SAMPLE PROBLEMS FOR FINAL EXAM

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1 Dec 1, Calculus III for CS Fall 001 SAMPLE PROBLEMS FOR FINAL EXAM TIME: Dec 14, Friday, 11:30-:0 in the usual classroom Skiles 49. MATERIAL: Everything we covered: Sections 1,, 3, 4, 5, 6, 7 from the [Notes:Th] Sections 1,, 3, 4, 5, 6 from [Notes:Num] (see on the web if you have lost your copy). Section 9 (Appendix on determinants) is not needed, but you should be able to compute by and 3 by 3 determinants. Salas-Hille-Etgen: Section , , , , Handout on the Chain rule and Jacobian matrix, Newton s method and Handout on Hessian and second partial derivative test. ( ( See also the List of important problems below. The exam will be slightly biased towards the material we learned since the second midterm. However, the exam is cumulative. The following sample problem set contains problems ONLY from the last third of our syllabus. However, the exam covers EVERY- THING. Use the previous Sample Problems for Midterm I and II to review the earlier material. If you lost your copy, look at: 1

2 All definitions and theorems should be learned and can be asked. Review the basic differentiation rules. You MUST be able to differentiate basic functions without errors (polynomials, trigonometric functions and inverses, exp, log, product rule, quotient rule, chain rule, derivative of inverse function). Review the basic integration formulas (including integrals of elementary functions, integration by parts, integration by substitution). The exam is closed-book, only pen, pencil, eraser and a small (non-programmable) calculator are allowed. You can use a cheat-sheet of size A4, one-sided, where you can put any formula you wish in advance (like for Midterm II. just twice bigger sheet). When solving the problems, include all side calculations, and clearly cross out those parts which you consider irrelevant or wrong. Unjustified results are not accepted, you have to show your work. The solution should contain sufficient details (side calculations, explanatory words or signs indicating which equation follows from the other one etc.) so that anyone could reproduce your argument based upon your writing. Showing the final result only is not enough to get credit. The final result should be reasonably simplified if possible. Results like or x+3 4x 6 will not be considered complete, you have to write = 1 4 or x+3 4x 6 = x 3. Usual OFFICE HOURS are on MWF 11:00-1:30 next week. Before the test, I will have an office hour on Dec 13, Thursday 4:00 7:00 PM and on Dec 14, Friday 10:00 11:30 PM. If these times are in conflict with your other schedules, let me know; other times are also possible. The graded finals will be available on Saturday morning from 10:00-11:00 AM. I can send your final grade by if you ask for it by from your own Gatech account before Saturday 10:00 AM. I will leave for Europe at 11:00 AM Saturday so every question should be settled before that. Do not jump onto the solutions immediately, try to solve the problems first by yourself. If you find any mistake, typo etc. in this problem set, please let me know.

3 List of important basic problems The following is a list of basic problems, which I expect everyone to be able to solve. The exam is not restricted to only these problems, but these will play a major role. Basic geometry of lines and planes in space. Problems of type Problem 1 and on Sample Midterm I. Full solution of Ax = b with Gaussian elimination and its application (finding basis in the null-space, column-space; dimension, deciding linear dependence, indepedence). Problem 3 (i)-(v) and Problem 4 from Sample Midterm I. Change of basis matrix and matrix of linear transformations in general vectorspaces and in R n. Problem 5 and 6 from Sample Midterm I. QR-decomposition with Gram-Schmidt. Projections onto nullspace and columnspace. Least square solutions. Problem 3 (vi)-(xii) from Sample Midterm I. (this problem did not work out well at the first midterm and is very important...) Finding eigenvalues and eigenvectors of a square matrix. (e.g. Problem 1 from Sample Midterm II.) Finding the spectral decomposition of a symmetric matrix. Solving systems of linear ordinary differential equations. (Problem from Sample Midterm II. This question again did not work out well...) Finding the singular value decomposition of a matrix (Problem 3 from Sample Midterm II.) Computing condition number and its application in error magnification. Importance of orthogonal matrices. (e.g. Problem 4 from Sample Midterm II.) Finding LU-decomposition by bookkeeping Gaussian elimination. (Problem 4 from Sample Midterm II.) Finding QR-decomposition with Householder and Givens. Sample Midterm II.) (Problems 5, 6, 7 from Jacobi and Gauss-Seidel iterations. Including: prediction of the number of steps needed (this latter part did not work out well...) e.g. Problem 9 from Sample Midterm II. Power method for eigenvalues (Problem 10 from Sample Midterm II.) QR-iteration for diagonalizing symmetric matrices (Problem 11 from Sample Midterm II.) 3

4 Calculus of curves. Tangent and principal unit normal vectors. Arclength. (Problems 1 3 below) Computing partial derivatives and gradients with its application (directional derivatives, tangent lines to level curves, tangent planes to graphs etc.). Problems 4 9 and below. Finding the Jacobian and applying chain rule for vector valued functions in several variables. (Problems 10 1 below) Newton s method in several variables. Finding all roots of complex equations. (Problem 13) Finding local and global extrema. Lagrange multiplier. (Problems 17 ) Evaluate multiple integrals. (Problem 3). 4

5 PROBLEMS PROBLEM 1. The curves r 1 (t) =(e t 1)i +sintj+ln(t+1)k and r (u) =(u+1)i+(u 1)j +(u 3 +1)k intersect at P (0, 0, 0). Find the angle of intersection. PROBLEM. Find the point(s) at which the curve r(t) =ti+t j+t 3 k intersects the plane 4x +y+z= 4 and find the angle of intersection. PROBLEM 3. Find the unit tangent vector, the principal normal vector and an equation in x, y, z of the osculating plane at the point on the curve r(t) =i+tj+t k corresponding to t =1. PROBLEM 4. Compute the arclength of the following curves a, b, r(t) =t 3 i+t j between 0 t 1 r(t) =(lnt)i+tj+t k between 1 t e PROBLEM 5. Find the gradient of the following functions f(x, y) =3x xy + y g(x, y) =(x+y)e x y h(x, y, z) =e x+y cos(z +1) PROBLEM 6. Find the directional derivative of ln(x + y )atthepoint(0,1) in the direction of 8i + j. 5

6 PROBLEM 7. Find the directional derivative of f(x, y, z) =x +xyz yz at (1, 1, ) in a direction parallel to the straight line x 1 = y 1= z 3 PROBLEM 8. a, Find the unit vector in the direction in which f(x, y, z) = x +y +z increases most rapidly at the point (1,, 1) and give the rate of change of f in this direction. b, Find the unit vector in the direction in which f(x, y, z) = x +y +z decreases most rapidly at the point (1,, 1) and give the rate of change of f in this direction. PROBLEM 9. a, Determine the path of the steepest descent along the surface z = x +3y starting from the point (1,, 13). b, Where does this path end? c, What is the projection of this path on the (x, y) plane? PROBLEM 10. Find the rate of change of f with respect to t along the given curve r(t) a, f(x, y) =x y,r(t)=e t i+e t j b, f(x, y, z) =y sin(x + z), r(t) =ti+ (cos t)j + t 3 k PROBLEM 11. Find the rate of change of u with respect to t in the following two problems: a, u = x +4 xy 3y where x = t 3, y = t 1 b, u = x tan y, wherex=s t,y=s+t. PROBLEM 1. Let f 1 (u 1,u )=u 1 u and f (u 1,u )=u 1+u u 1 (x 1,x,x 3 )=x 1 x x 3 6

7 a, Find the Jacobian of f = b, Find the Jacobian of f u c, What is f x 3? u (x 1,x,x 3 )=x +x 3 ( ) f1 f and the Jacobian of u = ( ) u1 u PROBLEM 13. Find a solution to the system of nonlinear equations x + y 3 = x + y = 3 with Newton s method starting from (x 0,y 0 )=(1,1). Run two steps of the iteration. Plug these approximate solutions to the original equations to see the error. PROBLEM 14. Find the normal vector and the tangent vector to the curve x 5 +y 5 = x 3 at the point (1, 1). Write up the equation of the normal line and the tangent line. (Recall that the equation of a line has two forms: normal form and parametric form. Write up both.) PROBLEM 15. a, Find an equation for the tangent plane of xy +z =1atthe point (1,,). b, Find the parametric equation of the normal line. c, Find the equation of the line in the tangent plane obtained in part a, which is parallel to the plane x + y + z = 3 and goes through (1,, ). PROBLEM 16. The curve r(t) =ti+3t 1 j t k and the ellipsoid x +y +3z =5 intersects at (, 3, ). What is the angle of intersection? PROBLEM 17. Find the critical points and local extreme values of the following functions. Decide whether the extreme point is local maximum or minimum. a, f(x, y) =x +xy +3y +x+10y+1 b, g(x, y) =xy + x 1 +8y 1 c, h(x, y) =(x y)(xy 1) d, k(x, y) =xy 1 yx 1 7

8 PROBLEM 18. Find the absolute maximum and minimum of the function f(x, y) = y(x 3) on the closed disk of radius 3 about the origin. PROBLEM 19. Find the absolute maximum and minimum of the function f(x, y) = (x+y ) on the closed square D = {(x, y) : 0 x 3, 0 y 3}. PROBLEM 0. Minimize x +y+4z on the sphere x + y + z =7. PROBLEM 1. Find the distance from the point (0, 1) to the parabola x =4y. PROBLEM. Show that all triangles inscribed in a circle of radius 1 the equilateral one has the largest product of the lengths of the sides. PROBLEM 3. Evaluate the following integrals on the given domain (e) (f) { } (a) xy 3 dxdy = (x, y) : 0 x 1,0 y x { (b) sin(x + y) dxdy = (x, y) : 0 x π,0 y π } { } (c) (x+3y 3 )dxdy = (x, y) : x +y 1 (d) ye x dxdy = {(x, y) : 0 y 1,0 x y } (x 4 +y ) dxdy = bounded region between y = x and y = x 3 e x dxdy = triangle between the x-axis, x = and x=y PROBLEM 4. Decide whether the following statements true (T) or false (F). If false, give a short argument. WARNING: Some of them are If... then statements. Such a statement is true only when the if part surely implies the then part. Such a statement is false when the then part is not satisfied even in one single case for which the if part is valid. (a) If a function f : R R has first partial derivatives at the origin, then f(0, 0) exists. 8

9 (b) If a function f : R R has continuous first partial derivatives in a neighborhood of the origin, then f(0, 0) exists. (c) If a function f : R R is continuous at the origin, then f(0, 0) exists. (d) If f(0, 0) exists for a function f : R R, then it is continuous at the origin. (e) Let U be an open set of R.Iff:U Rsatisfies f(x) = 0 for all x U, thenf is constant on U. (f) Let U be an open connected set of R. If f : U R satisfies f(x) = 0 for all x U, thenf=0onu. (g) Let U be an open connected set of R. If f : U R and g : U R are two functions whose gradients are equal at every point of U, then the two functions are equal. (h) If f : R R has a local extreme value at x 0 then f(x 0 )=0 (i) Let f : R R. If f(x 0 ) = 0, then f has a local extremum at x 0. (j) If the function f : R R has a local maximum at x 0 then x 0 is a critical point. (k) Let be a closed subset of R and f : Ra continuous function. Then f attains its maximum on (l) Let be an open subset of R and f : Ra continuous function. Then f attains its maximum on. (m) Let be a closed subset of R and f : Rbe a continuous function and let f(x 0 ) = 0 be the only stationary point in. Then f attains its maximum or minimum at x 0. (n) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If f x (x 0) > 0thenfhas a local minimum at x 0. (o) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If the discriminant D =(f xy ) (f xx )(f yy ) is positive at x 0,thenx 0 is a saddle. (p) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If the discriminant D =(f xy ) (f xx )(f yy ) is negative at x 0,thenx 0 is a local maximum. (q) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If all second partial derivatives vanish at x 0, then the function has no local extremum in x 0. 9

10 SOLUTIONS and PROBLEM 1. The curves r 1 (t) =(e t 1)i +sintj+ln(t+1)k r (u) =(u+1)i+(u 1)j +(u 3 +1)k intersect at P (0, 0, 0). Find the angle of intersection. SOLUTION TO PROBLEM 1. We have to compute the angle between the tangent vectors to these curves at the given point. First we determine the parameter values corresponding to P (0, 0, 0): it is clear that r 1 (0) = r ( 1) = P (0, 0, 0) (for example you solve e t 1=0andu+1=0). and i.e Then we compute the derivatives: r 1 (t) =et i+costj+ 1 t+1 k r (u) =i+uj+3u k u := r 1(0) = i +j+k and v := r ( 1) = i j +3k The cosine of the angle between these two vectors is hence θ = π/. cos θ = u v u v = =0 PROBLEM. Find the point(s) at which the curve r(t) =ti+t j+t 3 k intersects the plane 4x +y+z= 4 and find the angle of intersection. SOLUTION TO PROBLEM. The t-parameter value of the points of intersection is given by 4t +t +t 3 =4 10

11 This is a cubic equation, so no simple solution formula is available. One can apply Newton s method, but before that, it is advisable to check for simple (=rational) roots. The candidates for rational roots of t 3 +t +4t 4 = 0 are exactly the divisors of 4 (since the main coefficient is 1). i.e. ±1, ±, ±3, ±4, ±6, ±8, ±1, ±4. After a short trial you get that t = is a root (you might have seen this immediately). Dividing by the factor t wehave t 3 +t +4t 4 = (t )(t +4t+ 1) The quadratic equation t +4t+ 1 has no real roots, so the only intersection is at t =, at the point P (, 4, 8). The tangent vector to r(t) at this point is from r (t) =i+tj+3t k. 1 u:= r () = i +4j+1k= The normal vector to the given plane is n = (coefficients of the equation). The 1 angle θ between u and n is obtained from cos θ = u n u n = i.e. θ 1.15(rad). However, this is the angle between the tangent to the curve and the normal to the plane. The angle of intersection is π 1.15 = 0.4. (This depends a bit on the interpretation, but if the curve happens to lie within the plane, then we would rather say that the angle of intersection is 0 and not π/). PROBLEM 3. Find the unit tangent vector, the principal normal vector and an equation in x, y, z of the osculating plane at the point on the curve r(t) =i+tj+t k corresponding to t =1. SOLUTION TO PROBLEM 3. We have r (t)=j+tk hence T(t) := r (t) r (t) = j+tk 4t +4 = j+tk t +1 11

12 (always simplify as much as possible!!!) Thus at t =1, T(1) = j + k For the normal vector we need T (t) = k t +1 t tj+1k (t +1) 3/(j+tk)= (t +1) 3/ (Chain rule!! Imagine T(t) = 1 t (j + tk)). Evaluate it at the point +1 T (1) = j + k 3/ We have to normalize this vector. Before we divide by its norm, recall that we are allowed to multiply it by any positive number, it will not change the normalized vector. Hence we will rather normalize j + k, i.e. N(1) = j + k Finally, the osculating plane goes through the point r(1) = (1,, 1) and its normal vector is n := T(1) N(1) = (1, 0, 0). Hence the equation is ( x 1 ) n y =0, i.e. x 1=0 z 1 PROBLEM 4. Compute the arclength of the following curves a, b, r(t) =t 3 i+t j between 0 t 1 r(t) =(lnt)i+tj+t k between 1 t e SOLUTION TO PROBLEM 4. a, Compute r (t) =3t i+tj 1

13 Hence 1 r (t) dt = t 4 +4t dt = 1 0 t 9t +4dt Use substitution: u =9t +4,thendu =18tdt and don t forget to change the limits: 1 0 t 9t +4dt = udu= (13 3/ 4 3/) = 1 ( 13 ) b, We compute r (t) = 1 t i+j+tk Hence e 1 r (t) dt = = e 1 e 1 e t +4+4t dt = 1 [ (t 1 +t)dt = ln t + t ] e (t 1 +t) dt 1 = e PROBLEM 5. Find the gradient of the following functions f(x, y) =3x xy + y g(x, y) =(x+y)e x y h(x, y, z) =e x+y cos(z +1) SOLUTION TO PROBLEM 5. f(x, y) =(6x y, x +1) ( ) g(x, y) = e x y (1 + x + y), e x y (1 x y) h(x, y, z) =e x+y( ) cos(z +1), cos(z +1), zsin(z +1) PROBLEM 6. Find the directional derivative of ln(x + y )atthepoint(0,1) in the direction of 8i + j. SOLUTION TO PROBLEM 6. The gradient of f(x, y) =ln(x +y )is At the given point (x, y) =(0,1): f(x, y) = ( x y ) x +y, x +y f(0, 1) = (0, ) = j. 13

14 The unit vector in the given direction is u = 8i + j 8i + j = 8i + j 8 +1 = 8 i + 1 j The directional derivative f u(0, 1) = f(0, 1) u = 65. PROBLEM 7. Find the directional derivative of f(x, y, z) =x +xyz yz at (1, 1, ) in a direction parallel to the straight line x 1 = y 1= z 3 SOLUTION TO PROBLEM 7. at the given point f(x, y, z) =(x+yz,xz z, xy yz), f(1, 1, ) = (6, 0, ). The direction is given in the parametric form of a straight line. Although the parameter is not explicitly written, you should view the common value of the given three fractions as the parameter, say t. Hence the line is t = x 1 t = y 1 t = z 3 where t R. Expressing x, y, z we obtain the more familiar parametric form: x =t+1 y=t+1 z= 3t+ We need only the direction of this line, which is obtained from the coefficients of t, i.e. it is the vector (, 1, 3). The unit vector in this direction is u = i + j 3k i + j 3k = i + j 3k 14 or its opposite, u (there are two unit vectors parallel with a given line). Hence the directional derivative is f u(1, 1, ) = f(1, 1, ) u =

15 or its opposite, which is PROBLEM 8. a, Find the unit vector in the direction in which f(x, y, z) = x +y +z increases most rapidly at the point (1,, 1) and give the rate of change of f in this direction. b, Find the unit vector in the direction in which f(x, y, z) = x +y +z decreases most rapidly at the point (1,, 1) and give the rate of change of f in this direction. SOLUTION TO PROBLEM 8. a, Compute the gradient f(x, y, z) = xi+yj+zk x +y +z at the given point f(1,, 1) = i j +k 7 This gives the direction of biggest increase, but it is not a unit vector. We have to find its norm f(1,, 1) = 3 7 Hence the unit vector in the direction of biggest increase is i j +k 3 f f at the given point, i.e. The rate of change in this direction is the norm of the gradient, i.e. it is 3 7. b, The biggest decrease is in the opposite direction as the biggest increase and the rate of change is also the opposite. Hence the answer to part b, is i j+k 3 for the direction and 3 7 for the rate. PROBLEM 9. a, Determine the path of the steepest descent along the surface z = x +3y starting from the point (1,, 13). b, Where does this path end? c, What is the projection of this path on the (x, y) plane? SOLUTION TO PROBLEM 9 a, The path of steepest descent r(t) =x(t)i+y(t)j (with t being a real parameter along the path) has the property that its derivative is always 15

16 parallel with the gradient of the given function f(x, y) =x +3y, and it points in the opposite direction. Since f(x, y) =(x, 6y) the coordinate functions x(t),y(t) of the curve r(t) must satisfy the following relations x (t) = x(t) y (t)= 6y(t) In addition, the curve starts at the point x(0) = 1,y(0) = (you can always choose the parametrization in such a way that the given point correspond to t =0). Hencewehave to find a function x(t) such that x (t) = x(t)andx(0) = 1. The relation x (t) = x(t) is satisfied by all functions of the form x(t) =Ce t with arbitrary constant C, whichis fixed by the condition x(0) = 1. Hence C =1andx(t)=e t Similarly you find that y(t) =e 6t. The path of steepest descent is r(t) =(e t,e 6t ) with a parameter t 0. (If you go in the opposite direction, i.e. you run the path for t<0, or equivalently you consider the path (e t, e 6t )fort>0, then you get the path of steepest ascent.) b, After a very long time (t ) the particle on this path gets arbitrarily close to the origin. This is the other endpoint of the path. c, To see the projection of this curve onto the (x, y) plane, then you have to express y =e 6t in terms of x = e t, which is clearly y =x 3. Only that segment of this curve counts which is covered by the regime t 0, i.e. x (0, 1]. Hence the projection of this path is the piece of the graph of the function y =x 3 in the (x, y) plane between (1, ) and (0, 0). PROBLEM 10. Find the rate of change of f with respect to t along the given curve r(t) a, f(x, y) =x y,r(t)=e t i+e t j b, f(x, y, z) =y sin(x + z), r(t) =ti+ (cos t)j + t 3 k SOLUTION TO PROBLEM 10. We need to compute a, Here d dt f(r(t)) = f(r(t)) r (t) f(x, y) =(xy, x )=xyi + x j r (t)=e t i e t j 16

17 hence the rate of change is e t e t e t (e t ) e t = e t (Don t forget to express x, y in terms of t; the final result must contain only t, sincewe asked for the rate of change with respect to this variable.) b, Here hence the rate of change is f(x, y, z) =y cos(x + z)i +ysin(x + z)j + y cos(x + z)k r (t) =i (sin t)j +3t k cos tcos(t + t 3 ) costsin t sin(t + t 3 )+3t cos t cos(t + t 3 ) PROBLEM 11. Find the rate of change of u with respect to t in the following two problems: a, u = x +4 xy 3y where x = t 3, y = t 1 b, u = x tan y, wherex=s t,y=s+t. SOLUTION TO PROBLEM 11. a, Here u depends only on t (via x and y). The rate of change is du dt = u dx x dt + u dy y dt ( y ) ( x ) = 1+ 3t + x y 3 ( t ) =(1+t )3t +(t 3)( t )=3t +4+3t by b, Here u depends on both variables s, t. The rate of change with respect to t is given u t = u x x t + u y =(xtan y)s +(x sec y)t =s 4 ttan(s + t )+s 4 t 3 sec (s + t ) (Notice the difference between the notation du u dt and t. The first notation is used for simple derivative, when u depends only on the variable t. Ifudepends on several variables, we use partial derivative notation). the u t y t 17

18 PROBLEM 1. Let f 1 (u 1,u )=u 1 u and f (u 1,u )=u 1+u u 1 (x 1,x,x 3 )=x 1 x x 3 a, Find the Jacobian of f = b, Find the Jacobian of f u c, What is f x 3? u (x 1,x,x 3 )=x +x 3 ( ) f1 f and the Jacobian of u = ( ) u1 u SOLUTION TO PROBLEM 1. a, ( ) u u J f (u 1,u )= 1 u 1 1 ( ) 1 x3 x J u (x 1,x,x 3 )= 0 x 1 b, J f u (x 1,x,x 3 )= = [ ][ ] J f (u 1,u ) J u (x 1,x,x 3 ) = ( u u 1 u 1 1 )( 1 x3 ) x 0 x 1 ( x )( ) = +x 3 x 1 x x 3 1 x3 x (x 1 x x 3 ) 1 0 x 1 ( x +x 3 x 3 (x +x 3)+x (x 1 x x 3 ) x 3 +x 1 x x 3 ) (x 1 x x 3 ) x 3 (x 1 x x 3 )+x x (x 1 x x 3 )+1 c, f x 3 is just the third entry of the second row of J f u (x 1,x,x 3 ), i.e. f x 3 = x (x 1 x x 3 )+1 You could have obtained it of course without computing the full Jacobian: f x 3 = f u 1 u 1 x 3 + f u u x 3 =u 1 ( x )+1 1= x (x 1 x x 3 )+1 18

19 PROBLEM 13. Find a solution to the system of nonlinear equations x + y 3 = x + y = 3 with Newton s method starting from (x 0,y 0 )=(1,1). Run two steps of the iteration. Plug these approximate solutions to the original equations to see the error. SOLUTION TO PROBLEM 13. We have to find the root of ( x+y f(x, y) = 3 ) x +y 3 The Jacobian is ( 1 3y ) J f (x, y) = x 1 ( ) 1 Let x 0 = be the initial guess. Then we have to solve 1 ) J f (x 0 ) (x x 0 = f(x 0 ) i.e. ( ) 1 3 ( ( ) 1 ) ( 0 x = 1 1 1) since J f (1, 1) = ( ) ( ) 0 and f(1, 1) = 1 ( 1 1 ),andthe The solution will be the first approximation, x 1. Temporarily let u = x solution of ( ) ( u = 1 1) requires solving the system u 1 + 3u = 0 u 1 + u = 1 You can use your favorite method to do that, the result is u 1 =3/5, u = 1/5, hence ( ) ( ) ( ) ( ) 3/5 1 8/5 1.6 x 1 = + = = 1/5 1 4/5 0.8 This is the result of the first Newton iteration. In the second step we have ( 1 3(0.8) ) J f (x 1 )= = (1.6) 1 ( 1 )

20 and andwehavetosolve f(x 1 )= ( ) ( x 3. 1 ( ) 1.6 = ( ) 1.6 ) = 0.8 ( ) ( ) The solution of u u = u 1 + u = 0.36 is u 1 = 0.115, u = Hence x = This is second approximation. ( ) ( ) ( ) = Let s plug back x 0, x 1 and x to the original equation. This is the same as computing the value of f on these points; the true solution would give zero. Here we have f(x 0 )= ( ) 0 1 f(x 1 )= so after two steps we are already pretty close. ( ) f(x )= ( ) PROBLEM 14. Find the normal vector and the tangent vector to the curve x 5 +y 5 = x 3 at the point (1, 1). Write up the equation of the normal line and the tangent line. (Recall that the equation of a line has two forms: normal form and parametric form. Write up both.) SOLUTION TO PROBLEM 14. Represent the curve as a level curve of a function, hence choose f(x, y) =x 5 +y 5 x 3. Compute the gradient at the given point f(x, y) =(5x 4 6x )i+5y 4 j f(1, 1) = i +5j Hence the normal vector is n = i +5j (not normalized), the tangent vector is t =5i+j. Notice that the orientation of these two vectors are not determined, i.e. i 5j would also be a normal vector and 5i j is also a tangent vector. The equation of the tangent line is n (x x 0 ) = 0 where x 0 = (1,1), i.e. it is x +5y 4=0. 0

21 The equation of the normal line is t (x x 0 ) = 0 i.e. it is 5x + y 6=0. Be careful: if you use the normal equation v (x x 0 ) = 0 for these lines (where v is the vector orthogonal to the line), then you have to use v = n for the tangent line and v = t for the normal line. It sounds a bit confusing but think about it: the normal line is orthogonal to the tangent vector and the tangent line is orthogonal to the normal vector. You can equivalently use the parametric equation of a line. If e =(e 1,e ) is a vector along the line, then the equation has the form x = et + x 0, or in coordinates x = e 1 t + x 0 y = e t + y 0 where t is a parameter along the line. If you use this form, then you have to use the normal vector for the normal line, i.e. x = t +1 y=5t+1 and the tangent vector for the tangent line x =5t+1 y=t+1 You can check that the normal equation 5x + y 6 = 0 and the parametric equation (x, y) =( t+1, 5t+ 1) of the normal line really determine the same line: 5( t +1)+(5t+1) 6=0 Similarly you can check that the two given forms of the tangent lines are equivalent. PROBLEM 15. a, Find an equation for the tangent plane of xy +z =1atthe point (1,,). b, Find the parametric equation of the normal line. c, Find the equation of the line in the tangent plane obtained in part a, which is parallel to the plane x + y + z = 3 and goes through (1,, ). SOLUTION TO PROBLEM 15. a, View the given surface as the level surface of f(x, y, z) =xy +z (at level c = 1). f(x, y, z) =y i+xyj +4zk f(x 0 )= f(1,, ) = 4i +4j+8k The tangent plane is given by f(1,, ) (x x 0 )wherex 0 =(1,,), and it is 4(x 1) + 4(y ) + 8(z ) = 0 1

22 i.e. x + y +z=7. b, The parametric equation of the normal line is given as x = t f(x 0 )+x 0, hence it is x =4t+1, y =4t+, z =8t+ where t R (you can reparametrize it, and use s =4tas a new parameter to get slightly simpler formulas: x = s +1,y =s+,z=s+). c, The vector in the direction of the line in question must be orthogonal both to the normal vector of the plane x + y + z = 3, hence to (1, 1, 1) and must also be orthogonal to the gradient, i.e. to (4, 4, 8). Hence this vector is given by the vector product of these two vectors, i.e. it is e =(4, 4,0). The parametric equation of this line is x =4t+1, y = 4t+, z = PROBLEM 16. The curve r(t) =ti+3t 1 j t k and the ellipsoid x +y +3z =5 intersects at (, 3, ). What is the sine of the angle of intersection? SOLUTION TO PROBLEM 16. The given point correspond to the t = 1 parameter value. The tangent vector to this curve is the value of r (t) at this point, i.e. from r (t) =i 3t j 4tk r (1) = i 3j 4k The normal vector to the surface at this point is given by the gradient of f(x, y, z) = x +y +3z (whose level curve is the surface). f(x, y, z) =(x, y,6z) f(, 3, ) = (4, 6, 1) = 4i +6j 1k The angle θ between the normal vector to the surface and the tangent vector to the curve satisfies i 3j 4k cos θ = i 3j 4k 4i +6j 1k 4i +6j 1k = = (in order to simplify the calculation, you could have considered half of the gradient vector i + 3j 6k because only its direction matters). The angle of intersection is the angle between the tangent plane to the ellipsoid and the tangent vector to the curve, hence it is π 19 θ. Its sine is the same as cos θ = 7. 9

23 (You can easily compute the angles with a calculator by taking inverse cosine, but calculator is not allowed on the exam, hence I asked for the sine instead of the angle itself.) PROBLEM 17. Find the critical points and local extreme values of the following functions. Decide whether the extreme point is local maximum or minimum. a, f(x, y) =x +xy +3y +x+10y+1 b, g(x, y) =xy + x 1 +8y 1 c, h(x, y) =(x y)(xy 1) d, k(x, y) =xy 1 yx 1 SOLUTION TO PROBLEM 17. a, The function is differentiable everywhere, the critical points are the stationary points (where the gradient vanishes). Since we have to solve the system of equations f(x, y) =(x+y+, x+6y+ 10) x + y = x + 6y = 10 We easily find that (x, y) =(1, ) is the only critical point. To check its type, we use the second partial derivative test. We compute A = f x = B= f x y = C= f y =6 hence D = 6<0andA>0, so it is a local minimum. The value of the function at this point is 8. b, The function is differentiable everywhere on its domain, hence the only critical points are the stationary points. Since g(x, y) =(y x )i+(x 8y )j, this is zero only at (x, y) =( 1,4). The second partials at this point are f xx =x 3 =16 f xy =1 f yy =16y 3 = 1 4 hence D = fxy f xx f yy = 3. Since A>0, we have a local minimum, and its value is 6. c, All critical points are stationary points. Multiply out the given expression: h(x, y) = x y xy x + y h(x, y) =(xy y 1)i +(x xy +1)j 3

24 We have to solve xy y = 1 x xy = 1 This is not a linear equation, but it is fairly simple, so one can try to solve it with some pedestrian way. The easiest if you add the two equations, then you get x y = 0, i.e. (x y)(x + y) = 0. Hence either x = y or x = y. In the first case from the first equation we have y y = 1, i.e. x = y = ±1, in the second case the first equation becomes 3y = 1 which has no solution. Hence there are two critical points: (1, 1) and ( 1, 1). The second partials are h xx =y h xy =x y h yy = x and D = h xy h xx h yy is positive at both points. Hence both stationary points are saddle. d, Again, the function is differentiable at every point of the domain. k(x, y) = x +y +y x y i x xy j This gradient is never zero. The only possibility would be if the numerator x + y were zero, but that would mean x = y = 0 which is not in the domain of k. Hence no critical points, no stationary points, no local extrema. PROBLEM 18. Find the absolute maximum and minimum of the function f(x, y) = y(x 3) on the closed disk of radius 3 about the origin. SOLUTION TO PROBLEM 18. The function is differentiable, hence all critical points are stationary points. First compute the gradient to see if the stationary points are possible candidates for absolute minimum or maximum. Since f(x, y) =(y,x 3), the only stationary point is at (3, 0) which is not in the interior of the domain. Hence it does not count (recall local max. or min. must be in the interior of the domain). Next, we have to check the boundary points. We give two solutions. First solution: Parametrize the boundary, i.e. the circle of radius 3 about the origin. The most convenient parametrization is Therefore the function on the boundary is r(t) = (3 cos t, 3sint) t [0, π) f(r(t)) = 3 sin t(3 cos t 3) 4

25 Now we have to minimize and maximize the function g(t) =9sint(cos t 1) of one variable on the closed interval [0, π] (you can add the other endpoint, since g(0) = g(π) anyway). Compute g (t) =9cost(cos t 1) 9sin t = 9( cos t cos t 1) (notice the trick: using sin t +cos = 1 we expressed everything in terms of cos t). The equation g (t) =0is a quadratic equation for z =cost,i.e. wehavetosolvez z 1 = 0, which gives cos t = z =1, 1. There is one point corresponding to cos t =1,t=0,thisisA=(3,0) on the circle. There are two points corresponding to cos t = 1,namelyt= π 3, π 3, i.e. B =( 3,3 3 )andc=( 3, 3 3 ). We can easily compute that f(a) =0,f(B)= and f(c) = Hence B is the absolute minimum point and C is the absolut maximum point with values given above. Second solution: One can find the extremum points and values on the circle by Lagrange s method. Let g(x, y) =x +y, then the circle can be viewed as a constraint g(x, y) = 9 and we have to maximize and minimize f(x, y) under this condition. Since g(x, y) =(x, y), we have y = λx x 3=λy in addition to the constraint x + y = 9. From these two equations we get x 3=4λ x, i.e. 3 x = 1 4λ y = 6λ 1 4λ Putting this back into the constraint i.e. ( 3 1 4λ ) + ( 6λ 1 4λ ) =9 1+4λ =(1 4λ ) It is clearly an equation for c =4λ, hence first we solve 1 + c =(1 c),orc 3c=0, which gives c =0,3. If c =0,thenλ=0and(x, y) =(3,0) giving the point A in the first solution. If c =3,thenλ = 3 4, which gives two solutions λ = ± 3. It is easy to see that these give the points B and C above. Finally we have to compute the value of the function at these three points, as above. PROBLEM 19. Find the absolute maximum and minimum of the function f(x, y) = (x+y ) on the closed square D = {(x, y) : 0 x 3, 0 y 3}. SOLUTION TO PROBLEM 19. First we look for local extrema. The function is differentiable everywhere, hence we look for stationary points. Since f(x, y) =(x+y )(i + j), all points on the line y = xare stationary points. (Yes, it could happen that 5

26 there are infinitely many stationary points). The value of f is zero at all of these points. Since f 0 always (it s a square of something), all these points which fall in the iterior of D are local minima. Hence the local minima are (x, x) for all 0 <x< and these are clearly absolute minima as well. Since we exhausted all stationary points, there are no local maxima. To find the absolute extrema, one has to look at the boundary. It is clear that there are two more absolute minima on the boundary, (0, ) and (, 0) (the endpoints of the line y = x), which do not count as local minima since they are not interior points, but they count as absolute minima. For the absolute maxima in principle you should parametrize the boundary of the square and maximize four functions. However, there is a shorter way if you think a bit: (x+y ) is clearly the biggest if either x + y is the biggest or it is the smallest (in principle it could be negative). The biggest value of x + y is 6 at (x, y) =(3,3) and the smallest value is 0 at (0, 0). The point (3, 3) gives the absolute maximum of f with value 16, while the point (0, 0) does not give maximum (the value of the function is only 4). PROBLEM 0. Minimize x +y+4z on the sphere x + y + z =7. SOLUTION TO PROBLEM 0. Since the sphere is closed, bounded and f(x, y, z) = x+y+4z is continuous, the minimum exists. Let g(x, y, z) =x +y +z, then the sphere is a level curve of g. Compute f =(1,,4) g =(x, y,z) and from f = λ g we have 1 = λx, =λy and 4 = λz. Clearly λ 0 and after eliminating it we get express everything in terms of x as y =x,z=4x.fromg(x, y, z) =7 we get 1x = 7, i.e. x = ± 1 3, y = ± 3, z = ± 4 3 and the minimum of f(x, y, z) is clearly attained when you choose the minus signs. The value of the minimum is 7 3. PROBLEM 1. Find the distance from the point (0, 1) to the parabola x =4y. SOLUTION TO PROBLEM 1. The distance is attained at the point closest to the parabola from (0, 1). Such point clearly exists. Also, it is simpler the minimize the distance square than the distance itself (to get rid of the square root). Hence we minimize f(x, y) =x +(y 1) under the constraint g(x, y) =x 4y= 0. Compute f =(x, (y 1)) g =(x, 4) hence from f = λ g we have x =λxand y = 4λ. We can express λ =1 y from the second equation and insert into the first: x =x(1 y) orxy = 0. Hence either 6

27 x or y is zero, but then the other one is also zero from the constraint x =4y. So the closest point is (x, y) =(0,0) and the minimum distance square is f(0, 0) = 1. PROBLEM. Show that all triangles inscribed in a circle of radius 1 the equilateral one has the largest product of the lengths of the sides. SOLUTION TO PROBLEM. Let O be the center of the circle and A, B, C be the vertices of the triangle. The angles between the radii from O to the vertices are x, y, z. The constraint is g(x, y, z) =x+y+z=π. The lengths of the sides are sin(x/), sin(y/) and sin(z/) from basic trigonometry. We have to maximize f(x, y, z) =8sin x sin y sin z and clearly we can assume that none of x, y, z is zero. We compute f =4sin x sin y sin z ( cot x, cot y, cot z ) and g =(1,1,1). Hence from f = λ g weseethatcot x =coty =cotz. Since x, y, z are between 0 and π, the half angles are between 0 and π and in this interval the cot function is one-to-one. Hence it has well defined inverse, so from the cotangents being equal it follows that x/ = y/ = z/, i.e. the triangle is equilateral. PROBLEM 3. Evaluate the following integrals on the given domain (e) (f) { } (a) xy 3 dxdy = (x, y) : 0 x 1,0 y x { (b) sin(x + y) dxdy = (x, y) : 0 x π,0 y π } { } (c) (x+3y 3 )dxdy = (x, y) : x +y 1 (d) ye x dxdy = {(x, y) : 0 y 1,0 x y } (x 4 +y ) dxdy = bounded region between y = x and y = x 3 e x dxdy = triangle between the x-axis, x = and x=y SOLUTION TO PROBLEM 3. 7

28 (a) 1 x 1 [ y xy 3 dxdy = xy 3 4 ] x dy dx = x dx = 1 1 x 5 dx = (b) π/ π/ π/ sin(x + y) dxdy = sin(x + y) dydx = [ cos(x + y)] π/ 0 dx π/ [ ( = cos x cos x + π )] dx = 0 (c) 1 (x +3y 3 1 y 1 )dxdy = (x +3y 3 )dxdy = 6y 1 3 y dy =0 1 1 y 1 since the integrand is odd on a symmetric interval. (d) ye x dxdy = 1 y 0 0 ye x dxdy = 1 0 [ e y y(e y 1)dy = y ] 1 0 = e (e) (x 4 + y ) dxdy = = 1 0 ( 4x 6 1 x 0 3 x7 x9 3 (x 4 + y ) dydx = x 3 ) dx = 1 0 [x 4 y + y3 ] x 3 x 3dx [ 4x 7 1 x8 8 x10 ] 1 30 = (f) e x dxdy = x/ 0 0 e x dydx = 0 1 [ 1 xex dx = 4 ex] = e PROBLEM 4. Decide whether the following statements true (T) or false (F). If false, give a short argument. WARNING: Most of them are If... then statements. Such a statement is true only when the if part surely implies the then part. Such a statement is false when the then part is not satisfied even in one single case for which the if part is valid. (a) If a function f : R R has first partial derivatives at the origin, then f(0, 0) exists. (b) If a function f : R R has continuous first partial derivatives in a neighborhood of the origin, then f(0, 0) exists. 8

29 (c) If a function f : R R is continuous at the origin, then f(0, 0) exists. (d) If f(0, 0) exists for a function f : R R, then it is continuous at the origin. (e) Let U be an open set of R.Iff:U Rsatisfies f(x) = 0 for all x U, thenf is constant on U. (f) Let U be an open connected set of R. If f : U R satisfies f(x) = 0 for all x U, thenf=0onu. (g) Let U be an open connected set of R. If f : U R and g : U R are two functions whose gradients are equal at every point of U, then the two functions are equal. (h) If f : R R has a local extreme value at x 0 then f(x 0 )=0 (i) Let f : R R. If f(x 0 ) = 0, then f has a local extremum at x 0. (j) If the function f : R R has a local maximum at x 0 then x 0 is a critical point. (k) Let be a closed subset of R and f : Ra continuous function. Then f attains its maximum on (l) Let be an open subset of R and f : Ra continuous function. Then f attains its maximum on. (m) Let be a closed subset of R and f : Rbe a continuous function and let f(x 0 ) = 0 be the only stationary point in. Then f attains its maximum or minimum at x 0. (n) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If f x (x 0) > 0thenfhas a local minimum at x 0. (o) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If the discriminant D =(f xy ) (f xx )(f yy ) is positive at x 0,thenx 0 is a saddle. (p) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If the discriminant D =(f xy ) (f xx )(f yy ) is negative at x 0,thenx 0 is a local maximum. (q) Let f : R R have continuous second partial derivatives and let x 0 be a critical point. If all second partial derivatives vanish at x 0, then the function has no local extremum in x 0. SOLUTION TO PROBLEM 4. (a) F. The function even does not have to be continuous, for example let f(x, y) =0if 9

30 x =0ory=0andf(x, y) = 1 otherwise. (b) T. (c) F. Continuity does not imply any differentiability. (d) T. (e) F. The set has to be connected. Otherwise you can choose different values of f on different components of f. (f) F. f can be constant, it does not have to be zero. (g) F. The functions can differ by a additive constant. (h) F. The gradient may not exist (i) F. Saddle is a counterexample. (j) T. (k) T. (l) F. Think of the paraboloid f(x, y) =x +y ontheopenunitdisk. (m) F. The extrema can be on the boundary, For example think of the saddle x y on the closed unit disk. (n) F. You also need the discriminant to be negative. (o) T. (p) F. It can be local minimum. (q) F. Think of f(x, y) =x 4 +y 4 which has local minimum at the origin but all second partials vanish. 30