1 Problem 1. Consider the vector function r(t) = 〈sintcost,cos 2 t,sint ...

1Solution of Test 2: Math 211 - Multivariate Calculus- Spring 2003Problem 1. Consider the vector function r(t) = 〈sin t cos t, cos 2 t, sin t〉. Show that its curve liesin the intersection of the sphere of center (0, 0, 0) and radius 1, and the surface y + z 2 = 1.SOLUTION:It is enough to show that the components of the vector r(t) satisfy the equations of surfaces. Indeed,with x = sin t cos t, y = cos 2 t, z = sin t we have:y + z 2 = cos 2 t + sin 2 t = 1 ⇐⇒ the curve lies on the surface y + z 2 = 1.x 2 + y 2 + z 2 = (sin t cos t) 2 + (cos 2 t) 2 + (sin t) 2 = sin 2 t cos 2 t + cos 2 t cos 2 t + sin 2 t= cos 2 t(sin 2 t + cos 2 t) + sin 2 t = cos 2 t + sin 2 t = 1 ⇐⇒ the curve lies on the sphere x 2 + y 2 + z 2 = 1.Problem 2. Give a parametric representation (a vector function) for the curve of the intersectionof the paraboloid z = x 2 + y 2 and the plane x − 2y = 0.SOLUTION:The equation of the plane implies in particular that x = 2y.Then by substitution on the equation of the paraboloid we have: z = x 2 +y 2 = (2y) 2 +y 2 = 4y 2 +y 2 = 5y 2Therefore, if we put y = t, then a parametric representation of the curve of intersection is:x = 2t, y = t, z = 5t 2 or by the vector function I(t) = 〈2t, t, 5t 2 〉Problem 3. A Bat is trying to get out of a deep cave along a path with coordinates given bythe space curve r(t) = 〈3 cos(t 2 ), 3 sin(t 2 ), 4t 2 〉; (t in minutes and the coordinates in yards).1. Find the equation of the projection onto the xy-plane, and give a rough sketch.2. Find the equation of the projection onto the zy-plane, and give a rough sketch.3. Give the speed of the Bat at any time t.4. Describe completely the path of this Bat: curve, direction and variation of the speed.5. What is the change in the Bat’s altitude from t = 0 to t = 5 ?6. What is the total distance travelled by the Bat from t = 0 to t = 5? (Hint: Arc Length ).SOLUTION:1. For the projection of this curve onto the xy-plane, we look for the relationship between x and y only.Since x = 3 cos(t 2 ) and y = 3 sin(t 2 ), then clearly x 2 + y 2 = 9 , which is the equation of the circleon the xy-plane with center (0, 0) and radius 3.2. For the projection of this curve onto the zy-plane, we look for the relationship between z and y only.Since z = 4t 2 ⇒ t 2 = z 4 and y = 3 sin(t2 ), then y = 3 sin( z 4) , which is the equation of the sinefunction along the z-axis with period 8π.3. The velocity V (t) = r ′ (t) = 〈−6t sin t 2 , 6t cos t 2 , 8t〉. So, the speed v(t) = |V (t)| is:√(6t) 2 sin 2 (t 2 ) + (−6t) 2 cos 2 (t 2 ) + (8t) 2 = √ 36t 2 (sin 2 (t 2 ) + cos 2 (t 2 )) + 64t 2 = √ 36t 2 + 64t 2 = 10t .4. The Bat is flying along a helix with radius 3, upward (because z = 4t 2 ) and counter-clockwise (becausex = 3 cos t 2 , y = 3 sin t 2 ) with an increasing speed (because speed = 10t gets larger as t gets larger).5. The altitude is just the change in the z coordinate. We have:t = 0 =⇒ z = 4(0) 2 = 0 and t = 5 =⇒ z = 4(5 2 ) = 100.100 − 0 = 100 yards .Therefore the change of altitude is

26. The distance travelled is exactly the arc length between t = 0 to t = 5, which is given by:∫ 50∣∣r ′ (t) ∣ dt =∫ 5010tdt = [10 t2 2 ]5 0 = 125Problem 4. Consider r(t) = 〈3t + 2, t 2 , 4t〉. Find each of the following at the point P(2, 0, 0):1. The unit Tangent vector T at P2. The Normal vector N at P3. The equation of the Normal Plane at P4. The curvature K at P .SOLUTION:First, the point P corresponds to t = 0.r(t) = 〈3t + 2, t 2 , 4t〉, then r ′ (t) = 〈3, 2t, 4〉 and |r ′ (t)| = √ 3 2 + (2t) 2 + 4 2 = √ 25 + 4t 21. T (t) = r′ (t)|r ′ (t)| = 1√ 〈3, 2t, 4〉. At P , T = T (0) = 1 〈3, 0, 4〉25 + 4t 2 52. The normal vector is given by N = T ′ (0)|T ′ (0)| . First we need to first T ′ (t).T ′ 1(t) = ( √ 〈3, 2t, 25 + 4t 2 4〉)′ = −12 4t(25 + 4t2 ) −321〈3, 2t, 4〉 + √ 〈0, 2, 0〉. So, for t = 0 we have:25 + 4t 2T ′ (0) = 0 + √ 1 〈0, 2, 〉 = 〈0, 2 25 5 , 0〉 and ∣ ∣T ′ (0) ∣ = 2 5 . Therefore N = T ′ (0)|T ′ = 〈0, 1, 0〉(0)|3. The normal plane has T = 1 5〈3, 0, 4〉 as a normal vector and passes through the point P (2, 0, 0).1therefore its equation os given by:5[3(x − 2) + 0(y − 0) + 4(z − 0)] = 0 =⇒ 3x + 4z = 64. The curvature is given by K = |T ′ 2(0)||r ′ (0)| = 55 = 2 25 .Problem 5. Consider the function f(x, y) = √ 8 − x 2 − y 2 + 2x.1. Find and sketch the domain of this function. (Hint: completing the square)2. Describe in words the graph of this function3. Describe and give a rough sketch of the level curves for k = 0, 1, 2, 3.SOLUTION:1. f(x, y) exists if 8−x 2 −y 2 +2x ≤ 0. Since 8−x 2 −y 2 +2x = 0 ⇐⇒ 8 = x 2 +y 2 −2x ⇐⇒ 9 = (x−1) 2 +y 2 ,which is the circle on the xy-plane of center (1, 0) and radius 3, then the domain is the inside of thiscircle with the circle included.2. Since z = √ 8 − x 2 − y 2 + 2x, then z 2 = 8−x 2 −y 2 +2x ⇒ x 2 +y 2 −2x+z 2 = 8 ⇒ (x−1) 2 +y 2 +z 2 = 9,which is (because z is positive) the upper-half of the sphere of center(1, 0, 0) and radius 3.3. With z = k we have (x − 1) 2 + y 2 + k 2 = 9 =⇒ (x − 1) 2 + y 2 = 9 − k 2k = 0 =⇒ (x − 1) 2 + y 2 = 9 =⇒ circle of center (1, 0) and radius 3.k = 1 =⇒ (x − 1) 2 + y 2 = 8 =⇒ circle of center (1, 0) and radius √ 8.k = 2 =⇒ (x − 1) 2 + y 2 = 5 =⇒ circle of center (1, 0) and radius √ 5.k = 3 =⇒ (x − 1) 2 + y 2 = 0 =⇒ the point (1, 0) only.