234 Review Sheet 2 Solutions

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1 4 Review Sheet Solutions. Find all the critical points of the following functions and apply the second derivative test. (a) f(x, y) (x y)(x + y) ( ) x f + y + (x y)x (x + y) + (x y) ( ) x + ( x)y x + x y The critical points occur where the gradient is zero. The second coordinate gives y (x x ), and pluggin this into the first equation gives x + ( x) (x x ). Simplifying we get x + x + x x(x + x + ). Thus x or x 4 ( ± 9 8) 4 ( ± ), by the quadratic formula. So our critical points are (, ), ( ), (, )., 8 f xx 6x y f xy x f yx f yy So for the second derivative test at (, ), ( quadratic forms respectively:, 8 Q x y y ( x y) y ), (, ) we get the following Q 5 8 x + x y y 7 8 x ( y x) Q x + x y y 4 x ( y x) Where all simplifications are done by completing the square. (, ) is a saddle point by a quadratic form uv indefinite, (, 8 ) is a local maximum by u v negative definite, and (, ) is a saddle point by u v indefinite. (b) f(x, y) x + y f ( x y ) The critical points occur where the gradient is zero, so our only critical point is (, ). f xx 6x f xy f yx f yy 6y So the second derivative test gives us the quadratic form Q, so the test fails.

2 (c) f(x, y) (x x + )(y + 5y ) ( ) (x )(y f + 5y ) (x x + )(y + 5) The critical points occur where the gradient is zero. The first coordinate tells us that either x or y ( 5 ± 9) by the quadratic formula. The second coordinate tells us that either x (± 5) or y 5. This gives us five critical points: (, 5 ) (, ( + 5), ( 5 ± ) ( 9), ( 5), ( 5 ± ) 9) Continuing with the second derivative test f xx (y + 5y ) f xy (x )(y + 5) f yx f yy (x x + ) So for the second derivative test at our critical points we get the following quadratic forms (in the order the critical points are presented above): Q 9 4 x y Q ± 5 9 x y Q 5 9 x y Thus by u v negative definite and uv indefinite, we have ( maximum and the rest are saddle points. (d) f(x, y) e (x +y ) ( ) xe f (x +y ) ye (x +y ) 5 ) is a local The critical points occur where the gradient is zero. Because exponential functions are never zero, the only critical point is at (, ). f xx ( + 4x )e (x +y ) f xy 4xye (x +y ) f yx f yy ( + 4y )e (x +y ) So for the second derivative test we get the quadratic form Q x + y positive definite. Thus (, ) is a local minimum. (e) f(x, y) sin(xy) f ( ) y cos(xy) x cos(xy)

3 Critical points occur where the gradient is zero. The first coordinate tells us that either x or cos(xy). If x then in the second coordinate we get y (by cos() ). If cos(xy) then both coordinates are zero. We get infinitely many critical points of the following three forms: (, ) All solutions to xy We proceed with the second derivative test. f xx y sin(xy) (k + )π f xy cos(xy) xy sin(xy) f yx f yy x sin(xy) ( ) Recall that sin (k+)π ( ) k. Then we get the following quadratic forms corresponding to our collections of critical points (in the order presented), where (x, y ) is a critical point with x y (k+)π. Q x y Q y ( ) k x (k + )π ( ) k x y x ( ) k y ( ) k ( ) y x (k + )π + y x y + x y y ( ( ( ) k ( (k + )π x y x + y) y + y ( ) k ( y x + ( ) k ( y x + (k + )π y ) (k + )π y y y) + (k + ) π ) ) 4y 4 y ( ) (k+)π y y (k + ) π 4y 4 y The first is clearly indefinite, so (, ) is a saddle point. The second is of the form ±u and is semidefinite, so the second derivative test fails for all the other critical points. (f) f(x, y) ln(x + y) f ( x+y x+y This is never zero, so there are no critical points.. Maximize the following functions along the unit circle x +y. For the rest of these questions denote g(x, y) x +y. The first case to check is points where g, but these are exactly where x y, i.e. (, ). This does not lie on the constraint set, so for each of the functions below we can assume g. )

4 (a) f(x, y) x y ( ) x f y ( ) λx λ g λy Thus we have x(x λ) and y( y λ). If x then y ± by the constraint. If x λ then x λ. For the second, if y then x ± by the constraint and if y λ then y λ. Thus in the case where x, y we can plug into the constraint to find x + y ( ) ( λ λ λ Thus λ ±. So we have six possible points: (, ±), (±, ), (±, ). In our function we get that f(, ±) ) f(±, ) ± ( f ±, ) ± ± The maximum value of the function on this constraint is, and it is obtained at the points (, ) and (, ). (b) f(x, y) x f ( ) ( ) λx λ g λy Thus either y or λ. But λ contadricts the first coordinate, so our only possible points are (±, ). f(±, ) ±, so the function is maximized on this constraint at (, ) with a value of. (c) f(x, y) xy f ( ) y x ( ) λx λ g λy So we have y λx and x λy 4λ x. Therefore either x or λ ±. In the first case we get the points (, ±), and in the second ( case we ) get y ±x, so ) plugging into the constraint would give us four point, ±, (, ±. So pluggint these into our function, we find that it has a maximum ) value of with respect to this constraint, obtained at the points (±, ±. (d) f(x, y) x + x y f ( ) x + y ( ) λx λ g λy 4

5 In the second coordinate we get either y or λ. In the first case we get points (±, ) on our constraint. In the second, we plug in this value to the fisrt coordinate to get the equation x + x, i.e. x 4. Plugging into 7 the constraint tells us y ± 8 anyways. So our possible points are (±, ) ) and ( 4, ±. Plugging them into our function tells us the function is has a 7 8 maximum value of on the constraint obtained at the point (, ).. Maximize the following functions along the level set x 4 + y 4 + z 4. For the rest of these questions denote g(x, y) x 4 + y 4 + z 4. The first case to check is points where g, but these are exactly where 4x 4y 4z, i.e. (,, ). This does not lie on the constraint set, so for each of the functions below we can assume g. (a) f(x, y, z) x λ4x f λ4y λ g λ4z The first coordinate tells us that λ, thus in the second and third coordinates we find y z. Then from our constraint we see the only possible points are (±,, ). Thus our function has a maximum value of on the constraint at (,, ). (b) f(x, y, z) x + y + z x λ4x f y λ4y λ g z λ4z This gives us x( λx ) y( λy ) z( λz ). By the constraint at least one of x, y, z is nonzero. We get the six points (,, ±), (, ±, ), (±,, ) from when two of x, y, z are zero. If only one is zero, say without loss of generality x, then y λ z. Then from the constraint we get y 4. This gives us ) ) ) the twelve points (, ( ) a 4, ( )b 4, (( ) a 4,, ( )b 4, (( ) a 4, ( )b 4, for a, b either or.. Lastly, if none of x, y, z are zero then x y z λ and in the constraint we get x4. This gives us eight points ) (( ) a 4, ( )b 4, ( )c 4 for a, b, c either or. The first six points has value in f. The next six have value in f. And the last two have value. Thus f has maximum value on the constraint at all of the eight points in the last case. (c) f(x, y, z) xyz yz λ4x f xz λ4y λ g xy λ4z 5

6 Thus xyx 8λx 4 8λy 4 8λz 4. If λ then yz xz zy, i.e. two of x, y, z are zero. So we get the points (±,, ), (, ±, ), (,, ±). If λ then x 4 y 4 z 4. Plugging into the constraint ) tells us x 4. So we get the eight points (( ) a 4, ( )b 4, ( )c 4 for a, b, c either or. Pluggin in our points to f tells us that f has a maximum value of ( ) /4 on the constraint at four of the eight points above with a b c or exactly two of a, b, c are. (d) f(x, y, z) x + y + z f λ4x λ4y λ g λ4z So we have x y z λ. Plugging into the constraint we get that x4. ) This gives us two points (± 4, ± 4, ± 4. So the maximum value of f is ( ) 4 /4 4 on the constraint at the point, 4, Consider the following functions f(x, y), do each of the following in order Find the level set f(x, y) Using the implicit function theorem find all points on the level set such that, around those points, the graph of the level set is the graph of a function y g(x). At every such point (x, y ), find g (x) (NOTE g(x) may be different at different points. There may not be an explicit formula for g(x), use the implicit function theorem for this question) (a) f(x, y) x + y The level set is x + y the line y x. f y so by the implicit function theorem the level set can be represented by a function y g(x) everywhere (in fact y x works) By the implicit function theorem, g (x) fx f y (b) f(x, y) x + y The level set is x + y the point (, ). f y y which is zero on the entire level set, so there are no such points. Not applicable. This is a special example, as the level set is just a point and not a curve. However, in every neighborhood of the point (,), the graph of any continuous function y g(x) is a curve and not a point. (c) f(x, y) x y The level set is x y, the set of points the line y x or y x. f y y which is zero when y. So It is all points such that y ±x By the Implicit function theorem g (x) fx f y x y x y. 6

7 5. Consider the following domains: A {(x, y) x, y } R B the region between the graphs y x 4 and y x R C the region inside the right semi-circle of radius R D {(x, y, z) x, y, z } R E the unit sphere R F the region between the xy-plane and the surface z 5 x y R We can desribe each of these domains in a form that can be more easily integrated: A {(x, y) x, y } R B {(x, y) x y x 4, x } R C {(r, θ) r, θ π } {(r, θ) r, π D {(x, y, z) x, y, z } R E {(ρ, θ, φ) ρ, θ π, φ π} R F {(r, θ, z) r 5, θ π, z 5 r } R θ π} R Compute the following integrals using Cartesian, polar, cylindrical, or spherical coordinates as necessary: ( (a) (x + y)da (x + y)dxdxy x + ) dx (b) (c) A A e x+y da B y da e x e y dxdy x 4 x (e )e y dy (e ) y dydx (x (x ) )dx (x 8x 6 + x 4 6x + )dx ( x 8 7 x7 + 5 x5 6 ) 4 x4 + x ( ) + 7

8 (d) C e x +y da π/ π e r rdrdθ + π/ ( π + π π ) e r rdr π 4 eu du π (e4 ) e r rdrdθ (e) sin(πx) sin(πy)dv D 4 sin(πx) sin(πy)dxdydz ( cos(π) + cos()) sin(πy)dydz dz 4 sin(πy)dydz (f) E π ρ + dv π π ρ + ρ sin φdρdφdθ u / sin φdudφ π 4π + π ( ) sin(φ)dφ 4π ( )(cos() cos(π)) 8π ( ) 8

9 (g) F (x + y + )dv π 5 5 r π 5 5 (r + )rdzdrdθ (r + )r(5 r )dr π ( r 5 + 4r + 5r)dr ( π ) π π (h) F (x + y + z )dv π 5 5 r π 5 π 5 π π 5 (r cos(θ) + r sin (θ) + z )rdzdrdθ 5(r cos(θ) + r sin (θ))r(5 r ) + 4 (5 r ) 4 rdrdθ (5r r 4 ) cos(θ) + (5r r 5 ) sin (θ) + 4 (5 r )rdrdθ ( 5 r 5 r5 ) π π ( 5 cos(θ) + 4 r4 ) 6 r6 sin (θ) 4 (5 r ) 5 5 dθ cos(θ) + sin(θ) + 5 θ sin(θ) θ π 5 5 cos(θ) dθ 6. Using integration, find the area of A, B, C given in question 4. 9

10 This is done by integrating : A B C da da da x 4 dxdy x dydx x 4 x + dx 5 x5 x + x π/ π/ π + π dθ + dx π rdrdθ + π π/ ( π π ) π/ dθ rdrdθ 7. Using integration, find the volume of D, E, F given in question 4.

11 This is done by integrating : D E dv dv dydz π π π π dxdydz sin(φ)dφ ρ sin φdρdφdθ π (cos() cos(π)) F dv 4π π 5 5 r 5 π (5 r )rdr ( 5 π 5 ) 4 5 π 5 π rdzdrdθ 8. A farmer needs to build a silo to hold his grain and the end of harvest season. A silo has the shape of a cylinder with a half-sphere for the roof. The farmer knows that he hervests about 9π cubic meters of grain every year, and wants to build the cheapest silo possible. If the brick walls cost $ per square meter and the tin roof costs $4 per square meter, what should the radius and height of his silo be? How much will it cost? (Hint: the surface area of a sphere is 4πR where R is the radius.) The silo needs to have a fixed volume of V 9π. If the silo has radius r and height (of the cylindrical part) h, then we have 9π πr h + πr. The function describing the cost is given by f(r, h) (πrh)+4(πr ) coming from the formula for surface area of the side of a cylinder and of a sphere respectively. So we have a Lagrange multiplieresque problem, minimizing f with respect to the constraint g(r, h) πr h+ πr 9π. We first check the case when g. We have ( ) πrh + πr g πr is zero implies r, which does not lie on the constraint set (g(, h) 9π). So

12 we have the remaining case ( ) 4πh + 6πr f 4πr ( ) λ(πrh + πr ) λπr λ g The second coordinate tells us that either r or r 4 λ. But we already know r is not on the constraint set, so plugging in r 4 λ into the first coordinate gives 4πh + 6π 4 ( λ λ π 4 ) 6 h + π λ λ hλ + 6 hλ + 8 h 8 λ Now plugging these both into our constrint gives 9π πr h + πr π 6 8 λ λ + π 64 λ 8π 4 λ λ λ 4 8 So we only get one point (r, h) (, ). Our constraint set is a curve, so this point is either a global minimum or a global maximum (if it were only local, there would be another point to represent where the function switched directions to go below or above the point again respectively). As h we must have r to keep g(r, h) 9π. In fact r h 9. Then we see that rh which implies lim h f(r, h). This point cannot be a maximum, so it must be a global minimum. 9. Your living room is given by the domain D {(x, y, z) x 5, y, z } measured in feet. You bought a new heater for the winter, which hangs from the cieling. If it sits at the point (x, y, z ), then the temperature at any other point in the room is given by T (x, y, z) e x+y+z x y z What is the average temperature of the room when the heater is placed at an arbitrary point (x, y, z ) on the cieling? Where should the heater be placed to maximize the average temperature? The average temperature is computed by the following integral: Ave(T ) T (x, y, z)dv V ol(d) D 5 e x+y+z e x e y e z dzdydx ex+y+z (e 5 )(e )(e )

13 We want to maxime Ave(T ) with respect to the point (x, y, z ). This can be done in two ways, the first is to find it s critical points in D and on it s boundary then test them all with the second derivative test. But the second, and far easier in this case, is to notice that e x+y+z is increasing in all three variables. So the maximum average temperature occurs at the point (5,, ) in the upper corner of the room. This point is, in fact, on the cieling so this is where we should hang the heater.. You own a bowl in the shape of the surface z x + y which is 5 inches tall. How much ice cream can fit in the bowl without going over the top (in cubic inches)? We can find this volume by taking a triple integral over the inside of the bowl B. We can describe B as B {(r, θ, z) r 5, θ π, r z 5} So the volume is B dv π 5 5 π 5 r rdzdrdθ 5r r dr ( π 5r ) 5 4 r4 ( 5 π 5 ) 4 5π. Suppose you are in a spaceship nearing a black hole. The gravitational force of the black hole is given by G(x, y, z) x + y + z For x + y + z >. All points such that x + y + z are inside the event horizon of the black hole and can never escape. To avoid getting turned into spaghetti, you want to move away from the black hole as quickly as possible in a way that the gravitational force decreases the most. If your spaceship is sitting at the point (,,), what dircetion should you pilot your spaceship in? You want to pilot your ship in the direction which G has the largest rate of decrease. This is G. G x y (x + y + z ) z So you want to pilot your spaceship in the direction G(, 6, ) 4. 6

14 . Set up but do not evaluate an integral for finding the volume of the ellipsoid x +y + z. We can do this in all three coordinate systems for R first Cartesian coordinates x x y x E {(x, y, z) x, y, V ol(e) dv E x x y x y x dzdydx y x z } Second cylindrical coordinates. Converting the equation of the ellipseoid into cylindrical coordinates gives us r ( + sin (θ)) + z, so that we have E (r, θ, z) θ π, r r, ( + sin (θ)) r z ( + sin (θ)) + sin (θ) V ol(e) dv π E +sin (θ) r (+sin (θ)) r (+sin (θ)) rdzdrdθ Lastly spherical coordinates, again we conver the equation of the ellipsoid to get ρ (cos (θ) sin (φ) + sin (θ) sin (φ) + cos (φ)) ρ (sin (φ)( + sin (θ)) + cos (φ)) ρ ( + sin (θ) sin (φ) + cos (φ)) And so we have E (ρ, θ, φ) θ π, φ π, ρ + sin (θ) sin (φ) + cos (φ) V ol(e) dv E π π +sin (θ) sin (φ)+ cos (φ) ρ sin(φ)dρdφdθ. Suppose you purchase a fog machine for Halloween, and put it inside your living room. Suppose your living room is a rectangle with corner at the origin, width in the x direction equal to ft, width in the y direction equal to 5 ft, and height equal to ft. You leave the fog machine running for a while and find that your room is now filled with fog at a denstiy of µ(x, y, z) e z x grams per cubic foot. 4

15 (a) What is the total mass of fog in your living room? Total mass is given by the integral of the density function µ(x, y, z)dv room 5 5 e zxdzdydx ( e )xdydx 5( e )xdx 5( e ) ( e ) (b) What is the center of mass of the fog? 5

16 We must compute each coordinate of the center of mass (X, Y, Z) seperately. X xµ(x, y, z)dv M room ( e ) ( e ) 5 5 5x dx () Y yµ(x, y, z)dv M room ( e ) ( e ) xdx Z zµ(x, y, z)dv M room ( e ) ( e ) ( e ) + 9e ( e ) + 9e ( e ) + 9e e e z x dzdydx ( e )x dydx xdx e z yxdzdydx ( e )yxdydx ze z xdzdydx ( ze z + e z dz) ( e ( 9) + )xdydx xdydx So the center of mass of the fog is at the point (X, Y, Z) ( 4, 5, +9e e ). 6

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