MATH 2400 Final Exam Review Solutions
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1 MATH Final Eam eview olutions. Find an equation for the collection of points that are equidistant to A, 5, ) and B6,, ). AP BP + ) + y 5) + z ) 6) y ) + z + ) y y z 6z y y + + z + z + 6y z 9. Alternatively, this is the plane with normal vector AB through the midpoint of A and B.. Using a computer, graph a contour plot of f, y) y and some flow lines of the vector field F, y. What appears to be true? Prove your conjecture. Note that it appears that every flow line lies on a level curve. That is, for any flow line, given by rt), we have that f rt)) k for some constant k. Note that on d dt f rt))) f r f F y,., y Thus, f is constant on any flow line. Therefore, every flow line lies on a level curve. However, note that is a level cure of f, but the vector field is not defined on the entire line. o, not every level curve is a flow line.
2 Alternatively, we can solve the differential equation dy dy d dt dy d + y d dt dy + y d d y ) d y + k f, y) y k. y. Find an equation of the plane through the points,, ),,, 5 ), and,, ). If we compute the vector between successive points, we get v,,, v,,. Then î ĵ ĵ v v,, 5 o, one form of the plane would be + y ) + 5 z + ).. For each of the following its, calculate the it if it eits, otherwise show the it does not eist. a) b) sin y,y),) + y sin y,y),) + y sin y,y),) y,y),) r r cos θ sin θ + sin θ r r sin θ + sin θ r r. Thus,,y),) sin y,y),) + y. y + y 6 y + y r r cos θ sin θ r cos θ + r sin θ y : y :,y),),y),) y + y 6 y + y 6,),) y,y),) + 6 y y y ) + y 6 y y 6 y 6 y. c) y Thus, does not eist.,y),) + y6 yz,y,z),,) + y + z yz,y,z),,) + y + z ρ cos θ sin θ sin φ cos φ ρ ρ ρ ρ yz Thus,,y,z),,) + y + z.
3 5. Two tugs are pulling a boat, one pulling in the direction 6 North of East and is half as strong as the other tug. In what direction should the stronger tug pull in order for the boat to move due East? If the weaker tug is moving at a speed of mph, at what speed is the boat traveling? If unattached, the weaker tug would be moving at a speed of sin6 ) 5 mph North. If the stronger tug is pulling at an angle θ outh of East, its unattached vertical speed would be sin θ mph. ince we need the boat to travel due East, we have sin θ 5, and so, ) θ sin Then, the horizontal speed of the boat, which is the combined )) ) horizontal speed of the tugs, would be cos6 ) + cos sin + ) 5 + ).77 mph. 6. Find the acute angle between two diagonals of a cube. The four vectors along the diagonals are,,,,,,,,,,,. Note that the dot product between any two of these vectors would be ±, and the length of each vector is ). o, the acute angle measurement is cos cos ). y) 7. Let f, y) esin + y ). ompute f, ). Hint: There is an easy way. / e Let g) f, ) + ). o, f, ) g ). 8. Find the tangent plane to the following surfaces at the given point: a) z e y sin y at π, π, ). z e y sin y, ey sin y + e y cos y z π, π ) e ), +, o, the tangent plane is z π) ) y π y. b) ru, v) u, u v, v at, y, z),, ). A plane is defined by a normal vector and a point. ecall r u and r v are tangent to the surface, and so, r u r v is perpendicular to the surface. Note r u u,,, r v, v, v v,,, and so, r u r v v, u, u. Now, how does, y, z),, ) correspond to u, v)? z gives us u v, and so u ±, v ±. Also, y u v u, which gives us u. o, a normal vector is,,, giving us a tangent plane ) + y + ) + z ), or + y + z. 9. Two legs of a right triangle are measured at 8cm and 5cm, each with a maimum error of.cm. Estimate the maimum error in computing the area and the hypotenuse.
4 A y da y d + d dy A + dy ) y A A + y ) A y + y ) y H + y dh H. 8)5) 8 +. ). cm 5 + y d + y + y dy d + y dy) H + y y) H H + y y ) 6 8.) + 5.)) cm.. Parameterize the line that is tangent to z + y and + y + z 9 at the point,, ). ecall that the D-gradient of a surface gives a vector perpendicular to the surface. If we cross the two gradients, we will get a vector that is perpendicular to both gradients, and so, tangent to both surfaces. The line starting at our point, pointing along the vector, will be the line we want.. alculate z + y z), y,,, + y + z 9) 8, y, z 8,,,,, y,,, 5, 8, 6 and z y for a) the surface defined by + z sinyz). rt),, + t 5, 8, 6. d) d + z sin yz) ) d + sinyz) dz + cosyz) yz d + z dy + y dz) sinyz) y cosyz)) dz + yz cosyz)) d + z cosyz) dy dz z z d + y dy + yz cosyz) sinyz) + y cosyz) d z cosyz) sinyz) + y cosyz) dy b) z y e t dt. z ey ) y z y ey ) y. A function f, y) is called homogeneous of degree n if for every t >, ft, ty) t n f, y). how that if f, y) is homogeneous of degree n, then f + y f y nf. t n f, y) ft, ty) d dt tn f, y)) d ft, ty)) dt nt n f, y) f t, ty) + f y t, ty) y t : nf, y) f, y) + f y, y) y.
5 . Let z f + ln y). alculate z, z y, and z y. z f + ln y ) z f + ln y ) + f + ln y ) z y y f + ln y ) z yy y f + ln y ) + y f + ln y ) z y z y y f + ln y ). ompute the Taylor series centered at,) of f, y) convergence, and on what set does the series converge? y + y ) + y ) n n. What is the radius of y The series will converge where + y + y <. This is the open disk of radius, centered at the origin. Note: You should now be more comfortable with the terminology radius of convergence. { y if, y), ) 5. Let f, y) +y. Is f continuous? Differentiable? if, y), ) Note that away from the origin, f is a quotient of functions that are differentiable, hence, f is differentiable ecept possibly at,).,y),) o, f is continuous at,). f, y) r cos θ sin θ r r r cos θ sin θ r f, ). r r f, ) h fh, ) f, ) h h h h + h h h. h imilarly, f y, ). o, the best linear approimation to f at,) is L, y) ) + y ) +. o, E, y) f, y) L, y) f, y).,y),) E, y) + y,y),) Thus, f is not differentiable at,). y + y r cos θ sin θ r r 6. Find and classify all critical points of f, y) y + y. f y f y y + f 8 f y y f yy r cos θ sin θ cos θ sin θ θ π ±, ± ) D, y) 8) ) y) 6 y < for, y), ) o, ±, ±) are saddle points..
6 7. Find the absolute ma/min of f, y) + y over the disk + y. Note that f is a continuous function, restricted to a compact set. o, we are guaranteed that f will attain a global ma and min on the closed disk. Furthermore, they must occur at either critical points on the interior, or on the boundary. Interior: f f y y ), We shall apply the method of Lagrange multipliers to the boundary. f, y λ, y λ y y y ± y y ± 5 ± ) ), < f, ) < f, ) 6 < f ) 5, ± 8. hallenging!) onsider the ellipsoid 9 + y + z with a point P on the surface in 5 the first octant specifically, where each coordinate is greater than zero). Then the coordinate planes and the tangent plane at P define a tetrahedron in the first octant. Find the point P that will minimize the volume of the resulting tetrahedron. First, an outline of the solution. We will find the tangent plane at an arbitrary point, and find the volume of the resulting tetrahedron. This then defines a volume function, dependent on the base point. From there, we will apply the method of Lagrange multipliers to the volume function in order to find the minimum. onsider the point P a, b, c) with a, b, c >. ince have that the tangent plane to the surface at P is 9 + y + z 5 ) 9, y, 5 z, we 9 a a) + by b) cz c) [ 5 a 9 + b y + c )] a 5 z 9 + b + c 5 9 a + y b + z 5 c is abc 6. ecall: For a, b, c >, the volume in the first octant under the plane a + y b + z c 9 a o, the volume of our tetrahedron is 5 b c 5. Thus, for any point P, y, z) satisfying 6 abc 9 + y + z 5 and, y, z >, the volume of the resulting tetrahedron is V, y, z) 5 yz.
7 Notice that we are not working over a closed set, however, as, y or z approach, that volume approaches infinity. o, if we restrict the value to say + y + z and, y, z. 9 5 we are only removing points with a corresponding larger volume, and so, will not change the minimum should it eist). However, V is continuous on our new set, which is compact. o, we are guaranteed a minimum on that set. To find the minimum, we shall proceed by Lagrange multipliers. ) V, y, z) λ 9 + y + z 5 V, y, λ z 9, y 5, z V λ 9 y z 5 ubstituting into our constraint, we get 9. Evaluate the following integrals: 9 P, y, z),, ) 5. a) 6 y e d dy b) 6 y e d dy e dy d e d [ ] e e6. y da where is the region in the first quadrant bounded by y 6, y and c) y dy d d. z dv, where G is the solid in the first octant bounded by + y and y + z. d) G r sin θ r cos θ r d dr dθ 6 cos θ sin θ cos θ dθ. r cos θ r sin θ) dr dθ sin y) cos + y) da where is the region bounded by y, y, and + y π. u + v y v u y y + y π u v u v π, y) u, v)
8 sin y) cos + y) da v sin u du dv + cos v cos v dv ln + ) π. e) f) + y dy d + y π +y + y + z ) dz dy d. + y dy d. r r dr dθ π. π 6 ρ ) ρ sin φ dρ dφ dθ π π ) [ ] [ cos φ] π 6 9 ρ9 5π ). 7. Hard) Three identical cylinders with radii intersect at the same point at right angles. Find the volume of their intersection. onsider the cylinders + y, + z, y + z. Using symmetry, we can restrict out attention to z, y, giving us 6 y +y da r r cos θ dr dθ [ r cos θ ) ] cos θ cos θ sin θ + cos θ dθ sec ) cos θ sin θ dθ cos θ sec θ sin θ ) cos θ + sin θ dθ [ tan θ ] π cos θ cos θ ) + + ) 8. Find the surface area of the portion of the cylinder + y 9 above the y-plane, and below + y + z. ). dθ
9 rθ, z) cos θ, sin θ, z r θ sin θ, cos θ, r z,, r θ r z cos θ, sin θ, 9 cos θ + 9 sin θ + cos θ sin θ) d dz dθ π. cos θ sin θ dθ... For the following problems, all closed curves are oriented counterclockwise when viewed from above, and surfaces are oriented outward/upward unless other wise stated. d + y dy + z dz, : rt) sin t, cos t, t, t π. d + y dy + z dz sin t cos t + sin t cos t sin t) + t ) t) dt t 5 dt π6. z cos z) d+z dy + cos z) + y ) dz, where is the intersection of z +y +5 and y from to. z cos z) d + z dy + cos z) + y ) dz,,),, ) yz + sin z) ) d r )) + sin) ) ) sin) y ) d + + ) dy where is the boundary of the region enclosed by y and y y ) d + + ) dy Green s y) dy d + 6 d 7. y,, y d r where is the intersection of z y and + y. Let be the surface z y over + y, which will have boundary curve.
10 y,, y d r tokes curl y,, y d, y,, y, da 6. +y r r dr dθ π. y, yz, z d r where is the triangle,,),,,),,,). Let be the flat interior of, given by + y + z, y. y, yz, z d r tokes curl y, yz, z d y, z,,, da + y + z) da da 7., y, + y d r where is the boundary of the portion of the paraboloid z y in the first octant. Let be the portion of z y inside., y, + y d r tokes curl, y, + y d y,,, y, da da 8. curl sin z,, z + tan y) d where is the portion of z 9 y inside + y. By either applying tokes twice or divergence theorem, any surface with same orientation and boundary as will have the same surface integral. Let be the disk z 5 with + y.
11 curl sin z,, z + tan y) d Divergence +y +y curl sin z,, z + tan y) d + y, y + y,,, da da π. 9. curl e yz, y e z, z e y d where is the top half of the sphere + y + z a. Note that, the disk z with + y a, has the same boundary curve as. curl e yz, y e z, z e y d curl e yz, y e z, z e y d Divergence curl e yz, y e z, z e y,, da +y a y e z e yz) da. Find the flu of F through the surface. +y a z y e e ) da +y a da. +y a a) F ze y, ze y, z, is the portion of + y + z in the first octant, oriented downward. F d ze y, ze y, z,, dy d ze y +ze y z dy d + + y dy d ) + ) d 6. b) F, y, z, is the upper half of the sphere + y + z a. F d r r r d r d a d a πa πa. c) F, z, z, is the boundary of the solid bounded by z y and z.
12 F d Divergence +y r π 6 π. y div F dz da + z)r dz dr dθ r r ) + r ) ) dr d) F + sinyz), y e z, z, is the boundary of the solid bounded by + y, + z and z. F d Divergence +y + + z dz da + ) ) + ) da +y +y π. + 6 da r cos θ r cos θ + 6r dr dθ cos θ 8 cos θ + dθ e) F, 5,, is the portion of the cone z + y inside + y. What if were instead the portion of z + y inside + y? Notice that the cone and the paraboloid can both be continuously deformed to the disk,, : z, + y and div F, so the flu through all three surfaces will be equal. F d +y, 5,,, da +y da π.. For each of the following, determine if the statement is true or false. If true, make sure you can prove it or eplain why. if false, give a countereample. a) For every pair of differentiable functions of one variable, f and g, the line integral f) d + gy) dy is path independent. ince f and g are differentiable, they have antiderivatives, F, G. ince F ) + Gy)) f), gy), then f) d + gy) dy is path independent. b) F y, is an eample of a vector field. F y, z has three inputs, but only two outputs, and so, F cannot be a vector field.
13 c) + y d y dy is path independent. + y + y, y + y,,, y curl + y ) y + y ). o, + y, y is not conservative. Thus, + y + y d y dy is not path + y independent. d) If F d r for a simple closed curve, then F is conservative. onsider F,, y. Then curl F y,,, so F is not conservative. However, for : rt) cos θ, sin θ,, t π, a simple closed curve, F d r F rt)) r t) dt e) If f f then f y d f dy is path independent. Let be a simple closed curve. r t) dt. f y d f dy Green s f ) y f y) da f + f yy da f da da. f) There is a vector field F such that curl F, yz, z. ecall that for any vector field F, divcurl F. ince div, yz, z +z+, then, yz, z is not the curl of any vector field.
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