Calculus III (MAC )

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1 Calculus III (MAC2-) Test (25/9/7) Name (PRINT): Please show your work. An answer with no work receives no credit. You may use the back of a page if you need more space for a problem. You may not use any calculators. Page Points Score Total: Page of 8

2 Calculus III (MAC2-) Page 2 of 8. Describe each of the following regions in set notation (with inequalities). (a) ( points) The region between the yz-plane and the plane x = 2. Solution: The yz-plane is the same as the plane x =. So, { (x, y, z) R x 2 }. (b) ( points) The surface and inside of a cylinder that lies on or below the plane z = and on or above the disk in the xy-plane with center the origin and radius. Solution: The inequality x 2 + y 2 2 gives the surface and inside of an unbounded cylinder with axis through the origin and radius. Restricting to z gives us region in question: { (x, y, z) R z, x 2 + y 2 9 }. 2. (5 points) Find the center and radius of the sphere given by 2x 2 + 2y 2 + 2z 2 = 4x 6z +. Solution: We have 2x 2 4x + 2y 2 + 2z 2 + 6z =, x 2 2x + y 2 + z 2 + 8z = /2, x 2 2x + + y 2 + z 2 + 8z + 6 = / , (x ) 2 + (y ) 2 + (z + 4) 2 = 5/2, which is a sphere with center at (,, 4) and radius 5/2.. (5 points) If the vector v lies in the first quadrant and makes an angle π/ with the positive x-axis and v = 4, represent v in component form (in the standard basis for R 2 ). Solution: With θ = π/, we have v = v cos θi + v sin θj = 4(/2)i + 4( /2)j = 2(i + j).

3 Calculus III (MAC2-) Page of 8 4. Circle True or False. (a) ( points) TRUE False u v = u v if and only if u is a scalar multiple of v or vice versa. Solution: If either u or v is zero, the statement is obviously true. Otherwise, since u v = u v cos θ, for u v = u v we need cos θ =, implying θ = (modulo π) which means u and v are parallel; i.e., one is a scalar multiple of the other. (b) (2 points) TRUE False u (v w) = (u v) w for all u, v, and w in R. Solution: Note that u (v w) := det(u, v, w) and that each time we swap two columns the determinant changes sign; i.e., (u v) w = w (u v) = det(w, u, v) = ( ) det(u, w, v) = ( )( ) det(u, v, w) = u (v w). (c) (2 points) True FALSE For a differentiable vector function, the derivative of the length is the length of the derivative. Solution: The vector function could have constant length without being constant. For example, the vector function representing (motion on) a curve on a sphere has constant length (and hence the derivative of the length is zero) but the vector is changing direction and its derivative is not zero. (d) (2 points) True FALSE i + 2j + represents a vector in R. Solution: Addition of a scalar () and a non-scalar vector (i + 2j) is an undefined operation. (e) (2 points) TRUE False u v 2 + u v 2 = u 2 v 2 for all u and v in R. Solution: u v 2 + u v 2 = u 2 v 2 cos 2 θ + u 2 v 2 sin 2 θ = u 2 v 2. (f) (2 points) True FALSE Two lines in R either intersect or are parallel. Solution: This would be True in R 2. But in R two lines may neither intersect nor be parallel. Such two lines are called skew lines. (g) (2 points) True FALSE u (v w) is orthogonal to v and w. Solution: u (v w) is orthogonal to u and (v w). But the latter is orthogonal to v and w (and out of their plane). So, u (v w) is in the plane of v and w.

4 Calculus III (MAC2-) Page 4 of 8 5. ( points) Consider the vectors a = i + 4j and b = 5i + 8j. Represent b as the sum of two vectors one parallel and the other orthogonal to a. Solution: Method : Since proj b a is the part of b that is parallel to a, what is left of b, namely b proj b a, is going to be orthogonal to a. So, b = proj b a +(b proj b a) is the representation (decomposition) we are looking for. Since a 2 = = 25 and b a = ()(5) +(4)(8) = 47, we have proj b a a a = b u a u a = b a a = b a a a = 47 (i + 4j) 2 25 and b proj b a = (5i + 8j) 47 (i + 4j) = i j = 4 ( 4i + j). 25 Method 2: We can also solve this problem [ ] using linear algebra[ alone. ] Since we are in R 2, 4 it is easy to see that the vector v = is orthogonal to a =, since v a =. So, we [ 4 ] [ ] [ ] 4 5 are looking for the scalars x and y such that xa + yv = b, i.e., x + y =, which 4 8 is [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 4 x x x 5 = = =, 4 y y 2 25 y 47 9(47/25) 5 giving x = 47/25 and y = = (47/25) 5 = So, [ ] 5 = 47 [ ] + 4 [ ]

5 Calculus III (MAC2-) Page 5 of 8 6. ( points) Find the volume of the parallelepiped with adjacent edges P Q, P R, and P S, where P (2,, ), Q(, 5, ), R(, 4, ), and S(, 6, ). Solution: The edges are a = P Q = 4, b = P R =, and c = P S = 5. So, the volume is given by the scalar triple product a (b c) = det(a, b, c) = 8, since 4 5 = ( 5) 4( ) + ( 5 ) = = ( points) Find the distance between the parallel planes 6z = 4y 2x and 9z = x + 6y. Solution: We pick two arbitrary points P = (,, ) and Q = (,, /9) on the first and second plane, respectively. The magnitude of the component of a = P Q = along the /9 normal vector N = 2 (which is the same for both planes) gives the distance between the planes. Since N = = 4 and a N = /9 = /, we have comp a N = a u N = a N N = / = 4 4.

6 Calculus III (MAC2-) Page 6 of 8 8. ( points) Find symmetric equations for the line of intersection of the planes given by x 2y 2z = 5 and 2x + y + z = 6. Solution: Method : We can solve this problem using linear algebra alone. Multiplying the second equation by 2, we have the following system of two linear equations in three unknowns, which we can solve, to get infinitely many solutions, namely our line of intersection, using Gaussian elimination (i.e., forward elimination and back substitution): { x 2y 2z = 5 4x + 2y + 2z = 2 { x 2y 2z = 5 7x = 7 { 2y 2z = 8 x = { y + z = 4 x =. So, symmetric equations for the line of intersection are x = and z = 4 y, i.e., x =, y 4 = z. Parametric equations for this line are x = + t, y 4 = ( )t, and z = ()t, i.e., x =, y = 4 t, and z = t. Note that this is a line in the plane x =. Method 2: We can also use geometry combined with algebra. The line of intersection is in 2 both planes, so it s parallel to the cross product of their normals N = 2 and N 2 = : 2 i 2 N N 2 = j 2 = ( 2 ( 2))i ( ( 4))j + ( ( 4))k = 7(j k) = 7. k 2 We can find{ a point that is on{ both planes by looking { at their intersection on the z = x 2y = 5 x 2y = 5 x 2y = 5 plane, e.g.: 2x + y = 6 4x + 2y = 2, which gives x = and 7x = 7 y = 4. So, x = ()t, y 4 = ()t, z = ( )t, or, equivalently, x =, y 4 = z.

7 Calculus III (MAC2-) Page 7 of 8 9. ( points) Sketch the surface given by z = x 2 + 2y 2. (Your sketch needs to be correct qualitatively; it need not be to scale.) Solution: In the xz-plane (y = ), we have z = x 2, a parabola. In the yz-plane (x = ), we have z = 2y 2, a steeper parabola. In the plane z = 4, we have x 2 + 2y 2 = 4, i.e., x 2 /4 + y 2 /2 =, an ellipse with major axis along the x-axis and semi-length 2 and minor axis along the y-axis and semi-length 2. The surface is an elliptic paraboloid.

8 Calculus III (MAC2-) Page 8 of 8. (9 points) Find the domain of the vector function r(t) = t 2 t + 2 i + sin t j + ln(25 t 2 )k. t Solution: The functions giving the i, j, and k components have domains { t R t 2 }, { t R t }, and { t R 25 t 2 > } = { t R 5 < t < 5 }, respectively. Intersecting these sets, the domain of r becomes ( 5, 2) ( 2, ) (, 5).. ( points) Let r(t) = t j + 2 te2t k. Evaluate the integral Hint: Integration by parts and partial fractions may help. /2 r(t) dt if it exists. Solution: We have and /2 t 2 dt = /2 /2 ( /2 ( t)( + t) dt = t + /2 ) dt + t = ( 2 ln t + 2 ) /2 ln + t /2 te 2t dt = /2 2 te2t = (ln(/2) ln(/2)) 2 = 2 (ln ln 2 (ln ln 2)) = 2 ln /2 = /2 2 te2t /2 4 e2t = 4 e 4 (e ) = 4. 2 e2t dt So, /2 r(t) dt = /2 t 2 dt j + /2 te 2t dt k = 2 ln j + 4 k.