Math 21a Final Exam Solutions Spring, 2009

Transcription

1 Math a Final Eam olutions pring, 9 (5 points) Indicate whether the following statements are True or False b circling the appropriate letter No justifications are required T F The (vector) projection of 3, 7, 9 onto,, 3 is equal to the (vector) projection of 3, 7, 9 onto 5,, 5 olution: This is True, since 5,, 5 5,, 3 T F The angle between the vectors, 3, 7 and 4, 6, is obtuse (greater than π ) olution: This is True We calculate the cosine of the angle θ between u, 3, 7 and v 4, 6, b cos(θ) u v ()( 4) + ( 3)(6) + (7)() 5 u v u v u v < ince cos(θ) <, the angle θ must be obtuse T F If F P, Q is a vector field with the propert that Q P, then Green s Theorem implies that F dr for an curve olution: This is False Green s Theorem implies that this integral is zero for an closed path, but not for an path at all T F The tangent plane to + 4z at the point (,, ) is + 4z olution: This is True The normal to the level surface g(,, z) + 4z at the point (,, ) is g(,, ) ince g,, 8z, the normal is n g(,, ), 4, 8 The tangent plane is therefore, 4, 8,, z or 4 + 8z or + 4z This is the equation given T F The two curves, parameterized b r (t) t, t, and, parameterized b r (t) 4t, t, both pass through the point (, ) At this point, the curvature of is greater than the curvature of olution: This is False Here is a graph showing these two parameterized curves:, t, t Intuitivel, the curve is curving more at the origin, so has a greater curvature Here s a rigorous wa of thinking of this The osculating circle for at the origin is much larger than the same circle for But the radius for the osculating circle is the inverse of the curvature κ, so this is saing that κ > κ Thus κ < κ Put another wa, the curvature for is less than that of, not greater than

2 Math a Final Eam olutions pring, 9 (4 points) Match the sketch of each quadric surface with the appropriate equation You do not need to justif our choices (a) + + z (b) + z (c) z (d) + z I II III IV olution: One wa to do this is to look at traces Here s a little table of traces for our four equations, together with the corresponding sketch: Equation k trace k trace z k trace Quadric (a) + + z ellipses ( ) ellipses ( ) ellipses ( z ) IV (b) + z hperbolas (all ) hperbolas (all ) ellipses (all z) II (c) z ellipses (all > ) hperbolas (all ) hperbolas (all z) I (d) + z hperbolas (all ) hperbolas (all ) ellipses (all z > ) III The difference between II and III comes in the trace at z in equation (d) There the ellipse turns into onl a point Another difference is in the trace For (b) this is a hperbola but in (d) it produces two lines z ± (a degenerate hperbola)

3 Math a Final Eam olutions pring, 9 3 (6 points) ompute the integral Hint: hange the order of integration 3 3 d d + 4 olution: Here s a picture of the region over which we are integrating Notice that the coordinates are bounded below b the line and above b the hperbola between and That is, we can write as {(, ) :, } To change the order of integration, we write in terms of for both curves The line is simpl and the parabola is That is, the region can also be written as Thus we can re-write the integral as {(, ) :, } 3 3 d d d d + 4 Now we integrate: d ( ) ( ) d d Now the substitution u produces du 6( ) d, so we can continue 6 u du 6 ln ( ) (ln(4) ln(3)) 6

4 Math a Final Eam olutions pring, 9 4 (7 points) Find the volume of the solid enclosed b the clinders + z and + z z z olution: There are man was to set up the integral Let s do it as a triple integral, so if E is the solid, then we want E dv We can describe our solid as E {(,, z) : +z, +z } It s hard to write bounds for z in terms of and : we know that z and z, so we would have to write something like z the smaller of and, which would be a big mess We can get around this problem b making the z integral the outer integral Then we just have to sa that z goes from to ince z and z, so and both go from z to z (That is, the z constant trace is a square of side length z ) This gives an iterated integral z z z z d d dz z z z d dz 4( z ) dz 4z 4z3 3 z z 6 3 Another approach is to split the region up into equall-sized pieces and find the volume of a single, representative piece We ll cut the region b two planes, and The advantage here is that the height (the bounds for z) in each region depend on onl (or on onl ) Here s the region projected onto the -plane: On the shaded region z varies between and, while on the unshaded regions z varies between and We ll integrate over the region (this is one-quarter of the volume of the full region) Thus the full volume of the region is Volume 4 dz da 4 dz d d 6 3

5 Math a Final Eam olutions pring, 9 5 (8 points) uppose the gradient vector of a function f(,, z) at the point (3, 4, 5) is,, (a) ( points) Find the values of the partial derivatives f, f, and f z at the point (3, 4, 5) olution: ince the gradient vector of a function f(,, z) is the vector f f, f, f z, we have that f, f, and f z at the point (3, 4, 5) (b) (3 points) Find the maimum directional derivative of f at the point (3, 4, 5) and the unit vector in the direction in which this maimum occurs olution: ecall that the directional derivative u f f u is greatest in the direction of f Thus the directional derivative is greatest when u f f,, 3 3, 3, 3 The directional derivative in the direction of u is thus u f f u,,,, 3 3 (This is, of course, simpl f ) (c) (3 points) olution: f(,, z) is or, equivalentl, If f(3, 4, 5), estimate f(3, 4, 48) using linear approimation ecall that the linearization L(,, z) near (,, z) (a, b, c) of the function L(,, z) f(a, b, c) + f(a, b, c) a, b, z c L(,, z) f(a, b, c) + f (a, b, c)( a) + f (a, b, c)( b) + f z (a, b, c)(z c) Here (a, b, c) (3, 4, 5), f(3, 4, 5),,, and f(3, 4, 5) Thus L(,, z) +,, 3, 4, z + 5 The linear approimation of f(,, z) for (,, z) near (3, 4, 5) is simpl f(,, z) L(,, z) Thus we get f(3, 4, 48) L(3, 4, 48) +,, 3 3, 4 4, ,,,,

6 Math a Final Eam olutions pring, 9 6 (8 points) Let be the part of the elliptic clinder 4 + 9z 36 ling between the planes 3 and 3 and below the plane z Here are two views of this clinder, from different perspectives: z z (a) (4 points) Write an iterated integral which gives the surface area of You need not evaluate the integral, but our integral should be simplified enough so that all that is required is integration (that is, the integrand contains no dot or cross products or even vectors) olution: ecall that the scalar surface area element d is r u r v du dv, where r(u, v) is a parameterization of the surface o all we need is a parameterization There are several was to parameterize this Perhaps the most direct is as a graph z f(, ) We solve for z and find that r(, ),, has r r, so urface Area d d 9 3 Note the negative sign in the epression for z and also the general unpleasantness of computing with this epression A nicer epression involves parameterizing the elliptical clinder using sines and cosines: r(, θ) 3 cos(θ),, sin(θ) (with 3 3 and π θ π) From this we get i j k r r θ cos(θ),, 3 sin(θ), 3 sin(θ) cos(θ) and so r r θ 9 sin (θ) + 4 cos (θ) Thus π 3 urface Area 9 sin (θ) + 4 cos (θ) d dθ (b) (4 points) π Let F(,, z), e, z Evaluate the flu integral 3 3 F d if is oriented with normal vectors pointing upward (that is, so the normal vectors have positive z component) olution: We use the second parameterization from part (a), namel r(, θ) 3 cos(θ),, sin(θ), π θ π, 3 3 From the calculation above, r r θ gives us the wrong orientation (recall that π θ π, so sin(θ) ), so we use r θ r cos(θ),, 3 sin(θ) instead Thus F d F(r(, θ)) (r θ r ) da π 3 π 3 π 3 π 3 3 cos(θ), e, sin(θ) cos(θ),, 3 sin(θ) d dθ 6 d dθ 36π

7 Math a Final Eam olutions pring, 9 7 (4 points) Let be the curve of intersection of the clinder + 4 and the plane 3++z 6 (a) ( points) Parameterize Be sure to give bounds for our parameter olution: This intersection is an ellipse It is simplest to parameterize b noticing that its projection into the -plane is simpl the circle + 4 This is parameterized b cos(t) and sin(t); the z-coordinate on the ellipse can be found b solving for z in the plane: z 6 3 Thus we end up with the parameterization r(t) cos(t), sin(t), 6 6 cos(t) 4 sin(t) with t < π (b) (3 points) Write an integral that gives the arc length of You need not evaluate our integral, but our integral should be simplified enough so that all that is required is integration (that is, the integrand contains no dot or cross products or even vectors) olution: ecall that the incremental arc length element is ds r (t) dt (If we envision r(t) as describing the path of a particle, then r (t) is the speed of this particle while ds dt is the rate of change of the distance this particle has traveled These must agree) This is simpl Arc Length π π π π sin(t), cos(t), 6 sin(t) 4 cos(t) dt 4 sin (t) + 4 cos (t) + 36 sin (t) 4 sin(t) cos(t) + 6 cos (t) dt 4 sin (t) 4 sin(t) cos(t) + cos (t) dt sin (t) 6 sin(t) cos(t) + 5 cos (t) dt (c) (4 points) A bee is fling along the curve in a room where the temperature is given b f(,, z) z What is the hottest point the bee will reach? olution: Maimize f(r(t)) cos t + sin t + cos t 8 sin t cos t + sin t The derivative of this is sin t + cos t, which is when sin t cos t ± o, the hottest point is (,, 6 + ) where f(,, 6 + ) + (The other point, ( ),, 6 where f(,, 6 ) Alternativel, use Lagrange multipliers to maimize z subject to the constraints + 4 and z 6 This gives 5, 5, λ,, + µ 3,,, so µ, and,, λ,, Therefore, ±, as before (d) (5 points) Let F(,, z), e 3, cos z Evaluate the line integral F dr if is oriented counterclockwise when viewed from above olution: Use tokes Theorem: let be the part of the plane z 6 inside the clinder + 4, oriented with upward normals Parameterize this b r(, ),, 6 3, where the points (, ) lie in, the disk + 4 Then r r 3,, (which is the correct orientation for tokes ), and curl F,, Hence curl F d curl F(r(, )) (r r ) da,, 3,, da π r r dr dθ 8π ( + ) da

8 Math a Final Eam olutions pring, 9 8 (6 points) Evaluate the line integral ( 4 + 3) d + (5 3 ) d, where is the boundar of the square with vertices (4, ), (, 4), ( 4, ), and (, 4), traversed counterclockwise olution: The simplest wa to do this is to use Green s Theorem, which sas that P d+q d (Q P ) da, where is the oriented boundar of the region In our case, this means is the solid square (as shaded, above), P 4 + 3, Q 5 3, and so Q P Thus ( 4 + 3) d + (5 3 ) d da Area() ince is a square with side length 4, the area of is 3 and so the line integral is 64 Another approach is to just compute the line integral directl I ve added labels to the picture (above), so the four line segments are now labeled as through 4 We ll do the line integrals along each curve more or less without comment, starting with a parameterization and proceeding to the line integral Along : Use, 4 t, t (with t 4), so d, d dt, dt and ( 4 + 3) d + (5 3 ) d 4 [ (4 t) 4 + 3t ] ( dt) + [ 5(4 t) t 3] dt Along : Use, t, 4 t (with t 4), so d, d dt, dt and ( 4 +3) d+(5 3 ) d 4 [ ( t) 4 + 3(4 t) ] ( dt)+ [ 5( t) (4 t) 3] ( dt) Along 3 : Use, t 4, t (with t 4), so d, d dt, dt and 3 ( 4 +3) d+(5 3 ) d 4 [ (t 4) 4 + 3( t) ] dt+ [ 5(t 4) ( t) 3] ( dt) Along 4 : Use, t, t 4 (with t 4), so d, d dt, dt and 4 ( 4 + 3) d + (5 3 ) d 4 [ t 4 + 3(t 4) ] dt + [ 5t (t 4) 3] dt The total line integral around is the sum of these four line integrals That is, ( 4 + 3) d + (5 3 ) d , as before

9 Math a Final Eam olutions pring, 9 9 ( points) (a) (5 points) Find all critical points of f(, ) , and classif each critical point as a local minimum, local maimum, or saddle point olution: A critical point of f(, ) is a point where f f, f, Here f +, +, so we have ( + ) and + From the first equation, either (and so from the second equation) or (which means ± ) Thus we have three critical points: (, ) (, ) and (±, ) To classif these critical points, we use the second derivative test Using f +, f f, and f, we get f f f 4( + ) 4 Thus we have the following classifications: (b) (3 points) ritical Point (, ) (, ) (, ) f lassification addle Local Min addle Evaluate the line integral F dr, where be the straight line path from (, ) to (, ) and F(, ) is the gradient vector field of f(, ) (from part (a)) olution: We could parameterize the curve and compute the integral directl (and we do so below) It is simpler, however, to use the fundamental theorem of line integrals: F dr f dr f( end pt ) f( start pt ) f(, ) f(, ) 7 3 The straightforward-seeming approach is much more complicated We cab parameterize this path b r(t) t, t (with t ) Using this parameterization, our line integral is F dr F(r(t)) r (t) dt t + t, t + t, dt (4t + 3t ) dt 3 (c) (3 points) Here is the gradient vector field G(, ) of another function g(, ) (ots represent zero vectors) Find all critical points of g(, ) in the region shown, and classif each critical point as a local minimum, local maimum, or saddle point olution: ecall that a critical point of f(, ) is where f f, f, Thus a critical point is where the 3 vector appears as a dot, so in this region the critical points are at (, ) (, ) and (, ) We classif these points using the meaning of the gradient ecall that the gradient points in the direction in which f increases fastest The point (, ) is a minimum because f increases when we move awa this point in an direction The point (, ) is a saddle point because f either increases or decreases as we move awa from the point, depending on the direction 3

10 Math a Final Eam olutions pring, 9 ( points) The four points (,, ), (,, ), (,, ), and (,, ) are all co-planar and form the vertices of a quadrilateral Let be the boundar of directed as shown to the right, and let F k (a) (4 points) Find the equation of the plane that contains the quadrilateral olution: The plane contains the point (,, ) and is parallel to the vectors,, and,, Thus the normal is n,,,,,,, and the plane is,,,, z or + z z,,,,,,,, (b) (3 points) alculate F dr olution: While we could parameterize the four segments of curve, it is simpler to appl tokes Theorem and compute a flu integral instead We orient the surface to be compatible with the orientation of (that is, with the downward pointing normals) A simple parameterization is r(, ),, (using the equation of the plane z found in part (a)) Then is the surface parameterized b r for points (, ) in the region, shown below: Then r r,, is precisel the cross product found in part (a) (Notice this is the wrong orientation; we ll use r r,, instead) Using curl F,,, we get via tokes Theorem: F dr curl F d,,,, d d + d d 3 (c) (3 points) alculate curl F d olution: We ve alread computed this integral, assuming that has the orientation we gave it in part (b) If is given the opposite orientation (with upward-pointing normals), this would simpl change the sign of our computation in part (b) and we d get + 3

11 Math a Final Eam olutions pring, 9 ( points) (a) (4 points) Write shown to the right, b g () a g () f(, ) d d f(, ) da, where is the region as an iterated integral of the form olution: The line has equation tan( π 3 ) 3 or 3; this intersects the circle at the point (, ) (3 cos( π 3 ), 3 sin( π 3 )) ( 3, 3 3 ) Thus the integral can be written as 3/ 9 f(, ) da f(, ) d d (b) (4 points) Write π/6 f(, ) da as an iterated integral using polar coordinates olution: The region can be described in polar coordinates as those points with r 3 and π 3 θ π Thus f(, ) da π/ 3 π/3 f(r cos(θ), r sin(θ)) r dr dθ (c) ( points) olution: ompute cos( + ) da We use polar coordinates as in part (b) We get cos( + ) da π/ 3 π/3 π/ π/3 sin(9) cos(r ) r dr dθ sin(r 3 ) π/ dθ π/3 ( π π ) π 3 sin(9) sin(9) dθ

12 Math a Final Eam olutions pring, 9 ( points) Let W be the solid tetrahedron with vertices (,, ), (,, ), (,, ), and (,, ) This tetrahedron has volume 6 (a) (4 points) Let be the surface that forms the boundar of W, oriented with the outwardpointing normal Note that has four pieces ompute the flu of the vector field F(,, z) ( 3z)i + ( 3 + )j + (tan() + z)k out of the surface That is, compute F d olution: We use the ivergence Theorem There are two hints at wh this might be appropriate First, computing the flu directl would require that we parameterize each of the four pieces of separatel econd, and perhaps even more importantl, the vector field F is complicated and difficult to involve in an integral, while div F 3 The ivergence Theorem implies that F d W div F dv W 3 dv 3 Vol(W ), as we re told the volume of W is 6 (b) (3 points) ompute the flu of curl F out of the surface ; that is, compute curl F d olution: Again we appl the ivergence Theorem This time it s even simpler, as we recall that div (curl F) for an vector field F: curl F d W div (curl F) dv W dv (c) (4 points) Let be the part of that lies in the z-plane (where ), oriented in the same wa as Find the flu of F through ; that is, compute F d olution: We parameterize b r(, z),, z where (, z) lies in the region shown below: z The normal r z r,, is the proper orientation (outward from W ), so F d 3z, 3 +, tan() + z,, dz d 3z dz d 3 ( ) d 3 ( 3 + ) d 8