Calculus Quiz 2
Name: March 15, 20061. (5 points) Let r(t) = x(t)i + y(t)j + z(t)k be a differentiable vector-valued function such that r0(t) 6= 0 for all t ≥ 0. Show that the arc length function s defined by s(t) =
Z t
0
r¡dx dt
¢2 +¡dy
dt
¢2 +¡dz
dt
¢2 , for t ≥ 0, has an inverse t =φ(s).
Solution: Since ds
dt = kr0(t)k 6= 0 for all t ≥ 0, Mean Value Theorem implies that the function s is an 1 − 1 function and the inverse t =φ(s) exists.
2. (5 points) Consider the circular helix r(t) = 3 costi + 3 sintj + 4tk for t ≥ 0. Determine the arc length s as a function of t by evaluating the integral s =
Z t
0
r¡dx dt
¢2 +¡dy
dt
¢2 +¡dz
dt
¢2
. What is the inverse t =φ(s)?
Solution: By direct computation, we get s = Z t
0
√9 + 16 = 5t, and t = s 5.
3. (5 points) Let r(t) = etcosti + etsintj + etk. Find the curvatureκ of the path.
Solution: By direct computation, we get r0(t) =¡
et(cost − sint), et(cost + sint), et¢ , kr0(t)k =√
3et, and T (t) = 1
√3(cost − sint, cost + sint, 1).
Therefore,κ= kdT
dsk = kdT dt
1 ds/dtk =
√2
√3
√1 3et =
√2e−t
3 .
4. (5 points) Show that the curvature of a polar curve r = f (θ) is given byκ=|[ f (θ)]2+ 2[ f0(θ)]2− f (θ) f00(θ)|
([ f (θ)]2+ [ f0(θ)]2)3/2 . Hint: Express the curve as R(θ) = (x(θ), y(θ)) = (r cosθ, r sinθ), where r = f (θ).
Solution: By direct computation, we get R0(θ) = (x0, y0), T (θ) =p 1
(x0)2+ (y0)2(x0, y0), dT
dθ =
−(x0x00+ y0y00)
¡(x0)2+ (y0)2¢3/2(x0, y0) +
¡(x0)2+ (y0)2¢
¡(x0)2+ (y0)2¢3/2(x00, y00) = x0y00− x00y0
¡(x0)2+ (y0)2¢3/2¡
− y0, x0¢ ,
kdT dθk =
|x0y00− x00y0|
(x0)2+ (y0)2, andκ= kdT
dsk = kdT dθk
1 ds/dθ =
|x0y00− x00y0|
¡(x0)2+ (y0)2¢3/2. Now x = f cosθ, y = f sinθ, and
x0= f0cosθ− f sinθ, y0= f0sinθ+ f cosθ,
x00 = f00cosθ− 2 f0sinθ− f cosθ, y00= f00sinθ+ 2 f0cosθ− f sinθ, we getκ =|[ f (θ)]2+ 2[ f0(θ)]2− f (θ) f00(θ)|
([ f (θ)]2+ [ f0(θ)]2)3/2 .
