110.202 CALCULUS III 23 February 2000

Solutions to First Examination

10:00 a.m. Edition

1. (a) The vector d = i + 2j + 3k is parallel to the plane p. Since the points (3, 2, 5) and (1, 1, 1) = i + j + k lie in the plane, the difference vector 2i + j + 4k is parallel to the plane. Therefore to find a normal vector to the plane, take the cross product

N = d × (2i + j + 4k) = | \begin{matrix}i & j & k \\ 1 & 2 & 3 \\ 2 & 1 & 4 \end{matrix} | = 5i + 2j − 3k.

(Any nonzero multiple of this will do.) (b) Then the equation of the plane p is (r−i− j−k) • N = 0, where r = xi+yj+zk, which expands to 5(x−1)+2(y−1)−3(z−1) = 0, which simplifies to 5x + 2y − 3z = 4. (Again, any nonzero multiple of this will do.)

2. P is the point (1, 1, 0) = i + j. For (a), differentiate:

r′(t) = 2ti + (3t^{2} − 2)j + (4t^{3} − 2)k

and set t = 1, to get v = 2i + j + 2k as a tangent vector. Then ∥v∥ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3, so that the unit tangent vector is T = v/3 = (2/3)i + (1/3)j + (2/3)k. Then for (b), the parametric equation of the tangent line at P is

R(u) = i + j + uv = i + j + u(2i + j + 2k).

(Or we could replace v by T here.)

3. (a) First compute the derivative

r′(t) = 2t^{2}i + 2tj + k.

Its norm was cooked to work out nicely,

∥r′(t)∥ = \sqrt{4t^{4} + 4t^{2} + 1} = 2t^{2} + 1.

The relevant values of t are −1 ≤ t ≤ 2 (look at the k-component of r(t)). Then the arc length is

∫_{−1}^{2}(2t^{2} + 1)dt = [2t^{3}/3 + t]_{−1}^{2} = 16/3 + 2 − (−2/3 − 1) = 9.

(b) The direct distance QR is \sqrt{6^{2} + 3^{2} + 3^{2}} = \sqrt{54}.

4. (a) First compute ∥a∥ = \sqrt{4 + 1 + 1} = \sqrt{6} and ∥b∥ = \sqrt{1 + 1 + 1} = \sqrt{3}. The dot product is a • b = 2 − 1 + 1 = 2. If the angle between the vectors is θ,

cos θ = (a • b)/∥a∥∥b∥ = 2/\sqrt{3}\sqrt{6} = \sqrt{2}/3.

So θ = cos^{−1}(\sqrt{2}/3) (which does not simplify). In (b), the angle is simply π − θ = π − cos^{−1}(\sqrt{2}/3) = cos^{−1}(−\sqrt{2}/3).

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