Math 0 Prelim I Solutions Spring 010 1. Let f(x, y) = x3 y for (x, y) (0, 0). x 6 + y (4 pts) (a) Show that the cubic curves y = x 3 are level curves of the function f. Solution. Substituting y = x 3 in f(x, y) yields f(x, x 3 ) = x3 (x 3 ) x 6 + (x 3 ) = x 6 x 6 + x 6 = 1 +. This is a constant, not depending on x. Therefore the curve y = x 3 is part of the level curve f(x, y) = 1 +. (4 pts) (b) Draw the level curves of f which pass through (1, 1 ) and (1, 3). Solution. The curve y = x 3 passing through (1, 1 ) is y = 1 x3 and the curve y = x 3 passing through (1, 3) is y = 3x 3. So you might think that this is all there is to the answer, but there is actually something tricky going on here. The value = 1 for the curve y = 1 x3 gives f(1, 1 ) = 1+ = 5. However, the equation 1+ = 5 actually has another solution besides = 1, namely =, as one can see by simplifying the equation 1+ = 5 to a quadratic equation. This means that the level set f(x, y) = 5 consists of not one but two cubic curves, y = 1 x3 and y = x 3. Similarly for the level curve y = 3x 3 passing through (1, 3) we have the equation = 3 1+ 10 which has the solution = 1 in addition to the solution = 3, so the 3 level set f(x, y) = 3 10 consists of the two cubic curves y = 3x3 and y = 1 3 x3. You can easily draw these cubic curves, so we won t do this here. (6 pts) (c) Use these level curves to determine whether lim f(x, y) exists. (x,y) (0,0) Solution. The limit does not exist because the point (0, 0) can be approached via several different level curves, for example y = x 3 which is part of the level curve f(x, y) = 1 and y = x 3 which is part of the level curve f(x, y) = 5. As (x, y) approaches (0, 0) along the curve y = x 3 the value of f(x, y) is the constant 1 so the limit is 1, while if (x, y) approaches (0, 0) along the curve y = x 3 the limit is 5. Since there are different limits for different ways of approaching (0, 0), we conclude that lim f(x, y) does not exist. (x,y) (0,0)
Math 0 Prelim I Solutions Spring 010 (14 pts). Suppose that w = g(x/y, y/z) where g(u, v) is a differentiable function. Show that x x + y y + z z = 0. Solution. We have u = x/y and v = y/z. By the chain rule one then has Now substitute these into the formula: x x + y y + z x = g u u x + g v v x = 1 g y u + 0 g v y = g u u y + g v v y = x g y u + 1 g z v z = g u u z + g v v z = 0 g u y g z v z = x ( 1 y = g ) ( + y x g u y u + 1 g ) ( + z y g ) z v z v ( x y y x ) g ( y y u + z z y ) g z v = 0 3. Let f(x, y) be a differentiable function of two variables. Let u and v be unit vectors in the directions (1, 1) and (1, 1). Suppose you know that the directional derivatives D u f and D v f at the point (1, 5) are equal to 3 and, respectively. (9 pts) (a) Determine f at the point (1, 5). Solution. The unit vectors u and v are u = ( 1, 1 ) and v = ( 1, 1 ). Using the formula for the directional derivatives D v f = f v (everything is evaluated at the point (1, 5)) we obtain the system 1 f x + 1 f y = 3 1 f x 1 f y = which is equivalent to f x + f y = 6 f x f y = Therefore f x = 4 and f y = and f = (4, ). (5 pts) (b) Compute D w f at (1, 5) for w a unit vector in the direction of (3, 4). Solution. The unit vector w is equal to (3/5, 4/5) and the directional derivative is D w (f) = f w = (4, ) (3/5, 4/5) = 1/5 8/5 = 4/5
Math 0 Prelim I Solutions Spring 010 4. Consider the ellipsoid x + y + 3z = 0. (7 pts) (a) Find an equation for the tangent plane to the ellipsoid at the point (3,, 1). Solution. The surface S is a level surface for the function F(x, y, z) = x + y + 3z. The tangent plane to S at a point (x, y, z) is perpendicular to the gradient vector F = (x, 4y, 6z). Thus, the tangent plane at the point (a, b, c) has an equation a(x a) + 4b(y b) + 6c(z c) = 0. Substituting (3,, 1) for (a, b, c) we get that the tangent plane has an equation 6(x 3) + 8(y ) + 6(z 1) = 0 or 3x + 4y + 3z = 0 (7 pts) (b) Find all points (a, b, c) on the ellipsoid such that the tangent plane to the ellipsoid at (a, b, c) passes through the point (0, 0, 4). Solution. The tangent plane passes through the point (0, 0, 4) if a(0 a) + 4b(0 b) + 6c(4 c) = 0 which is equivalent to a + b + 3c = 1c The left side of this equation equals 0 since (a, b, c) is a point on the surface x + y + 3z = 0, so we get the equation 1c = 0, i.e., c = 5/3. Thus, the set of points is the intersection of the ellipsoid with the plane z = 5/3, which is an ellipse in a horizontal plane. 5. Let S be the surface x + y = ( z)(z + z + ). (7 pts) (a) Find all points on S where the tangent plane is horizontal (parallel to the xy plane). Solution. The surface S is a level surface for the function F(x, y, z) = x + y + (z )(z + z + ) = x + y + z 3 3z 4 The tangent plane is horizontal if F is perpendicular to the xy plane, i.e., F x = F y = 0. We have F = (x, y, 6z 6z). Thus the condition F x = F y = 0 becomes x = y = 0. The only point on S with this property is (0, 0, ) since the only solution of ( z)(z + z + ) = 0 is z =.
Math 0 Prelim I Solutions Spring 010 (7 pts) (b) Show that the set of points on S where the tangent plane is vertical (perpendicular to the xy plane) is the union of two circles. Solution. The tangent plane is vertical if F is parallel to the xy plane, i.e., F z = 0. We have F z = 6z 6z so F z = 0 only when z = 0 or z = 1. The intersections of the two planes z = 0 and z = 1 with the surface S are the circles C 1 : x + y = 4, z = 0 and C : x + y = 5, z = 1. (15 pts) 6. Find all critical points of the function f(x, y) = x(x + y 4) and determine their types. Hint: There is at least one critical point of each type. Solution. The partial derivatives of f are f x = 3x +y 4 and f y = xy. In order to find the critical points we need to solve the system f x = 0 and f y = 0. The second equation implies that either x = 0 or y = 0. The first case x = 0 yields y = 4, i.e., y = ±, which leads to critical points (0, ) and (0, ). The second case y = 0 leads to more critical points (/ 3, 0) and ( / 3, 0). The types of the critical points can be determined by the second derivative test. The Hessian of f is equal to [ ] 6x y Hf = with det(hf) = 1x 4y y x The points (0, ±) are saddle points because det(hf) = 16 < 0 at these two points. The point (/ 3, 0) is a local minimum because det(hf) = 16 > 0 and f xx > 0 at this point. Finally ( / 3, 0) is local maximum since det(hf) = 16 > 0 and f xx < 0 at this point. (15 pts) 7. Let f(x, y) = x y. Determine the maximum value of the directional derivative D u f(x, y) as (x, y) ranges over all points on the ellipse x + 4y = 1 and u ranges over all directions. (Your final answer should be a single number.) Solution. The directional derivative D u f at a fixed point (x, y) is maximal when u has the same direction as the gradient vector f and this maximum is equal to f. This allows us to reduce the problem to computing the maximum of h(x, y) = f = (x, y) = 4x + 4y as (x, y) ranges over all points on the ellipse x + 4y = 1.
Math 0 Prelim I Solutions Spring 010 The method of Lagrange multipliers gives that the extremal points are solutions of h = g for g(x, y) = x + 4y. This gives the system of equations 8x = x 8y = 8y x + 4y = 1 The first equation gives that either x = 0 or = 4, and if = 4 then the second equation gives that y = 0. Thus either x = 0 or y = 0. From the third equation we then get four critical points (0, 1/), (0, 1/), (1, 0), and ( 1, 0). The value of f at the first two points is 1 and its value at the second two points is. Thus the maximum value of f is. By what we said at the beginning this means that the maximal value of the directional derivative D u f is. (This is achieved at the points x = ±1, y = 0 with the directions u = (±1, 0).)