110.202 CALCULUS III 23 February 2000

Solutions to First Examination

11:00 a.m. Edition

11. The relevant values of t are 0 ≤ t ≤ 3 (look at the k-component of r(t)). Differentiate,

r′(t) = t^{2}i + 2tj + 2k.

Its norm has been cooked to work out nicely,

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

Then the arc length is

∫_{0}^{3}(t^{2} + 2)dt = [t^{3}/3 + 2t]_{0}^{3} = 9 + 6 = 15.

12. Call these points P_{1}, P_{2} and P_{3}. Then the difference vectors \overrightarrow{P_{1}P_{2}} = i + j + 3k, \overrightarrow{P_{1}P_{3}} = 2i + 3j + 2k (and \overrightarrow{P_{2}P_{3}} = i + 2j − k) are parallel to the plane p. (a) A normal vector to the plane is

N = \overrightarrow{P_{1}P_{2}} × \overrightarrow{P_{1}P_{3}} = | \begin{matrix}i & j & k \\ 1 & 1 & 3 \\ 2 & 3 & 2 \end{matrix} | = −7i + 4j + k.

(b) The equation of p is N •(r − \overrightarrow{OP_{1}}) = 0, where r = xi + yj + zk, which reduces to −7x + 4y + z = 7. (Check that the constant term makes P_{1} lie on p.) (Any nonzero multiple of this equation or N will do.)

13. (a) Compute the norms ∥a∥ = \sqrt{1 + 4 + 4} = 3 and ∥b∥ = \sqrt{1 + 1 + 9} = \sqrt{11}. The dot product is a • b = 1 + 2 + 6 = 9. So if the angle is θ,

cos θ = a • b/∥a∥∥b∥ = 9/3\sqrt{11} = 3/\sqrt{11}

and θ = cos^{−1}(3/\sqrt{11}). (b) Compute 2a−b = 2(i+2j+2k)−(i+j+3k) = i+3j+k. Its norm is \sqrt{1 + 9 + 1} = \sqrt{11}.

14. At P, t = 1 (from the k-component). (a) Differentiate, to get

r′(t) = (2t − 1)i + (−3t^{2} + 1)j + 2k.

Then r′(1) = i − 2j + 2k, with norm ∥r′(1)∥ = \sqrt{1 + 4 + 4} = \sqrt{9} = 3. So the unit tangent vector at P is T = r′(1)/3 = (1/3)i − (2/3)j + (2/3)k. (b) The tangent line to E at P = i + j + k is

R(u) = i + j + k + uT = i + j + k + u((1/3)i − (2/3)j + (2/3)k),

or we can replace T here by r′(1).

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