Math 48 Exam III Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture, all homework problems, all lab assignment problems, and all quiz problems.. Find the largest possible domain and corresponding range of each function. (a) f(x, y) = 2 x + y (b) f(x, y) = 36 9x 2 4y 2 2. Draw a contour map showing several level curves of each function. (a) f(x, y) = xy (b) f(x, y) = x 2 + 9y 2 3. Consider the function f defined by 3xy, if (x, y) (0, 0) f(x, y) = x 2 + y2 0, if (x, y) = (0, 0) Determine whether f is continuous at (0, 0). Explain your answer. 4. Find all first-order partial derivatives of each function. (a) f(x, y) = x2 y y2 x (b) f(x, y) = e xy2 (c) f(x, y) = y ln(xy) 5. Find all first-order partial derivatives of each function. (a) f(x, y, z) = x 3 y 2 z + x yz (b) f(x, y, z) = e y cos z sin x 6. Find all second-order partial derivatives of each function. (a) f(x, y) = x 2 y + x y (b) f(x, y) = sin(x + y) + cos(x y) 7. Find an equation of the tangent plane to f(x, y) = 4 x 2 2y 2 at (,, ). 8. Find the linearization of f(x, y) = y cos(x y) at (2, 2) and use it to approximate f(.9, 2.). Check the accuracy of your approximation using a calculator.
9. Let f(x, y) = e x sin y, where x(t) = t 2 and y(t) = 4t. Find df dt when t = 0. 0. Compute the directional derivative of f(x, y) = xy 2x 2 at the point P = (, 6) in the direction of the point Q = (3, ).. In what direction does f(x, y) = x 2 y 2 increase most rapidly at (5, 3)? What is the largest increase? 2. Find a unit vector that is normal to the level curve of the function at (, 2). f(x, y) = x 2 + y2 4 3. Find all local extrema and saddle points of f(x, y) = x 2 y 2 + 6x + 8y 2. 4. Find all local extrema and saddle points of f(x, y) = 2x 4 + y 2 2xy. 5. Find the absolute extrema of f(x, y) = x 2 y 2 + 4x + y on the rectangular region R = {(x, y) 4 x 0, 0 y }. 6. Find the absolute extrema of f(x, y) = x 2 + y 2 + x y on the disk D = {(x, y) x 2 + y 2 }. 7. Consider the linear system x (t + ) = 0.4x (t) + 0.2x 2 (t) x 2 (t + ) = 0.3x (t) + 0.x 2 (t) Determine the stability of X = 0. 8. Find all nonnegative equilibria of x (t + ) = 2x (t)[ x (t)] x 2 (t + ) = x (t)[ x 2 (t)] and determine their stability. 2
Solutions. Find the largest possible domain and corresponding range of each function. (a) f(x, y) = 2 x + y The expression for f is defined if the denominator is not zero. So the domain of f is D = {(x, y) x + y 0}. Note that x + y 0 means that point on the line y = x must be excluded from the domain. The range of f is { z z = 2 }, (x, y) D. x + y Every point (x, y) in the domain D satisfies either (i) 0 < x + y < in which case 0 < 2 x + y < or (ii) < x + y < 0 in which case < 2 x + y < 0. So the range of f is (, 0) (0, ). (b) f(x, y) = 36 9x 2 4y 2 The expression for f is defined if the quantity under the square root sign is nonnegative. So the domain of f is { } D = {(x, y) 36 9x 2 4y 2 0} = (x, y) x 2 4 + y2 9 which is a closed elliptical disk. The range of f is {z 36 9x 2 4y 2, (x, y) D}. Since z is a positive square root, z 0. Also 36 9x 2 4y 2 36 = 36 9x 2 y 2 6. So the range of f is {z 0 z 6} = [0, 6]. 3
2. Draw a contour map showing several level curves of each function. (a) f(x, y) = xy The level curves are defined by xy = k. If k 0, then x 0 and the level curves are defined by y = k x. This is a family of hyperbolas centered at (0, 0). If k = 0, then x = 0 or y = 0. Thus, the level curve for k = 0 consists of the coordinate axes x = 0 and y = 0. A contour plot showing the level curves for k = ±, ±4, ±8 is shown below. (b) f(x, y) = x 2 + 9y 2 The level curves are defined by x 2 + 9y 2 = k. This is a family of concentric ellipses with center (0, 0). A contour plot showing the level curves for k =, 9, 8 is shown below. 4
3. Consider the function f defined by 3xy, if (x, y) (0, 0) f(x, y) = x 2 + y2 0, if (x, y) = (0, 0) Determine whether f is continuous at (0, 0). Explain your answer. The function f is defined at (0, 0) as f(0, 0) = 0. However, we will show that the limit of the function as (x, y) (0, 0) does not exist. Along the x-axis (y = 0), Along the line y = x, lim (x,y) (0,0) lim (x,y) (0,0) 3xy x 2 + y 2 = lim x 0 0 x 2 = 0. 3xy x 2 + y = lim 3x 2 2 x 0 2x = 3 2 2 0. Therefore, the limit does not exist. Thus, f is discontinuous at (0, 0). 4. Find the first-order partial derivatives of each function. (a) f(x, y) = x2 y y2 x For simplicity, we rewrite the function as Using the Quotient Rule, we have (b) f(x, y) = e xy2 f(x, y) = x3 y 3. xy f x (x, y) = 3x2 (xy) y(x 3 y 3 ) = 2x3 y + y 4 = 2x3 + y 3 x 2 y 2 x 2 y 2 x 2 y f y (x, y) = 3y2 (xy) x(x 3 y 3 ) = 2xy3 x 4 = 2y3 x 3. x 2 y 2 x 2 y 2 xy 2 Using the Chain Rule, we have f x (x, y) = y 2 e xy2 f y (x, y) = 2xye xy2. 5
(c) f(x, y) = y ln(xy) Using the Chain Rule, we have f x (x, y) = y ( ) y = y xy x. Using the Product and Chain Rules, we have ( ) x f y (x, y) = ln(xy) + y = ln(xy) +. xy 5. Find all first-order partial derivatives of each function. (a) f(x, y, z) = x 3 y 2 z + x yz For simplicity, we rewrite the function as Therefore, we have f(x, y, z) = x 3 y 2 z + xy z. f x (x, y, z) = 3x 2 y 2 z + yz f y (x, y, z) = 2x 3 yz x y 2 z f z (x, y, z) = x 3 y 2 x yz 2. (b) f(x, y, z) = e y cos z sin x Treating y and z as constants, we have f x (x, y, z) = e y cos z cos x. Using the Chain Rule, we have f y (x, y, z) = cos ze y cos z sin x f z (x, y, z) = y sin ze y cos z sin x. 6
6. Find all second-order partial derivatives of each function. (a) f(x, y) = x 2 y + x y The first-order partial derivatives of f are Therefore, we find that f x (x, y) = 2xy + y f y (x, y) = x 2 + x 2 y. f xx (x, y) = 2y f xy (x, y) = 2x + 2 y f yx (x, y) = 2x + 2 y f yy (x, y) = x 4y 3/2 (b) f(x, y) = sin(x + y) + cos(x y) The first-order partial derivatives of f are f x (x, y) = cos(x + y) sin(x y) f y (x, y) = cos(x + y) + sin(x y). Therefore, we find that f xx (x, y) = sin(x + y) cos(x y) f xy (x, y) = sin(x + y) + cos(x y) f yx (x, y) = sin(x + y) + cos(x y) f yy (x, y) = sin(x + y) cos(x y) 7
7. Find an equation of the tangent plane to f(x, y) = 4 x 2 2y 2 at (,, ). Using the Chain Rule, we have f x (x, y) = f y (x, y) = x 4 x2 2y 2 2y 4 x2 2y 2. At the point (, ), the partial derivatives are f x (, ) = and f y (, ) = 2. Therefore, an equation of the tangent plane is z z 0 = f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) z = (x ) + 2(y + ) z = x + 2y + 3 x 2y + z = 4. 8. Find the linearization of f(x, y) = y cos(x y) at (2, 2) and use it to approximate f(.9, 2.). Check the accuracy of your approximation using a calculator. Using the Product and Chain Rules, we have f x (x, y) = y sin(x y) f y (x, y) = cos(x y) + y sin(x y). At the point (2, 2), the partial derivatives are f x (2, 2) = 0 and f y (2, 2) =. Moreover, z = f(2, 2) = 2. Therefore, the linearization of f at (2, 2) is L(x, y) = z 0 + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) L(x, y) = 2 + 0(x 2) + (y 2) L(x, y) = y. To approximate f(.9, 2.), we use the linearization Using a calculator, we find that f(.9, 2.) L(.9, 2.) = 2.. f(.9, 2.) = 2. cos( 0.2) 2.06. 8
9. Let f(x, y) = e x sin y, where x(t) = t 2 and y(t) = 4t. Find df dt Using the Chain Rule for functions of one parameter, when t = 0. df dt If t = 0, then x = y = 0 and = f dx x dt + f dy y dt = e x sin y(2t) + e x cos y(4). df dt = 0 + 4 = 4. 0. Compute the directional derivative of f(x, y) = xy 2x 2 at the point P = (, 6) in the direction of the point Q = (3, ). The vector which defines the direction is A unit vector in the direction of v is u = v = P Q = 2, 5. v v = 2, 5. 29 29 The gradient vector of f is y 4x f(x, y) = 2 xy 2x, x 2 2. xy 2x 2 At the point P = (, 6), f(, 6) = 2,. 4 Therefore, the directional derivative of f at P in the direction of Q is D u f(, 6) = f(, 6) u = 29 5 4 29 = 4 29. 9
. In what direction does f(x, y) = x 2 y 2 increase most rapidly at (5, 3)? What is the largest increase? The function increases most rapidly in the direction of the gradient vector: x f(x, y) = x2 y, y. 2 x2 y 2 At the point (5, 3), we have f(5, 3) = 5 4, 3. 4 The largest increase is given by f(5, 3) = 25 6 + 9 34 6 = 6 = 34 4. 2. Find a unit vector that is normal to the level curve of the function at (, 2). f(x, y) = x 2 + y2 4 The gradient vector is normal to the level curve: f(x, y) = 2x, y. 2 At the point (, 2), we have f(, 2) = 2,. A unit vector in the direction of f(, 2) is u = f(, 2) f(, 2) = 2 5,. 5 0
3. Find all local extrema and saddle points of f(x, y) = x 2 y 2 + 6x + 8y 2. We first locate the critical points by setting the partial derivatives equal to zero: The only critical point is (3, 4). f x = 2x + 6 = 0 and f y = 2y + 8 = 0. Next we compute the second partial derivatives: f xx = 2 f xy = 0 f yy = 2. The discriminant is D(x, y) = 4. Since D(3, 4) = 4 > 0 and f xx (3, 4) = 2 < 0, there is a local maximum at (3, 4). The local maximum value is f(3, 4) = 4. There are no local minima or saddle points. 4. Find all local extrema and saddle points of f(x, y) = 2x 4 + y 2 2xy. We first locate the critical points by setting the partial derivatives equal to zero: f x = 8x 3 2y = 0 and f y = 2y 2x = 0. It follows from the second equation that y = 6x. equation, we have Substituting this into the first 8x 3 2(6x) = 0 8x 3 72x = 0 8x(x 2 9) = 0 x = 0, ±3. The three critical points are (0, 0), (3, 8), and ( 3, 8). Next we compute the second partial derivatives The discriminant is f xx = 24x 2 f xy = 2 f yy = 2. D(x, y) = 48x 2 44. Since D(0, 0) = 44 < 0, (0, 0, 0) is a saddle point. Since D(±3, ±8) = 288 > 0 and f xx (±3, ±8) = 26 > 0, there are local minima at (3, 8) and ( 3, 8). The local minimum value is f(±3, ±8) = 62. There are no local maxima.
5. Find the absolute extrema of f(x, y) = x 2 y 2 + 4x + y on the rectangular region R = {(x, y) 4 x 0, 0 y }. Since f is a polynomial, it is continuous on the closed, bounded rectangular region R, there is both an absolute maximum and an absolute minimum. We first find the critical points by setting the partial derivatives equal to zero: f x = 2x + 4 = 0 and f y = 2y + = 0. The only critical point is ( ( ) 2, 2) and the value of f there is f 2, 2 = 5. 4 Next we look at the values of f on the boundary of R which consists of four edges. On E, we have y = 0 and To find the critical values of f on E, set f(x, 0) = x 2 + 4x 4 x 0. f (x) = 2x + 4 = 0. So ( 2, 0) is a critical point and f( 2, 0) = 4. On E 2, we have x = 0 and To find the critical values of f on E 2, set f(0, y) = y 2 + y 0 y. f (y) = 2y + = 0. So ( ( ) 0, 2) is a critical point and f 0, 2 =. 4 On E 3, we have y = and To find the critical values of f on E 3, set f(x, ) = x 2 + 4x 4 x 0. f (x) = 2x + 4 = 0. So ( 2, ) is a critical point and f( 2, ) = 4. On E 4, we have x = 4 and To find the critical values of f on E 4, set f( 4, y) = y 2 + y 0 y. f (y) = 2y + = 0. So ( ( ) 4, 2) is a critical point and f 4, 2 =. 4 Evaluating the function at each of the vertices on the boundary, we have f(0, 0) = 0, f( 4, 0) = 0, f(0, ) = 0, and f( 4, ) = 0. Therefore, the absolute maximum value is /4 and the absolute minimum value is 4. 2
6. Find the absolute extrema of f(x, y) = x 2 + y 2 + x y on the disk D = {(x, y) x 2 + y 2 }. Since f is a polynomial, it is continuous on the closed, bounded disk D, there is both an absolute maximum and an absolute minimum. We first find the critical points by setting the partial derivatives equal to zero: f x = 2x + = 0 and f y = 2y = 0. The only critical point is (, ( 2 2) and the value of f there is f, ) 2 2 =. 2 Next we look at the values of f on the boundary of D which consists is the circle x 2 + y 2 =. The circle can be parameterized using polar coordinates On the boundary of D, we have x = cos θ, y = sin θ, 0 θ 2π. f(cos θ, sin θ) = + cos θ sin θ, 0 θ 2π. To find the critical values of f on the boundary of D, set It follows that f (θ) = sin θ cos θ = 0. sin θ = cos θ sin θ cos θ = tan θ = tan θ =. Therefore, θ = 3π and θ = 7π give critical points and the endpoints of the interval 4 4 θ = 0, 2π give a point of interest. ) ( 2 ). At θ = 7π, (x, y) =. At θ = 0, 2π, 4 At θ = ( 3π, (x, y) = 2, 2 4 2 2 (x, y) = (, 0). Evaluating f at each of these points, we have f f ( 2 2 2, 2 ( 2 2 2, 2 ) ) = 2 = + 2 f(, 0) = 2., 2 2 2 The absolute maximum value is + 2 and the absolute minimum value is 2. 3
7. Consider the linear system x (t + ) = 0.4x (t) + 0.2x 2 (t) x 2 (t + ) = 0.3x (t) + 0.x 2 (t) Determine the stability of X = 0. Let X(t) = x (t), x 2 (t). The system can be written in matrix form as [ ] 0.4 0.2 X(t + ) = X(t). 0.3 0. To determine stability, we find the eigenvalues of the matrix A. That is, 0.4 λ 0.2 0.3 0. λ = 0 ( 0.4 λ)(0. λ) + (0.2)(0.3) = 0 λ 2 + 0.3λ + 0.02 = 0 (λ + 0.)(λ + 0.2) = 0. The eigenvalues are λ = 0. and λ 2 = 0.2. Since λ < and λ 2 <, it follows that 0 is locally stable. 4
8. Find all nonnegative equilibria of x (t + ) = 2x (t)[ x (t)] x 2 (t + ) = x (t)[ x 2 (t)] and determine their stability. To find the equilibria we need to solve Simplifying the first equation, we have x = 2x ( x ) x 2 = x ( x 2 ). 2x 2 x = 0 x (2x 2 ) = 0. Thus, x = 0 or x =. If x 2 = 0, then the second equation gives x 2 = 0. If x =, 2 then the second equation gives x 2 =. Therefore, the equilibria are 3 ( (0, 0) and 2, ). 3 The Jacobian matrix is At (0, 0), [ ] 2 4x 0 J(x, x 2 ) =. x 2 x J(0, 0) = [ ] 2 0. 0 The eigenvalues are λ = 2 and λ 2 = 0. Since λ >, it follows that (0, 0) is unstable. At ( 2, 3), J ( 2, ) [ ] 0 0 = 2 3. 3 2 The ( eigenvalues are λ = 0 and λ 2 =. Since λ 2 < and λ 2 <, it follows that, ) 2 3 is locally stable. 5