2 Functions of Several Variables - Solutions

2. Functions of Several Variables (from Stewart, Calculus, Chapter 14) Functions of Two Variables: f : D ! R, D ✓ R2 . The graph of the function f (x, y) is the set of all points (x, y, z) 2 R3 such that (x, y) 2 D and z = f (x, y). The graph of a function of two variables is a surface in R3 . Functions of Three Variables: f : D ! R, D ✓ R3 . The graph of a function f (x, y, z) of three variables would lie in R4 . For a visual representation, we often look instead at the level surfaces of f , i.e. the surfaces in R3 with equations f (x, y, z) = k for constants k. Continuity: A function f (x, y) is continuous at (a, b) if lim (x,y)!(a,b) Graph of f (x, y) f (x, y) = f (a, b). Partial Derivatives fx (x, y) = @f f (x + h, y) (x, y) = lim h!0 @x h f (x, y) and fy = @f f (x, y + h) = lim h!0 @y h Shortcut for Finding Partial Derivatives • To find fx , regard y as constant and di↵erentiate f (x, y) with respect to x. • To find fy , regard x as constant and di↵erentiate f (x, y) with respect to y. Ex. 1. Consider the function f (x, y) = (a) Show that f is continuous at (0, 0). (b) Find fx (0, 0) and fy (0, 0). ( 4x3 x2 +y 2 0 if (x, y) 6= (0, 0) if (x, y) = (0, 0). f (x, y) Ex. 2. Let f (x, y) = ( 3x2 y 2 2x4 +y 4 0 if (x, y) 6= (0, 0) if (x, y) = (0, 0). (a) Is f continuous at (0, 0)? Justify your answer. (b) Find fx (0, 0) and fy (0, 0). Level Curves and Level Surfaces Def. The level curves of a function f of two variables are the curves in the xy-plane with equation f (x, y) = k for any constant k. Level curves are the projection onto the xy-plane of traces of the surface in the planes z = k. (A contour map is a collection of level curves.) Def. The level surfaces of a function f (x, y, z) are the surfaces with equations f (x, y, z) = k. Directional Derivatives and the Gradient Vector Directional Derivatives Given a unit vector ~u = ha, bi, the vertical plane through (x0 , y0 ) in the direction of ~u intersects S in a curve C. The slope of the tangent line to C at the point (x0 , y0 ) gives the rate of change of f at (x0 , y0 ) in the direction of the unit vector ~u. This is called the directional derivative of f in the direction of ~u, and is denoted D~u f (x0 , y0 ). 2 1 0 z -1 -2 -2 x Note that the partial derivatives fx and fy are the directional derivatives of f in the directions of ~i and ~j, respectively. 0 -2 0 2 y 2 Thm. If f is a di↵erentiable function then f has a directional derivative in the direction of any unit vector ~u and D~u f = rf · ~u The Gradient Vector rf (x, y) = ⌧ @f @f , @x @y or rf (x, y, z) = ⌧ @f @f @f , , @x @y @z Properties of the Gradient: Let ✓ be the angle between a unit vector ~u and rf at a point P . D~u f = rf · ~u = |rf | cos(✓) 1. The maximum value of the directional derivative D~u f (P ) at a point P is |rf (P )| and it occurs when ~u has the same direction as the gradient vector rf (P ). So rf (P ) points in the direction of maximum rate of increase of f at P . 2. rf (x, y) is perpendicular to the level curves of a function f (x, y). Similarly, rf (x, y, z) is perpendicular to the level surfaces of a function f (x, y, z). Ex. 3. Find the directional derivative of the function f (x, y, z) = y 2 exyz at the point (0, 1, 1) in the direction of the vector h4, 2, 1i. Ex. 4. Let f (x, y) = 2x2 + xy 2 . Find a unit vector that points in the direction of the maximum rate of increase at the point (1, 2). What is the rate of change of f in this direction? Ex. 5. The temperature at the point (x, y, z) is T (x, y, z) = ⇡1 sin(⇡xy) + ln(z 2 + 1) + 60. (a) Let ~v = ~i + 2~j + 2~k. What is the rate of change of the temperature at the point (2, 1, 1) in the direction of ~v ? (b) Find a vector pointing in the direction in which the temperature increases most rapidly at the point (2, 1, 1). Ex. 6. A hiker is walking on a mountain path. The surface of the mountain is modeled by f (x, y) = 1 4x2 3y 2 . The positive x-axis points to the East direction and the positive y-axis points North. (a) Suppose the hiker is now at the point P (1/4, 1/2, 0) and heading North. Is she ascending or descending? (b) When the hiker is at the point Q(1/4, 0, 3/4), in which direction should she initially head to descend most rapidly? Tangent Planes Tangent Planes: The tangent plane to a surface S at a point P = (x0 , y0 , z0 ) is the plane containing the tangent line to any curve C on S passing through P . Equations of Tangent Planes If S is the graph of a function, i.e. z = f (x, y) is given explicitly as a function of x and y: z f (x0 , y0 ) = fx (x0 , y0 )(x x0 ) + fy (x0 , y0 )(y If S is the graph of a level surface F (x, y, z) = k and rF (x0 , y0 , z0 ) 6= 0: Fx (x0 , y0 , z0 )(x x0 ) + Fy (x0 , y0 , z0 )(y y0 ) y0 ) + Fz (x0 , y0 , z0 )(z z0 ) = 0. Rmk. This second equation is just the usual scalar equation of a plane, with ~n = rF . This equation can be used in the first case as well. Ex. 7. Let f (x, y) = y + sin(x/y). Find an equation of the tangent plane to the graph of z = f (x, y) at the point (0, 3, 3). Ex. 8. Consider the surface S given by the equation x2 y yz 2 + z = 1. (a) Find an equation of the tangent plane to S at the point (11, 0, 1). (b) Find two points on the surface S where the tangent plane at P is parallel to the yz-plane.