Practice Exam 1 Solutions
Plane equations and parameterizations are not unique. Your solution may not match the one given.
1. F, T, F, T, T
2. ukv = h8/11, 8/11, 24/11i, u⊥v = h3/11, 3/11, −2/11i from which we can see ukv = (8/11)v
and u⊥v · v = 0.
3. The graph of x2 has higher curvature at (0, 0).
4. c(t) = h1, 0, 1it + h0, 1, 1i.
5. This problem contained an error, making
√ the2 integral much harder than it should have been.
Try the problem with r(t) = hln(t), 4 2t, 8t i and it will be easier. You should get a length
of 24 + ln(2) in this case.
6. r(t) = h4 − 4 cos(t) − 16 sin2 (t), 4 cos(t), 4 sin(t)i
7. (a) Hyperbola, Circle, Hyperbola (b) Bottom left
8. (a) r ≤ z ≤
√
1 − r2 , 0 ≤ r ≤ 1/4, −π/2 ≤ θ ≤ π/2
(b) 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π/4, −π/2 ≤ θ ≤ π/2
√
9. Area(t) =
194 2
t
2
10. Vector form: h2, 1, 1i · hx − 2, y − 3, z − 2i = 0
Scalar form: 2(x − 2) + (y − 3) + (z − 2) = 0.