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Parametric curves: exercises

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This is an addendum for Parametric curves.

(1) Find the parametrization of the curve made of straight segments between points (1,0), (0,1), (-1,0), and (0,-1).

Parametrization of square.jpg

Let

 f(t) = tu + P, 0 ≤ t ≤ 1,

and in particular

 u = ( 0, 1) - ( 1, 0) = ( -1, 1).

Parametrization of the first edge is:

 f(t) = ( -1, 1) t + ( 1, 0),

which is not natural.

With time = 1 and distance = √2 (slowing down), we obtain

 1st edge: g1(s) = ( -1, 1) s / √2 + ( 1, 0), 0 ≤ s ≤ √2

which is natural. Further we have

 ‖ g′ ‖ = 1.

Analogously we obtain

 2nd edge:  g2(s) = ( -1, -1 ) ( s - √2 ) / √2 + ( 0, 1 )  for √2 ≤ s ≤ 2 √2,
 3rd edge:  g3(s) = ( 1, -1 ) ( s - 2 √2 ) / √2 + ( -1, 0 )  for 2 √2 ≤ s ≤ 3 √2
 4th edge:  g4(s) = ( 1, 1 ) ( s - 3 √2 ) / √2 + ( 0, -1 )  for 3 √2 ≤ s ≤ 4 √2.

Observe that g′2(s) ≠ g′1(s), etc at the points of gluing, so there is no derivative here. For example, vompute g′( √2 ): With Calc 1 we obtain

 g(s) = -3 / √2 + 1 = s / √2    if 0 ≤ s ≤ √2
 g′(s) = -( s - √2 ) / √2 = -( s - √2 ) / √2 + 1   if √2 ≤ s ≤ 2 √2.

From this we see that there is an angle. We then evaluate the derivative as a limit and show that is does not exist. From this it follows that g′(s) does not exist.

(2) Find the curvature of the curve:

 f(t) = ( 3 sin t, 4 cos t ).

Compute

 f′(t) = ( 3 cos t, -4 sin t )
 ‖ f′ ‖ = ( 9 cos2 t + 16 sin2 t )1/2 ≠ 1.

Find t = γ(τ) with

 γ′(τ) = 1 / ‖ f′(γ(τ)) ‖
    = 1 / ( 9 cos2 (γ(τ)) + 16 sin2 (γ(τ)) )1/2.

(2) Plot

     f(t) = ( t3 - t, 1 - t2 ),
     f1(t) = t3 - t,
     f2(t) = 1 - t2.

Parametric curve plotting example.jpg

Double spiral.jpg

(3) Plot

 f(t) = t ( sin t, cos t).

Then

 f(0) = ( 0, 0) and
     f(t) = 1 / t ( sin t, cos t ) → 0 as t → ∞, f(t) ≠ 0.

(4) Find the length of

     f(t) = ( t cos t, t sin t, 3t )
Spiral on cone.jpg

Re-write

     f(t) = t ( cos t, sin t, 3 ) = (x, y, z). 

Then these x, y, and z satisfy

     x2 + y2 = t2 = ( z / 3 )2 or
     x2 + y2 = ( z / 3 )2 

which is a cone - for each z, the cross section is a circle of radius z / 3. Next

     f′(t) = ( cos t, sin t, 3 ) + t ( -sin t, cos t, 0 )
          = ( cos t - t sin t, sin t + t cos t, 3 ),
     ‖ f′(t) ‖ = ( cos2 t - 2 t cos t sin t + t2 sin2 t + sin2 t + 2 t sin t cos t + t2 cos2 t + 9 )1/2
             = ( 10 + t2 )1/2.

Finally

     L = ∫0t ( 10 + t2 )1/2 dt.

(5) ==Orthogonality==

We know that if an object on a plane is affected only by forces within the plane will stay within the plane if its initial velocity is within the plane. Let's prove that.

Formally, we want to prove that given:

 (1) F′′(t) ⊥ Z for all t,
 (2) F′(0) ⊥ Z,
 (3) F(0) ⊥ Z,

where z is some vector, it follows that

 F ⊥ Z.

Re-write orthogonality conditions:

 < F′′(t), Z > = 0,
 < F′(0), Z > = 0,
 < F(0), Z > = 0 

The plan now is to show:

 < F(t), Z > = 0.

Step 1:

 F′(t) = ∫0t F′′(t) dt + F′(0)

Step 2: Show that

 < F′(t), Z > = 0, indeed
 < F′(t), Z > = < ∫0t F′′(t) dt + F′(0), Z >
      = < ∫0t F′′(t) dt, Z > + < F′(0), Z >
      = ∫0t F′′1(t) dt ⋅ Z1 + ∫0t F′′2(t) dt ⋅ Z2 + ∫0t F′′3(t) dt ⋅ Z3 + 0
      = ∫0t ( F′′1(t) Z) dt + ( F′′2(t) Z) dt + ( F′′3(t) Z) dt
      = ∫0t < F′′(t), Z > dt
      = ∫0t 0 dt 
      = 0.

Step 3: Repeat Step 1 but for F(t). Step 4: Repeat Step 2but for F(t).