Winter 2012 Math 255. Review Sheet for Second Midterm Exam Solutions. 1. Let r(t) = cost,sint,t. Find T(π), N(π), and B(π)...

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1 Winter 1 Math 55 Review Sheet for Second Midterm Exam Solutions 1. Let r(t) cost,sint,t. Find T(π), N(π), and B(π). We obtain T(π), 1, r (t) sint,cost,1, T(t) r r 1 sint,cost,1, N(t) T T cost, sint,, B(t) T N 1 sint, cost,1. 1, N(π) 1,,, B(π), 1, 1.. At what point do the curves r 1 (t) t,t,t and r (s) s+1,s + 1,e s intersect? Find cosθ, where θ is the angle of intersection. By solving t s+1, t s +1, t e s, we obtain t 1, s. (1) Therefore, the point of intersection is (1,1,1). We obtain r 1(t 1) 1,,, r (s ),,1. Theangleofintersectionistheangleθ betweenthesetwovectors. Weobtain cosθ 1,,,,1 5 1,,,, Math Math Math Math

2 . Consider the Frenet-Serret formulas dt/ds κn, dn/ds κt+τ B, and db/ds τn. Here, τ(s) is the torsion. (a) Using the definition of the curvature κ dt/ds, derive the first formula. (b) Deduce the second formula from the first and third formulas. (a) We have κ dt ds dt/dt ds/dt, N T T. Therefore, dt ds T s T s N κn, where we used s r (t) >. (b) Using N B T, we calculate as dn ds d db dt (B T) T+B ds ds ds τn T+κB N. Noting that B T N and T N B, we obtain dn ds τb κt. 4. Prove (a) r s T + κ(s ) N, (b) r r κ(s ) B, (c) r [s κ (s ) ]T + [κs s + κ (s ) ]N + κτ(s ) B, and finally (d) τ (r r ) r / r r. (a) By definition, we have We obtain r r T s T. r s T+s T. By Formula 1 of Prob. dt/ds T /s κn, we have (b) Using (a), we have r s T+κ(s ) N. r r s T [ s T+κ(s ) N ]. Math Math Math Math

3 Sine T T and T N B, we obtain r r κ(s ) N. (c) Using (a), we obtain r s T+s T +κ (s ) N+κs s N+κ(s ) N. Formulas 1 and of Prob. read dt ds T s κn, dn ds N s κt+τb. Therefore, r s T+s ( κs N ) +κ (s ) N+κs s N+κ(s ) [ s ( κt+τb) ] [s κ (s ) ]T+[κs s +κ (s ) ]N+κτ(s ) B. (d) Using (b), we have r r κ(s ) B κ(s ). Using (b) and (c), we obtain (r r ) r κ(s ) B {[s κ (s ) ]T+[κs s +κ (s ) ]N+κτ(s ) B } κ(s ) B κτ(s ) B [ κ(s ) ] τ. Therefore, we obtain τ (r r ) r r r. 5. Find the tangential and normal components of the acceleration vector for the position vector r(t) (1+t)i+sintj+costk. We calculate v and a as v r i+costj sintk, a r 9sintj 9costk. By definition, we have v vt, Math Math Math Math

4 where v v. Hence, a v v T+vT. We have T κvn because κ T /v and N T / T. We obtain a v v T+κv N. We can calculate the tangential component v as v T a v a v (i+costj sintk) ( 9sintj 9costk) i+costj sintk. We can calculate the normal component κv as κv r r r v v a v (i+costj sintk) ( 9sintj 9costk) v 7i+9costj 9sintk i+costj sintk Find the limit lim (x,y) (1,) xycos(x y) using the ε-δ definition. We begin to suspect that the limit is (1)()cos((1) ). Let ε >. We want to find δ > such that xycos(x y) < ε whenever < (x 1) +(y ) < δ. () When (x 1) +(y ) < δ, we have x 1 < δ, y < δ. Let us use cosθ 1 sin θ, θ x y. 4 Math Math Math Math

5 We obtain xycos(x y) (1 sin xy θ ) xy θ xysin If δ/ π/, we have sin θ sin ( (x 1) (y ) We will choose δ π/. Note that xy + xy sin θ. ) < sin δ. xy [(x 1)+1][(y )+] (x 1)(y )+(x 1)+(y )+. Therefore, xycos(x y) < δ +δ +(+δ +δ )sin δ f(δ). Hence, we can choose δ such that f(δ) ε is satisfied if f(π/) > ε. We can choose δ π/ if f(π/) ε. With this δ, Eq. () is satisfied. 7. Find the limit lim (x,y) (,) (x +y ) ln(x +y ). lim (x,y) (,) (x +y ) ln(x +y ) lim r + r6 lnr 6 lim lns s s, where s 1/r 6. We can use L Hospital s Rule. lim (x,y) (,) (x +y ) ln(x +y ) lns lim s s lim 1/s s 1. 5 Math Math Math Math

6 8. Find z/ x and z/ y for yz xln(y +z). Note that x and y are independent variables, and z is a function of x and y. By differentiating the equation implicitly with respect to x treating y as a constant, we obtain Therefore, we obtain y z x z ln(y +z)+ x y +z x. () z (y +z)ln(y +z) x y(y +z) x. By differentiating the equation implicitly with respect to y treating x as a constant, we obtain z +y z y x ( 1+ z ). y +z y Therefore, we obtain z y x z(y +z) y(y +z) x. 9. Find partial derivative f trsrs for f(r,s,t) r tln ( rs e t). We obtain f t r ln ( rs e t) +r t, f tr rln ( rs e t) +r +rt, then Therefore, f trs 4r s, f trsr 4 s. f trsrs 4 s. 1. Determine whether each of the following advection diffusion equation u t u xx +u x. (a) u e t+x, (b) u e t+x, (c) u e t x, (d) u e t x. (a) not solution, (b) solution, (c) solution, (d) not solution. 6 Math Math Math Math

7 11. Findanequationofthetangentplanetothesurfacez e x y ln(1+ x ) at the point (1, 1,ln). Let us define f(x,y) e x y ln(1 + x ). The tangent plane is expressed as z ln f x (1, 1)(x 1)+f y (1, 1)(y +1). Since [ f x (x,y) e x y xln(1+x )+ x ] 1+x, f y (x,y) e x y ( y)ln(1+x ), the equation is obtained as (1 ln)x+(ln)y z +5ln The dimensions of a quarter (United States coin) are measured to be 4.4 mm (diameter) and 1.8 mm (thickness). Each measurement is correct to within. mm. Use differentials to estimate the largest possible error when the volume of the quarter is calculated from these measurements. (Use π.14.) Let d and w be the diameter and thickness. The volume is V πd w/4. The differential is calculated as dv V V dd+ d w dw π dwdd+ π 4 d dw. (4) We are given that d. and w.. To find the largest error in the volume, we therefore use dd. and dw. together with d 4.4 and w 1.8. V dv (4.4)(1.8)(.)+ 4 (4.4) (.) (.14)(4.4)(1.4) Math Math Math Math

8 1. Provethatiff(x,y)isdifferentiableat(a,b), thenf(x,y)iscontinuous at (a,b). Since f is differentiable at (a,b), we have f(a+ x,b+ y) f(a,b)+f x (a,b) x+f y (a,b) y+ε 1 x+ε y, (5) where ε 1 and ε as ( x, y) (,). Hence, That is, f is continuous at (a,b). lim f(a+ x,b+ y) f(a,b). ( x, y) (,) 14. Let f(x,y) x y/(x 4 + y ) if (x,y) (,) and f(,). Is the function f differentiable at (, )? Explain. We let (x,y) approach the origin along the parabola y x. We have x 4 lim x f(x,x ) lim x x 4 +x 4 1 f(,). Therefore, f is not continuous at the origin and is not differentiable. 15. Find dw/dt for w xye z, x t, y t, z lnt. dw dt w dx x dt + w dy y dt + w dz z dt ye z +xe z (t)+xye z1 t 4t. 16. Find dy y4 for dx x +xcosy y +lnx. LetF(x,y)beF(x,y) y4 x +xcosy y lnx. ThenwehaveF(x,y). dy dx F/ x y4 F/ y cosy + 1 x x xsiny 1. 4y x 8 Math Math Math Math

9 17. Suppose that the equation F(x, y, z) implicitly defines each of the three variables x, y, and z as functions of the order two: z f(x,y), y g(x,z), x h(y,z). If F is differentiable and F x, F y, and F z are all nonzero, show that z x y x y z 1. By differentiating F(x,y,f(x,y)) with respect to x, we obtain F F z z x x +. By differentiating F(x,g(x,z),z)) with respect to z, we obtain F y y z + F z. By differentiating F(h(y,z),y,z) with respect to y, we obtain F x x y + F y. Thus, by the implicit function theorem, we obtain ( z x y x y z F )( x F )( y F ) z 1. F z F x F y 18. Near a buoy, the depth of a lake at the point with coordinates (x,y) is z 1 + x y + y, where x, y, and z are measured in meters. A fisherman in a small boat starts at the point (4,) and moves toward the buoy, which is located at (,). Is the water under the boat getting deeper or shallower when he departs? Explain. Let us define f(x,y) 1+x y +y. Note that we have f x x and f y y +. We obtain the directional derivative in the direction of u 4 5, 5 4, as ( D u f(x,y) f x (x,y) 4 ) ( +f y (x,y) ) 1.6x+1.8y When we start at (x,y) (4,), we have D u f <. Therefore, the water under the boat is getting shallower. 19. If g(x,y) x +y +xy+x y, find the gradient vector g(1, 1) and use it to find the tangent line to the level curve g(x,y) 4 at the point (1, 1). Sketch the level curve, the tangent line, and the gradient vector. The gradient vector is calculated as g x+y+,y+x and so g(1, 1),. Noting that the tangent line is perpendicular to the gradient vector, we write a vector equation of the tangent line as x 1,y +1,. 9 Math Math Math Math

10 We obtain y x. To sketch the level curve, let us first understand what g(x,y) 4 looks like by defining X (x + y)/ and Y (x y)/. We have Y (1/ )X +, so g(x,y) 4 is a parabola. The level curve, tangent line, and gradient vector are sketched as follows.. Show that the function f(x,y) (xy) 1/ is continuous and the partial derivatives f x and f y exist at the origin but the directional derivatives in all other directions do not exist. The function f(x,y) is continuous at (,) if lim f(x,y) f(,). (x,y) (,) That is, (xy) 1/ is continuous at the origin if for every ε > there exists δ > such that (xy) 1/ < ε whenever < x +y < δ. 1 Math Math Math Math

11 Since (x y), we have xy x +y < δ. Thus, ( ) (xy) 1/ δ 1/ <. For every ε, we can choose δ such that ε (δ /) 1/. Therefore, (xy) 1/ is continuous at the origin. The partial derivatives are calculated as f x (,) f y (,). Let us calculate the directional derivative in the direction of u a,b (a, b ). D u f(,) lim h [ha(hb)] 1/ [()] 1/ h lim h (ab) 1/ h 1/. Therefore, the directional derivatives in the direction of such u do not exist. 1. Find the local maximum and minimum values and saddle point(s) of the function f(x,y) xye x y. We find f x y(1 x)e x y, f y x(1 y)e x y, f xx y(x )e x y, f yy 4x(y 1)e x y, f xy f yx (1 x)(1 y)e x y. ( The function has critical points at (,) and 1, 1 ). Let us use the second derivatives test. We obtain D f xx f yy (f xy ) [ 4xy(x )(y 1) (1 x) (1 y) ] e x 4y. Therefore, we have ( D(,) 1, D 1, 1 ) e Math Math Math Math

12 Note that ( f xx 1, 1 ) 1 e. By the second( derivatives test, we see that the point (,) is a saddle point and the point 1, 1 ) is a local maximum.. A rectangular house is designed to minimize heat loss. The east and west walls lose heat at a rate of 9units/m per day, the north and south walls at a rate of 8units/m per day, the floor at a rate of 1units/m per day, and the roof at a rate of 5units/m per day. Each wall must be at least m long, the height must be at least m, and the volume must be exactly 1m. Find the dimensions that minimize heat loss. Is it possible to design a house with even less heat loss if the restrictions on the lengths of the walls were removed? Let L denote the heat loss. We have the following conditions. L (9+9)xz +(8+8)yz +(1+5)xy. x, y, z 4, xyz 1. Since the volume is fixed, we can eliminate z. Let us find the critical point(s). L(x,y) 6xy + 16 x y L x 6y 16 x, L y 6x 18 y. Using the above two equations and noting that 1 4/xy, we obtain x 4, y 15, z 5. The obtained x and y are less than. Let us find the extreme values of L on the boundary. L(x,) 1x+ 16 x 1 +9, x. Math Math Math Math

13 This function L(x,) (x ) has the minimum value at x. L(,y) 1y , y. y This function L(,y) (y ) has the minimum value at y. When x y, we have z.5. Therefore, the heat loss is minimized when x y m, z.5m, and the minimum heat loss is L(,) 41. At the critical point, we obtain L( 4 ( ) 4,15) 6 (15)+ 16 4/ This number is smaller than The plane x y + 11 z intersects the cone z x +y in an ellipse. Find the highest and lowest points on the ellipse. Let us define f(x,y,z) z, g(x,y,z) 5 x y + 11 z, h(x,y,z) z x y. With the Lagrange condition f λ g +µ h and two constraints g and h, we have 5 λ xµ, (6) λ yµ, (7) 1 11 λ+zµ, (8) 5 x y + 11 z, (9) z x y. (1) From Eqs. (6), (7), and (8), we obtain x 5λ 4µ, y λ µ, z 11λ. 4µ 1 Math Math Math Math

14 By plugging them into Eq. (9), we obtain Therefore, µ 11 56λ. 8 x 5λ 11 56λ, y 4λ 11 56λ, z 4 λ 11 56λ. By using these expressions in Eq. (1), we obtain 8λ 11λ+1. We obtain If λ 1/4, we have λ 1 4, 1 7. x 5 6, y 1, z 1. If λ 1/7, we have x 5 1, y 4 1, z 7. ( ) 5 Therefore, the highest point is 6, 1, 1 andthelowestpointis 4. Evaluate the double integral (1 ) x +y da, D {(x,y) D 1 x 1, 1 x y 1 x } by using an iterated integral. The double integral is expressed as V 4 (1 ) 1 x +y da D 1 1 x 1 1 x 1 x (1 x +y ) dydx. (1 x +y ) dydx ( ) 5 1, 4 1, Math Math Math Math

15 ( Weintroducey xsinht,dy xcoshtdt. Wehavesinh 1 (a) ln a+ ) a +1 for a > because if we put sinh 1 a lnu, then a (u 1 u )/ and u a+ a +1. Thus, v 1 x x +y dy sinh 1 ( 1 x /x) x ln ( ) 1+ 1 x x x 1+sinh tcoshtdt 1+cosht dt. Note that sinht sinht 1+sinh t. [ ( v x 1+ ) ] 1 x ln 1 x + x x. We obtain V 1 [ 1 x x ln( 1+ )] 1 x dx. x Let us first calculate the first integral. 1 1 x dx 1 π/ π/ π 4. 1 sin θcosθdθ 1+cosθ,dθ [ θ sinθ ] π/ To calculate the second integral, note that [ ( d x dx ln 1+ )] ( 1 x x 1+ ) 1 x ln x x x 1 x. Therefore, [ ( x ln 1+ )] 1 1 x x 1 ( x 1+ ) 1 x ln x 15 1 dx x 1 x dx. Math Math Math Math

16 Since, 1 x [ 1 x dx x 1 1 x ] x dx π 1, we have ( 1 x 1+ ) 1 x ln dx π x 1. Thus we obtain [ 1 1 x V x 1+ ln( )] 1 x dx x ( π 4 1) π π. 5. Evaluate the double integral (1 ) x +y da, D {(x,y) D x +y 1} by first identifying it as the volume of a solid. The solid is a cone. The volume is 1 π(1)/ π/. 6. Evaluate the integral D (1 ) x +y da where D {(x,y) x +y 1} by changing to polar coordinates. The region D is expressed as D {(r,θ) θ π, r 1}. V (1 ) x +y da 1 6 D π 1 π π. [ r π (1 r)rdrdθ r dθ ] 1 dθ 16 Math Math Math Math

17 7. Calculate the double integral R sin(x + y)da, R {(x,y) x π/, y π/} R sin(x+y)da π/ π/ π/ sin(x + y)dydx [ 1 ] yπ/ cos(x+y) dx π/ y 1 [cos(x+π) cos(x)]dx 1 [ sin(x+π) sin(x) ] π/ Find the volume of the solid that lies under the hyperbolic paraboloid z +x y and above the square R [1,] [ 1,1]. [volume] (+x y )dydx [ y +x y y ] 1 1 dx (+x )dx [ x+ x ] Find the average value of f(x,y) ln(x + y + 1) over the rectangle R [,] [,]. 17 Math Math Math Math

18 The average value f ave is calculated as Hence, f ave 1 6 ln(x+y +1)dydx. f ave 1 6 [(x+y +1)ln(x+y +1) y] y y dx 1 [(x+4)ln(x+4) (x+1)ln(x+1) ]dx 6 1 [ (x+4) ln(x+4) (x+4) (x+1) 6 4 ln + 9ln 4. ln(x+1)+ (x+1) 4 ] x. Let f(x,y) xy(x y )/(x +y ) for (x,y) (,) and f(,). Show that Fubini s theorem does not hold and 1 1 f(x,y)dxdy f(x,y)dydx. We have 1 1 [ x ] x y f(x,y)dxdy (x +y ) dy [ 1 1 y y (y +4) dy ] 1 x 18 Math Math Math Math

19 On the other hand, 1 f(x,y)dydx 1 5. [ xy (x +y ) x (x +1) dx [ ] 1 4(x +1) ] y1 y dx Therefore, Fubini s theorem does not hold. 1 1 x 1. Evaluate the iterated integral (x y )dydx x 1 1 x x (x y )dydx 1 1 ] y1 x [xy y dx yx ( 1 +x x + ) x dx [ x +x x x4 ] Evaluate the double integral D (x+y)da, D is bounded by y x 1/ and y x 19 Math Math Math Math

20 1 x 1/ x (x+y)dydx 1 ] yx 1/ [xy + y dx yx ( ) 1 x 4/ + x/ x4 x6 [ 7 x7/ + ] 1 1 x5/ x5 5 x dx. Evaluate the double integral D xy da, D is the triangular region with vertices (,), (,), and (,4) x/ xydydx+ 4 x+8 xy dydx [ x y] yx/ dx+ y 4 4 [ x y] y x+8 y ( 9 x dx+ x 16x +x ) dx [ ] x 4 [ x ] 4 x +16x. dx 4. Find the volume of the solid under the surface z x+y and above the region bounded by x y and x y. Math Math Math Math

21 1 y 5. Show that y ( x+y ) dxdy D 1 1 [ y [ x ] x y +xy ( y +y5/ y6 y5 xy dy y7/ y7 14 y ) dy ( x +y 4x y +5 ) / da π/, where D is the disk with center (,1) and radius. Note that for (x,y) D, we have (x ) +(y 1) 4. The double integral is estimated as ( x +y 4x y +5 ) / da ] 1 [ (x ) +(y 1) ] / da D D D (4) / da (4π)4 / π. 6. Find the volume of the solid inside the sphere x + y + z 9 and outside the cylinder x +z 1. We consider polar coordinates in the x-z plane. The volume is obtained as π 1 9 r rdrdθ π 64 π. [ 1 ( 9 r ) /] dθ 1 1 Math Math Math Math

22 4 4x x 7. Evaluate the iterated integral (x +y )dydx Note that y 4x x implies (x ) +y x x (x +y )dydx π/ 4cosθ π/ [ r 4 4 r rdrdθ ] r4cosθ r π/ 4 cos 4 θdθ 4 π/ [ 4 8 θ + sinθ 4 1π. dθ ( 8 + cosθ + cos4θ ) 8 + sin4θ ] π/ dθ 8. Show the Gaussian integral e x dx π. ( e dx) x π π e x dx e y dy e (x +y ) dydx e r rdrdθ [ 1 e r ] π 1 π. Therefore, e x dx π. Math Math Math Math

23 9. Find the mass m and center of mass ( x,ȳ) of the lamina that occupies the Cardioid D {(r,θ) θ π, r 1 + sinθ} when the density function is ρ(x,y) x +y. m 1 1 ρ(x,y)da D π 1+sinθ π 5π. [ r π rrdrdθ ] r1+sinθ dθ r ( sinθ cos(θ) 1 4 sin(θ) [ 5 θ 15 4 cosθ 4 sin(θ)+ 1 1 cos(θ) ] π ) dθ x 1 m 5π 5π π π. xρ(x,y)da D π 1+sinθ π π π [ r 4 cosθ 4 r cosθrdrdθ ] r1+sinθ r cosθ(1+sinθ) 4 dθ dθ ( 5 cosθ 8 +7sinθ 7 cos(θ) sin(θ)+ 1 ) 8 cos(4θ) dθ Math Math Math Math

24 ȳ 1 m 5π 5π π π 1. yρ(x,y)da D π 1+sinθ π π π [ r 4 sinθ 4 r sinθrdrdθ ] r1+sinθ r sinθ(1+sinθ) 4 dθ dθ ( 5 sinθ 8 +7sinθ 7 cos(θ) sin(θ)+ 1 ) 8 cos(4θ) dθ Thus, the mass is m 5π/ and the center of mass is ( x,ȳ) (,1.5). 4. Suppose X and Y are random variables with joint density function f(x,y) / π e (x 1) e y for < x <, y <. Find P(Y ln), and the expected values of X and Y. P(Y ln) π ln e y dy ln ] [ e y e ln 1 9. ln e (x 1) / e y dydx 4 Math Math Math Math

25 µ 1 xe (x 1) / e y dydx π 1 xe (x 1) / dx π 1 { } (x 1)e (x 1) / dx+ e (x 1) / dx π 1 { } te t / dt+ e t / dt π 1 { [ ] } e t / π + π 1 π π 1. µ ye (x 1) / e y dydx π π ye y dy π { [ y e y] + 1 } e y dy { [ 1 1 ] } e y Findtheareaofthesurfacewhichisthepartofthespherex +y +z 9 that lies within the cylinder x +y y. ( The domain D is the circle x + y ( ) ; we have r sinθ ) in polar coordinates. The sphere is expressed as z ± 9 x y. There- 5 Math Math Math Math

26 fore, we obtain z x x 9 x y, z y The surface area is calculated as ( ) z ( ) z π 1+ + da D x y 6 π π y 9 x y. sinθ 9 r rdrdθ [ ] sinθ 9 r dθ 18 (1 cosθ ) dθ { π π/ 18 dθ π cosθdθ + π/ { } 18 [θ] π [sinθ] π/ +[sinθ] π π/ 18(π 1 1) 18(π ). cosθdθ } 4. Evaluate the triple integral E x e y dv, where E is bounded by the parabolic cylinder z 1 y and the planes z, x, and x / 1 y 1/ x e y dzdydx 1 1/ [ x 4 4 e 1/ 1/ ] 1 x e y (1 y )dydx [ e y ( y +4y ) ] 1/ 1/ ) ( 1 1 ( +e 1/ 1 +1 ). 6 Math Math Math Math