≈ .354

Curvature κ ≈ .354 and radius of curvature ρ = 1κ ≈ 2.83.

Graph at end.

P3: For the plane curve y = 4x3/2

(a) Find the curvature function κ(x) (use the formula κ(t) =|f ′′(x)|

[1 + (f ′(x))2]3/2).

f ′(x) = 6x1/2

f ′′(x) = 3x−1/2

κ(x) =|f ′′(x)|

[1 + (f ′(x))2]3/2

=

∣

∣

∣

3√x

∣

∣

∣

[1 + (6√

x)2]3/2

=

3√x

(1 + 36x)3/2

=3√

x(1 + 36x)3/2

(b) What is the curvature at the point (1, 4)?

κ(1) =3√

1(1 + 36(1))3/2=

3

373/2≈ .013

(c) What is the radius of curvature at (1, 4)?

ρ ≈ 1

.013≈ 75.0

(d) Plot the curve (however you like; in MVT there’s a 2D function plotter for y = f(x))and sketch the tangent circle at (1, 4).

Attached.