Math 321 Final Exam. 1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.

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1 Math 31 Final Exam Instructions 1 Do NOT write your answers on these sheets Nothing written on the test papers will be graded Please begin each section of questions on a new sheet of paper 3 Do not write answers side by side 4 Please do not staple your test papers together 5 Limited credit will be given for incomplete or incorrect justification 1 Limits Show that the following its do not exist (a) (3) (x,y) (,) y + We consider the it along two paths First consider the it along the line (x, ) (x,) (,) Next consider the it along the line (,y) (,y) (,) + y + y (,y) (,) y 1 The two its are not the same; therefore, the it does not exist + (b) (3) (x,y) ( 1,) + We consider the it along an arbitrary line through the point (-1,) Such a line has the equation y a(x + 1)+ Substituting this into the it statement produces the following + +[a(x +1)+ (x,y) ( 1,) + (x,y) ( 1,) +[a(x +1)+ + a (x +1) +4a(x +1)+4 (x,y) ( 1,) + a (x +1) +4a(x +1)+4 ( + a) +(a +4a)x +(a +4a +4) (x,y) ( 1,) (1 + a) +(a +4a)x +(a +4a +4) ( + a) (a +4a)+(a +4a +4) (1 + a) (a +4a)+(a +4a +4) This is dependent on the slope of the line Therefore, the it does not exist 1

2 Derivatives Calculate the following derivatives [ (a) (4) f(x, y, z) + +, x + +, x + + Calculate Df Df δf 1 δx δf δx δf 3 δx δf 1 δf δf 3 δf 1 δf δf 3 + ) () ( + + ) y x 4 (y) ( + + )y y 4 () ( +y z 4 3 ( + + )() y x 4 y 3 ( + + )(y) y 4 3 ( + + )() z 4 ( + ) y x 3 ( + ) y 3 ( + ) z 3 (b) (4) f(x 1,,x 3 )[e +x 3,e x 1+x 3,e x 1+ g(t 1,t )[t 1,t,t 1t Using the chain rule, calculate D(f g)(t) t [t 1,t Df(x) e +x 3 e +x 3 e x 1+x 3 e x 1+x 3 e x 1+ e x 1+ Dg(t) t 1 t t t 1 D(f g)(t) Df(g(t))Dg e t +t 1t e t +t 1t e t 1 +t 1t e t 1 +t 1t e t 1 +t e t 1 +t t 1 t t t 1 t e t +t 1t t e t +t 1t + t 1 e t +t 1t t 1 e t 1 +t 1t + t e t 1 +t 1t t 1 e t 1 +t 1t t 1 e t 1 +t t e t 1 +t t e t +t 1t (t + t 1 )e t +t 1t (t 1 + t )e t 1 +t 1t t 1 e t 1 +t 1t t 1 e t 1 +t t e t 1 +t

3 3 Taylor Polynomials (a) (4) Find P (x, y) the second order Taylor polynomial at c (a, b) forf(x, y) cos( + ) f [ sin( + )(), sin( + )(y) [ sin(x Hf + ) 4 cos( + ) 4xy cos( + ) 4xy cos( + ) sin( + ) 4 cos( + ) P (x, y) cos(a + b )+ [ a sin(a + b )(x a)+[ b sin(a + b )(y b)+ [ sin(a + b ) a cos(a + b )(x a) + [ 4ab cos(a + b )(x a)(y b)+ [ sin(a + b ) b cos(a + b )(y b) (b) (3) Calculate P (x, y) atc (, ), c (, π), c ( π, π) For c (, ) P (x, y) 1 For c (, π) P (x, y) 1+π(y π) For c ( π, π) P (x, y) 1 π(x π) 4π(x π)(y π) π(y π) (c) (3) Is P at (, ) a good approximation of the function at (, )? Is P at ( π, π) a good approximation of the function at (, )? Is there any problem with using c (, π)orc ( π, π)? Explain The first option is not good, because the appoximation is a constant and therefore not good most places The third option would give a good approximation at (,), because is close to π However, we cannot use it easily, because we need to know a precise estimate of π This adds additional complications 3

4 4 Extrema (a) (4) Find all extreme points and identify as maximum, minimum, and saddle points for the function f(x, y) xye x+y f [y(xe x+y + e x+y ),x(ye x+y + e x+y [y(x +1)e x+y,x(y +1)e x+y Hf [ y((x +1)e x+y + e x+y ) (x +1)(ye x+y + e x+y ) (x +1)(ye x+y + e x+y ) x((y +1)e x+y + e x+y ) [ (x +)ye x+y (x +1)(y +1)e x+y (x +1)(y +1)e x+y x(y +)e x+y First we solve f Because an exponential never equals zero, we have the system of equations y(x +1) x(y +1) By the first equation either y or x + 1 must be zero Suppose that y For the second equation to be true x Suppose that x + 1 this forces y + 1 in the second equation Thus the two solutions are (, ) and ( 1, 1) Next we consider the Hessian at each point Hf(, ) Hf( 1, 1) [ 1 1 [ e e This indicates that (,) is a saddle point, and (-1,-1) is a local maximum 4

5 (b) (4) Solve the constrained optimization problem below f(x, y) xy g(x, y) 3y 6 y λ x λ( 3) y x 3 x 3 y ( 3 ) y 3y 6 6y 6 y 1 x 3 Thus the solution is ( 3, 1) (c) (4) Solve the constrained optimization problem below f(x, y) xy g(x, y) + 4 y λ x λy y y + 4 x ± y ± Thus the solutions are (, ), (, ), (, ), (, ) 5

6 5 Line Integrals (a) (4) Evaluate the line integral x F ds for F(x, y, z) [y + z,x + y + z,x + y and x(t) [t,t,t,t [, x F ds [t + t, t + t, t + t [t, 1, tdt t +t 3 +t + t +t +t 3 4t 3 +6t + tdt t 4 +t t 34 (b) (6) Prove that the vector field F(x, y, z) [yz, xz +z,xy+y +1 is conservative Find the scalar potential function i j k curl(f) δ δ δ δx yz xz + z xy + y +1 [ δ(xy + y +1) δ(xz + z) δ(xy + y +1), δx [(x +1) (x +1),y y, z z [,, δ(yz) δ(xz + z), δ(yz) δx Thus the vector field is conservative Next we find the scalar potential function f(x, y, z) yz dx xyz + g(y, z) δf xz + g (y, z) xz + z Thus g (x, y) z, and g(x, y) yz + h(z) f(x, y, z) xyz + yz + h(z) δf xy + y + h (z) xy + y +1 Thus h (z) 1, and h(z) z + C Finally f(x, y, z) xyz + yz + z + C 6

7 6 Thought Questions (a) () For any vectors a, b R 3 proj a (a b) proj a (a b), because the cross product is orgthogonal to both vectors (b) () a b (b a) Why? The cross product is a determinant whose rows are determined by the vectors Swapping the vectors is swapping rows, which negates the determinant [ ( ) (c) () F(x, y, z) +, x + +5, cos + is continuous Why? +5 + is a polynomial, and all polynomials are continuous + +5 ( + is a ratio of two continuous functions, and the denominator is never ; therefore, it is continuous ) cos is a composition of two continuous functions; therefore it is continuous +5 7