Examples from Section 13.1
#28.
At what points does the helix r(t) = ⟨ sint, cost, t ⟩ intersect the sphere x2 + y2 + z2
= 5?
Well, since x = sint and y = cost,
So
x2 + y2 = 1, always.
x2 + y2 + z2 = 5
1 + z2 = 5
z2 = 4,
but since z = t
t2 = 4,
so t = ± 2, giving us the points (sin2, cos2, 2) and (-sin2, cos2, -2)
#38:
Find the vector function that represents the curve of intersection of the paraboloid
z = 4x2 + y2 and the parabolic cylinder y = x2
What do we have here?
We know z = 4x2 + y2 and y = x2,
x = t,
y = t 2,
z = 4t2 + t4
so if we let x = t, we have
r(t) = ⟨ t, t2, 4t2 + t4 ⟩ and we are done.
OR, we could let x = T5 ,
then:
x = T5 ,
y = T10 ,
z = 4T10 + T20 , which gives
r(T) = ⟨ T5, T10, 4T10 + T20 ⟩
There are many (∞) possible (correct) answers.
#42
Two particles travel along the space curves:
r1(t) = ⟨ t, t2, t3 ⟩ and r2(t) = ⟨ 1 + 2t, 1 + 6t, 1 + 14t ⟩
Do the particles collide? Do their paths intersect?
Do they collide?
Is there a value of t such that x1 = x2, y1 = y2, and z1 = z2?
For x1 = x2, we need t = 1 + 2t which yields t = -1, but that value of t does not
satisfy the other 2 equations (y1 = y2, and z1 = z2). So the particles do not collide.
Do their paths intersect?
Rewrite the curves as: r1(t) = ⟨ t, t2, t3 ⟩ and r2() = ⟨ 1 + 2, 1 + 6, 1 + 14 ⟩
Can we satisfy: t = 1 + 2, t 2 = 1 + 6, and t3 = 1 + 14 simultaneously?
Yes, t = 1,  = 0 works as does t = 2,  = ½. So the paths intersect, but the
particles do NOT collide. The points of intersection are:
(1,1,1) (t=1,  = 0) and (2, 4, 8) (t = 2,  = ½)