# SOLUTIONS. f x = 6x 2 6xy 24x, f y = 3x 2 6y. To find the critical points, we solve

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1 SOLUTIONS Problem. Find the critical points of the function f(x, y = 2x 3 3x 2 y 2x 2 3y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Partial derivatives To find the critical points, we solve f x = 6x 2 6xy 24x, f y = 3x 2 6y. f x = = x 2 xy 4x = = x(x y 4 = = x = or x y 4 = f y = = x 2 + 2y =. When x = we find y = from the second equation. In the second case, we solve the system below by substitution x y 4 =, x 2 + 2y = = x 2 + 2x 8 = = x = 2 or x = 4 = y = 2 or y = 8. The three critical points are (,, (2, 2, ( 4, 8. To find the nature of the critical points, we apply the second derivative test. We have At the point (, we have A = f xx = 2x 6y 24, B = f xy = 6x, C = f yy = 6. f xx = 24, f xy =, f yy = 6 = AC B 2 = ( 24( 6 > = (, is local max. Similarly, we find since and since The function has no global min since and similarly there is no global maximum since (2, 2 is a saddle point AC B 2 = (2( 6 ( 2 2 =< ( 4, 8 is saddle AC B 2 = ( 24( 6 (24 2 <. lim f(x, y = y,x= lim f(x, y =. x,y=

2 Problem 2. Determine the global max and min of the function f(x, y = x 2 2x + 2y 2 2y + 2xy over the compact region x, y 2. We look for the critical points in the interior: f = (2x 2 + 2y, 4y 2 + 2x = (, = 2x 2 + 2y = 4y 2 + 2x = = y =, x =. However, the point (, is not in the interior so we discard it for now. We check the boundary. There are four lines to be considered: the line x = : f(, y = 3 + 2y 2 4y. The critical points of this function of y are found by setting the derivative to zero: the line x = : y (3 + 2y2 4y = = 4y 4 = = y = with f(, =. f(, y = 2y 2. Computing the derivative and setting it to we find the critical point y =. The corresponding point (, is one of the corners, and we will consider it separately below. the line y = : f(x, = x 2 2x. Computing the derivative and setting it to we find 2x 2 = = x =. This gives the corner (, as before. the line y = 2: f(x, 2 = x 2 + 2x + 4 with critical point x = which is again a corner. Finally, we check the four corners The values of the function f are (,, (,, (, 2, (, 2. f(, = 3, f(, =, f(, 2 = 3, f(, 2 = 7. From the boxed values we select the lowest and the highest to find the global min and global max. We conclude that global minimum occurs at (, Problem 3. Using Lagrange multipliers, optimize the function global maximum occurs at (, 2. f(x, y = x 2 + (y + 2

3 subject to the constraint 2x 2 + (y 2 8. We check for the critical points in the interior The second derivative test shows this a local minimum with f x = 2x, f y = 2(y + = (, is a critical point. f xx = 2, f yy = 2, f xy = f(, =. We check the boundary g(x, y = 2x 2 + (y 2 = 8 via Lagrange multipliers. We compute Therefore In the first case x = we get with values f = (2x, 2(y + = λ g = λ(4x, 2(y. 2x = 4xλ = x = or λ = 2 2(y + = 2λ(y. g(, y = (y 2 = 8 = y = + 3 2, 3 2 f(, = ( , f(, 3 2 = ( In the second case λ = 2 we obtain from the second equation 2(y + = y = y = 3. Now, g(x, y = 8 = x = ±. At (±, 3, the function takes the value f(±, 3 = (± 2 + ( = 5. By comparing all boxed values, it is clear the (, is the minimum, while (, is the maximum. Problem 4. Consider the function w = e x2 y where x = u v, y = uv 2. Using the chain rule, compute the derivatives u, v.

4 We have x = 2xy exp(x2 y = 2u v ( uv 2 exp u 2 v uv 2 = 2 ( u v 3/2 exp v ( u y = x2 exp(x 2 y = u 2 v exp v x u = x v, v = u 2 v Thus Similarly, Problem 5. y u = u 2 v 2, u = x x u + y y u = 2 y v = 2 uv 3. v ( u u 2 v exp v ( u v 3/2 exp v = 2 ( u v exp ( u v v exp = ( u v v exp v. u = x x u + y y u = u ( u v 2 exp. v (i For what value of the parameter a, will the planes ax + 3y 4z = 2, x ay + 2z = 5 be perpendicular? (ii Find a vector parallel to the line of intersection of the planes (iii Find the plane through the origin parallel to (iv Find the angle between the vectors (v A plane has equation For what values of a is the vector normal to the plane? (i The normal vectors to the two planes are x y + 2z = 2, 3x y + 2z =. z = 4x 3y + 8. v = (,, 2, w = (, 3,. z = 5x 2y + 7. (a,, 2 n = (a, 3, 4, n 2 = (, a, 2. u 2 v 2 = The planes are perpendicular if n, n 2 are perpendicular. We compute the dot product n n 2 = = a + 3 ( a + ( 4 2 = = 2a 8 = = a = 4.

5 (ii The vectors normal to the two planes are n = (,, 2, n 2 = (3,, 2. The line of intersection will be perpendicular to both n, n 2. But so is the cross product. Thus the line of intersection will be parallel to the cross product n n 2 = (,, 2 (3,, 2 = (, 4, 2. (iii The second plane must have the same normal vector hence the same coefficients for x, y, z. Since it passes through the origin, the equation is z = 4x 3y. (iv We compute the angle using the dot product cos θ = (v The plane has the equation v w v w = 2 = x 2y z = 7 = 5 2 x + y + 2 z = 7 2 hence a normal vector is ( 5 2,, 2. Comparing with the vector we are given, we see that Problem 6. a = 5 2. (i Compute the second degree Taylor polynomial of the function f(x, y = e x2 y around (,. (ii Compute the second degree Taylor polynomial of the function f(x = sin(x 2 around x = π. (iii The second degree Taylor polynomial of a certain function f(x, y around (, equals 4x 2 2(y 2 + 3x(y. Can the point (, be a local minimum for f? How about a local maximum? (i After computing all derivatives and substituting, we find the answer + 2(x (y + 3(x (y 2 2(x (y. (ii We have f( π =. The first derivative is The second derivative is The Taylor polynomial is f x = 2x cos x 2 = f x ( π = 2 π cos π = 2 π. f xx = 2 cos x 2 2x sin x 2 = f xx ( π = 2. 2 π(x π (x π 2 = x 2 + π.

6 (iii From the Taylor polynomial we find f x (, = f y (, = so (, is a critical point. We can find the second derivatives 2 f xx(, = 4, 2 f yy(, = 2, f xy (, = 3. By the second derivative test Problem 7. AC B 2 = ( 8( >, A = 8 < = (, is a local maximum. (i The temperature T (x, y in a long thin plane at the point (x, y satisfies Laplace s equation Does the function T xx + T yy =. T (x, y = ln(x 2 + y 2 satisfy Laplace s equation? (ii For the function f(x, y = sin(x 2 + y 2 ln(x 4 y 4 + tan(xy is it true that (i We compute Therefore, T x = T y = f xyxyy = f yyxyx? 2x x 2 + y 2 = T xx = 2y2 2x 2 (x 2 + y 2 2 2y x 2 + y 2 = T yy = 2x2 2y 2 (x 2 + y 2 2. T xx + T yy = 2y2 2x 2 (x 2 + y x2 2y 2 (x 2 + y 2 2 =. (ii The two derivatives are equal as the order in which derivatives are computed is unimportant. Problem 8. Consider the function f(x, y = x2 y 4. (i Carefully draw the level curve passing through (,. On this graph, draw the gradient of the function at (,. (ii Compute the directional derivative of f at (, in the direction u = ( 4 5, 3 5. Use this calculation to estimate f((, +.u. (iii Find the unit direction v of steepest descent for the function f at (,. (iv Find the two unit directions w for which the derivative f w =.

7 (i The level is f(, =. The level curve is f(x, y = f(, = = x 2 = y 4 = x = ±y 2. The level curve is a union of two parabolas through the origin. The gradient ( 2x f = y 4, 4x2 y 5 = f(, = (2, 4 is normal to the parabolas. (ii We compute ( 4 f u = f u = (2, 4 5, 3 = 4. 5 For the approximation, we have f(, = and f((, +.u f(, +.f u = +. 4 =.4. (iii The direction of steepest decrease is opposite to the gradient. We need to divide by the length to get a unit vector: v = f ( (2, 4 = f =, (iv Write We have w = (w, w 2. f w = f w = (2, 4 w = 2w + 4w 2 = = w = 2w 2. Since w has unit length w 2 + w 2 2 = = ( 2w w 2 2 = = w 2 = ± 5. Therefore ( 2 w = ±,. 5 5 Problem 9. Consider the function f(x, y = ln(e 2x y 3. (i Write down the tangent plane to the graph of f at (2,. (ii Find the approximate value of the number ln(e 4. (.2 3. (i Using the chain rule, we compute 2e 2x e 2x f x = 2 ln(e 2x y 3 = ln(e 2x y 3 = f x(2, = = = ln e

8 Similarly, 3y 2 e 2x y 3 e 2x f y = 2 ln(e 2x y 3 = 3 2y ln(e 2x y 3 = f y(2, = 3 2 = 3 ln e 4 2 We compute The tangent plane is f(2, = ln e 4 = 4 = 2. z 2 = 2 (x (y = z = 2 x y + 4. (ii The number we are approximating is f(2.5, = = 3 4. Problem. Suppose that Calculate z = e 3x+2y, y = ln(3u w, x = u + 2v. z v, z. By the chain rule Similarly, Problem. z v = z x x v = 3e3x+2y 2 = 6e 3x e 2y = 6e 3u+6v e 2 ln(3u w = 6e 3u+6v (3u w 2. (i Find z such that (ii Calculate the series z = z y y = 2e3x+2y = 2e 3u+6v (3u w 2 3u w = 2e3u+6v e 2 ln(3u w 3u w 3u w = 2e3u+6v (3u w. + z + z 2 + z = (i This is a geometric series with step z. Its sum equals = 3 = z = 3 = z = 3 2. z

9 (ii This is a finite geometric series with terms and initial term /3 and step 2/3. The sum equals 3 ( 2 ( =. 3 3 Problem 2. The probability density function for the outcome x of a certain experiment is p(x = Ce x, for x. (i What is the value of the constant C? (ii What is the cumulative distribution function? (iii What is the median of the experiment? (iv What is the mean of the experiment? (v What is the probability that the outcome of the experiment is bigger than? (i The pdf must integrate to hence p(xdx = = C (ii The cdf is obtained by integrating the pdf: P (x = e x = = C e x x= = = C( = = C =. x p(tdt = x (iii To find the median, we set the pdf to /2: e t dt = e t t=x t= = e x. P (T = 2 = e T = 2 = e T = 2 = T = ln 2. (iv The mean is computed by the integral mean = xp(x dx = The last integral was found by integration by parts xe x = xe x x= + e x dx = + xe x =. e x dx = e x x= =. (v The probability the outcome is at most is P ( = e. The probability that x is Problem 3. P ( = e. Consider the function f(x, y = 5 (x + 2 y 2. (i Draw the cross section corresponding to x =. (ii Draw the contour diagram of f showing at least three levels. (iii Draw the graph of f. (iv What is the equation of the tangent plane to the graph of f at (,,? (i The cross section is the parabola z = y 2.

10 (ii The level curve for level c is f(x, y = c = (x y 2 = 5 c. The level curves are circles of centers (, and radius 5 c. For instance, picking levels c =, c = 2, c = 4 we get circles of radii 2, 3, of center (,. (iii The graph of f is a downward pointing paraboloid with vertex at (,, 5. (iv The partial derivatives The tangent plane is Problem 4. Find the point on the plane f x = 2(x + = f x (, = 4 f y = 2y = f y (, =. z = 4(x + (y = z = 4x x + 3y + 4z = 29 that is closest to the origin. You may want to minimize the square of the distance to the origin. We minimize the square of the distance to the origin subject to the constraint We use Lagrange multipliers Then Since f(x, y, z = x 2 + y 2 + z 2 g(x, y, z = 2x + 3y + 4z = 29. f = (2x, 2y, 2z, g = (2, 3, 4. f = λ g = 2x = 2λ, 2y = 3λ, 2z = 4λ = x = λ, y = 3λ 2, z = 2λ. 2x + 3y + 4z = 2λ + 9λ 2 The closest point is (2, 3, 4. Problem λ = 29λ 2 = 29 = λ = 2 = x = 2, y = 3, z = 4. Find the critical points of the function f(x, y = 2x 3 + 6xy + 3y 2 and describe their nature. We set the first derivatives to zero: We solve The critical points are (, and (,. We compute the second derivatives f x = 6x 2 + 6y = = x 2 + y = f y = 6x + 6y = = x + y =. y = x 2 = x = x 2 = x = x = or x =. A = f xx = 2x, B = f xy = 6, C = f yy = 6.

11 For the critical point (, we have AC B 2 = < = (, is saddle point. For the critical point (, we have AC B 2 = >, A > = (, is local minimum.

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