Tuesday February 11 Solutions Partial derivatives and differentiability 2 1 Consider f x y x y 2 e x Compute the following f x f y 2 f x y 2 f y x What is the relationship between your answers for the last two This is an instance of Clairaut s Theorem and holds for most functions Solution So we see that 2 2 2 f f 2x 2 y 2 e x y 2 e x 2x ye x x y 2 2 2 2 f f 2x ye x 4x 2 ye x 2ye x x y x y x 2 2 2 2 2 f f 2x 2 y 2 e x y 2 e x 4x 2 ye x 2ye x y x y x y 2 f x y 2 f y x 2 Shown are some level curves for the function f R2 R Determine whether the following partial derivatives are positive negative or zero at the point P fx fy f xx fyy fx y f yx Solution a If we fix y and allow x to vary the level curves indicate that the value of f increases as we move through P in the positive x direction so f x is positive at P b If we fix x and allow y to vary the level curves indicate that the value of f is neither increasing nor decreasing as we move through P in the positive y direction so f y is zero at P c f xx x f x so if we fix y and allow x to vary f xx is the rate of change of f x as x increases Note that at points to the right of P the level curves are closer together in the x direction than at points to the left of P demonstrating that f increases more quickly with respect to x to the right of P So as we move through P in the positive x direction the positive value of f x increases hence f xx is positive d f y y y f y so if we fix x and allow y to vary f y y is the rate of change of f y as y increases The level curves are closer together in the y direction at points above P than at those below P demonstrating that f increases more quickly with respect to y above P So as we move through P in the positive y direction the positive value of f y increases hence f y y is positive e f x y y f x so if we fix x and allow y to vary f x y is the rate of change of f x as y increases The level curves are closer together in the x direction at points above P than at those below P demonstrating that f increases more quickly with respect to x for y values above P So as we move through P in the positive y direction the positive value of f x increases hence f x y is positive 3 The wind chill index W f T v is the perceived temperature when the actual temperature is T and the wind speed is v Here is a table of values for W a Use the table to estimate f T and f v at T v 20 40 Solution f T 20 40 lim h 0 f 20 h 40 f 20 40 h which can be approximated by considering h 5 and h 5 as follows f 15 40 f 20 40 27 34 7 5 5 5 f 25 40 f 20 40 41 34 7 f T 20 40 5 5 5 f T 20 40 Averaging these values we estimate f T 20 40 to be 7 5 Similarly f v 20 40 lim h 0 f 20 40 h f 20 40 h which can be approximated by considering h 10 and h 10 1 f 20 50 f 20 40 35 34 10 10 10 f 20 30 f 20 40 33 34 1 f v 20 40 10 10 10 f v 20 40 Averaging these values we estimate f v 20 40 to be 1 10 b Use your answer in a to write down the linear approximation to f at 20 40 Solution The linear approximation of f at 20 40 is f T v f 20 40 f T 20 40 T 20 f v 20 40 v 40 7 1 34 T 20 v 40 5 10 c Use your answer in b to approximate f 22 45 Solution 4 Consider f x y 1 7 f 22 45 34 22 20 45 40 37 3 5 10 p 1 x 2 y 2 a What is the domain of f That is for which x y does the function make sense Solution D x y R2 x 2 y 2 1 b Describe geometrically the surface which is the graph of f Solution It is the upper half of a unit sphere c Find the tangent plane to the graph at x y 1 2 1 2 x 1 x 2 y 2 Solution f x x y p and f y x y p y 1 x 2 y Then f x 21 12 2 p 2 2 and f y 12 12 p 2 2 So the equation of the tangent plane at 12 12 is 1 1 1 1 1 1 1 1 z f f x x f y y 2 p2 2p 2 2 2 p2 2 2 2 2 1 1 z x y 2 2 2 2 2 or z p 2 2 x y 2 p d Consider the vector v 1 2 1 2 1 2 which goes from 0 to the point on the graph where we just found the tangent plane What is the angle between v and a normal vector to the tangent plane p p Solution From part c a normal vector to the tangent plane is n 22 22 1 Let be the angle between v and n then p p p 1 2 1 2 1 2 2 2 2 2 1 v n q 1 cos qp p p v n 2 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 so cos 1 1 0 5 Consider f x y p 3 x 3 y 3 a Compute f x 0 0 Note this partial derivative exists Solution f x 0 0 lim h 0 f 0 h 0 f 0 0 h 3 0 1 3 0 h lim lim 1 h 0 h 0 h h h b Is f differentiable at 0 0 Solution No f is not differentiable at 0 0 Consider that if we look along the line y x p p 3 3 the function is f x x x 3 x 3 2x Suppose on the contrary that f is differentiable at 0 0 Then the linear approximation would be f x x …
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