Multivariable Calculus Midterm I Solutions

Math 211, Multivariable Calculus, Fall 2011 Midterm I Solutions 1. Find the equation of the plane passing through the points (1, 1, 0), (0, 2, 2) and (0, 1, 4). Solution: If the points are labelled A, B and C, then we have ~ = h−1, 1, 2i, AB ~ = h−1, 0, 4i. AC A vector perpendicular to the plane containing these points is therefore given by n = h−1, 1, 2i × h−1, 0, 4i = h4, 2, 1i The equation of the plane is therefore given by 4(x − x0 ) + 2(y − y0 ) + (z − z0 ) = 0 where (x0 , y0 , z0 ) is a point on the plane. Taking the point (1, 1, 0), this gives us 4(x − 1) + 2(y − 1) + z = 0 or 4x + 2y + z = 6. 2. Find the equation of the tangent line to the curve r(t) = ht2 − 2t, t2 + 4ti at the point (3, −3). Solution: We have r0 (t) = h2t − 2, 2t + 4i. We are interested in the tangent vector at the t-value for which r(t) = h3, −3i, that is t2 − 2t = 3, t2 + 4t = −3. Subtracting these two equations we get −6t = 6 and so t = −1. Therefore, a tangent vector to the curve at (3, −3) is given by r0 (−1) = h−4, 2i. The equation of the tangent line is therefore l(u) = h3, −3i + uh−4, 2i. 3. Find the arc length of the curve r(t) = hcos t, sin t, t3 − ti between the points (1, 0, 0) and (−1, 0, π − π 3 ). Solution: We need to find the t-values for the endpoints of the arc we want to find the length of. If r(t) = h1, 0, 0i, then cos(t) = 1, sin(t) = 0, t3 − t = 0. The first equation tells us that t = 2nπ for some integer n. But the solutions to the third equation are t = −1, 0, 1 so we must have t = 0. For the other endpoint we want r(t) = h−1, 0, π − π 3 i, that is cos(t) = −1, sin(t) = 0, t3 − t = π − π 3 . The second equation tells us that t = nπ for some integer n. Clearly t = −π also satisfies the other equations. (One could then show that the other solutions to the third equation are not multiples of π so this is the only solution. But the question implied that there is a unique t-value for the given point so you didn’t need to do that.) To find the arc length, we need to first find the speed: r0 (t) = h− sin(t), cos(t), 3t2 − 1i and so q sin2 t + cos2 t + (3t2 − 1)2 p = 1 + (3t2 − 1)2 √ = 9t4 − 6t2 + 2. |r0 (t)| = Therefore, the arc length between the given endpoints is Z 0√ 9t4 − 6t2 + 2 dt. −π Note that if you integrate from t = 0 to t = −π, you would get a negative answer. The arc length itself should always be positive (as opposed to the values of the arc length function which could be negative). Therefore you would need to take the absolute value of that integral to get the right answer. 4. (a) (9 points) Find the curvature of the curve r(t) = h1 − t2 , 3 + t2 , 2t2 i. (b) (1 point) What does your answer to part (a) tell you about this curve? Solution: (a) We have r0 (t) = h−2t, 2t, 4ti and r00 (t) = h−2, 2, 4i. Therefore r0 (t) × r00 (t) = h8t − 8t, −8t + 8t, −4t + 4ti = h0, 0, 0i. Therefore the curvature is κ(t) = |r0 (t) × r00 (t)| = 0. |r0 (t)3 (Strictly speaking, this is only true for t 6= 0. At t = 0, the curvature is not defined.) (b) This curve has zero curvature everywhere and so is a straight line. (Strictly speaking, we can deduce that each section on which r0 (t) is nonzero is a straight line. It is possible for the curve to change direction at a point where r0 (t) = 0. You could then get a corner between two straight line segments. This curve is actually a half-line: it approaches the point (1, 3, 0), stops there and then goes back in the direction it came.) 5. (a) (5 points) Find the point of intersection of the curve r(t) = he2t , 2et + 1, et i with the plane P given by 2x + y − 2z = 3. (b) (5 points) Is the tangent line to the curve r(t), at the point of intersection you found in part (a), perpendicular to the plane P , or not? Solution: (a) For a point of intersection we must have 2(e2t ) + (2et + 1) − 2(et ) = 3 so 2e2t + 1 = 3 so 2e2t = 2 so e2t = 1 so 2t = 0 so t = 0. Therefore, the point of intersection is r(0) = h1, 3, 1i. (b) We have r0 (t) = h2e2t , 2et , et i so a tangent vector at the point of intersection is given by r0 (0) = h2, 2, 1i. For the tangent line to be perpendicular to the plane P , we would need this vector to be parallel to the normal vector n = h2, 1, −2i to P . It’s enough to see that r0 (0) is not a constant multiple of n (that’s the only way vectors can be parallel). If this were true we would have h2, 2, 1i = ch2, 1, −2i or 2 = 2c, 2 = c, 1 = −2c which is impossible. Alternatively, you could take the cross product of r0 (0) and n. Since that cross product is not the zero vector, they are not parallel. Either way, we see that the tangent line is not perpendicular to the plane.