18.022 Recitation Handout (with solutions)
20 October 2014
1. (3.2.17 in Colley) Use the formula
κ=
kv × ak
k v k3
to show that if f is C2 on an interval [ a, b] then the curvature of the graph y = f ( x ) is
κ=
| f 00 ( x )|
(1 + ( f 0 ( x ))2 )3/2
.
Solution. We parametrize the graph as usual by x(t) = (t, f (t)) for t ∈ [ a, b]. Then v(t) = (1, f 0 (t))
and a(t) = (0, f 00 (t)). Taking the norm of v × a gives | f 00 (t)|, and kvk3/2 = (1 + f 0 (t)2 )3/2 . Substituting into the given formula gives the desired expression for κ.
2. Let f : R2 → R3 be a map defined by f (x) = (|x|2 , 1, |x|) for x ∈ R2 . Find the total derivative
Df.
Solution. The total derivative is the matrix of partial derivatives, which is


2x1 2x2
0 .
D f ( x1 , x2 ) =  0
x1
|x|
x2
|x|
3. Sketch the curve x(t) = (t cos t, t sin t) and find its unit tangent vector.
Solution. The unit tangent vector is given by x0 (t)/|x0 (t)|. We calculate
x0 (t) = (cos t − t sin t, sin t + t cos t),
the squared norm of which is
cos2 t − 2t sin t cos t + 2t sin t cos t + t2 sin2 t + t2 cos2 t = 1 + t2 .
So the unit tangent vector is
cos t − t sin t sin t + t cos t
√
, √
1 + t2
1 + t2
.
4. Let f ( x, y) = log( x2 + y2 ) for ( x, y) ∈ R2 \ {(0, 0)}. Show that ∇ · (∇ f ) = 0.
Solution. The gradient of f is
∇f =
x
y
, 2
2
2
x + y x + y2
.
The divergence of this vector field is
∂
x
∂
y
y2 − x 2
x 2 − y2
∇ · (∇ f ) =
+
=
+
= 0.
∂x x2 + y2
∂y x2 + y2
( x 2 + y2 )2 ( x 2 + y2 )2