Unit #22 - The Chain Rule, Higher Partial Derivatives & Optimization Section 14.7 Some material from Calculus, Single and MultiVariable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. TEST PREPARATION PROBLEMS 1. f(x,y) = 3x 2 y +5xy 3 f x = 6xy +5y 3 f y = 3x 2 +15xy 2 f xx = 6y f xy = 6x+15y 2 f yx = 6x+15y 2 f yy = +30xy 3. f(x,y) = (x+y) 3 f x = 3(x+y) 2 f y = 3(x+y) 2 f xx = 6(x+y) f xy = 6(x+y) f yx = 6(x+y) f yy = 6(x+y) 5. f(x,y) = e 2xy f x = e 2xy 2y f y = e 2xy 2x f xx = e 2xy (2y)(2y) f xy = e 2xy (2y)(2x)+e 2xy (2) f yx = e 2xy (2x)(2y)+e 2xy (2) f yy = e 2xy (2x)(2x) 1
9. f(x,y) = sin(x 2 +y 2 ) f x = cos(x 2 +y 2 )2x f y = cos(x 2 +y 2 )2y f xx = sin(x 2 +y 2 )(2x)(2x) +cos(x 2 +y 2 )(2) f xy = sin(x 2 +y 2 )(2y)(2x) f yx = sin(x 2 +y 2 )(2y)(2x) f yy = sin(x 2 +y 2 )(2y)(2y) +cos(x 2 +y 2 )(2) 10. f(x,y) = 3sin(2x)cos(5y) f x = 6cos(2x)cos(5y) f y = 15sin(2x)sin(5y) f xx = 3sin(2x)(4)cos(5y) = 12sin(2x)cos(5y) f xy = 6cos(2x)( sin(5y))(5) = 30cos(2x)sin(5y) f yx = 15cos(2x)(2)sin(5y) = 30cos(2x)sin(5y) f yy = 15sin(2x)( cos(5y))(5) = 75sin(2x)cos(5y) 11. f(x,y) = (x y +1) 2 To compute the two-variable second-degree Taylor polynomial, we need all the first and second partial derivatives of f(x, y). f = (x y +1) 2 f(0,0) = 1 f x = 2(x y +1) f x (0,0) = 2 f y = 2(x y +1)( 1) f y (0,0) = 2 f xx = 2 f xx (0,0) = 2 f xy = 2 f xy (0,0) = 2 f yy = 2 f yy (0,0) = 2 Using the Taylor polynomial formula (which is an extension of our linearization/tangent plane formula), at the point (x,y) = (0,0), f(x,y) f(0,0)+f x (0,0)(x 0)+f y (0,0)(y 0) + 1 2 f xx(0,0)(x 0) 2 +f xy (0,0)(x 0)(y 0)+ 1 2 f yy(0,0)(y 0) 2 = 1+2x 2y + 2 2 x2 2xy = 1+2x 2y +x 2 2xy +y 2 2
Note that, because our original function was a quadratic in x and y, this Taylor approximation is exactly the same function we started with (try expanding (x y+1) 2 to see this). 15. f(x,y) = e x cos(y) f = e x cos(y) f(0,0) = 1 f x = e x cos(y) f x (0,0) = 1 f y = e x sin(y) f y (0,0) = 0 f xx = e x cos(y) f xx (0,0) = 1 f xy = e x sin(y) f xy (0,0) = 0 f yy = e x cos(y) f yy (0,0) = 1 Using the Taylor polynomial formula (which is an extension of our linearization/tangent plane formula), at the point (x,y) = (0,0), f(x,y) f(0,0)+f x (0,0)(x 0)+f y (0,0)(y 0) + 1 2 f xx(0,0)(x 0) 2 +f xy (0,0)(x 0)(y 0)+ 1 2 f yy(0,0)(y 0) 2 = 1+x+0y + 1 2 x2 0xy = 1+x+ 1 2 x2 1 2 y2 19. 20. (a) f x (P) is positive. Moving to the right leads to higher contours. (b) f y (P) is zero. Moving upwards in y stays on the same contour. (c) f xx (P) is positive. x slope is positive (from (a)), and because the contours are getting closer, the positive slopes are getting steeper, so the slopes are increasing. (d) f yy (P) is zero. Slopes aren t changing at all in the y direction. Either the y slopes are always zero, so don t change when we increase x, or our x slopes are constant when we move in y. 3
(a) f x (P) is positive, since increasing x leads to higher contours. (b) f y (P) is zero (as with #19). (c) f xx (P) is negative, since the contours get further apart, so the positive slopes are getting less steep, or decreasing. (d) f yy (P) is zero (as with #19). (e) f xy (P) is zero (as with #19). 21. (a) f x (P) is negative, since increasing x leads to lower contours. (b) f y (P) is zero. (c) f xx (P) is negative. The contours are getting closer together, so negative slopes are getting more steep, so slopes are decreasing. (d) f yy (P) is zero. 22. (a) f x (P) is negative. (b) f y (P) is zero. (c) f xx (P) is positive. x slopes are negative, but getting less steep, so their numeric values are increasing. (d) f yy (P) is zero. 23. (a) f x (P) is zero. 4
(b) f y (P) is negative. Moving upwards in y leads to lower contours. (c) f xx (P) is zero. (d) f yy (P) is negative. Moving upwards in y gives steeper negative slopes. 24. (a) f x (P) is zero. (b) f y (P) is positive. (c) f xx (P) is zero. (d) f yy (P) is negative. Slopes are positive, but getting less steep as we move upwards in y. 25. (a) f x (P) is positive. Moving to the right leads to higher contours. (b) f y (P) is positive. Moving upwards in y leads to higher contours. (c) f xx (P) is zero. Contours are evenly spaced, so slopes aren t changing, so their rate of change is zero. (d) f yy (P) is zero. Contours are evenly spaced, so slopes aren t changing, so their rate of change is zero. Contours are evenly spaced, so slopes aren t changing, so their rate of change is zero. 26. (a) f x (P) is negative. 5
(b) f y (P) is negative. (c) f xx (P) is zero. Contours are evenly spaced, so slopes aren t changing. (d) f yy (P) is zero. 27. (a) f x (P) is positive. Moving right goes to higher contours. (b) f y (P) is negative. Moving upwards in y leads to lower contours. (c) f xx (P) is negative. x slopes are positive and getting flatter as we move right. (d) f yy (P) is negative. y slopes are negative and getting steeper as we move upwards. (e) f xy (P) is positive. Consider either the x slopes are positive and getting steeper as we move upwards (so the values of the slope are getting numerically larger), or the y slopes are negative and getting flatter as we move right (so going from large negative values closer to zero, or increasing). 28. (a) f x (P) is negative. (b) f y (P) is positive. (c) f xx (P) is positive. x slopes are negative and get flatter as we move to the right. (d) f yy (P) is positive. y slopes are positive and getting steeper. (e) f xy (P) is negative. Consider either x slopes, which are negative, get steeper as you move upwards, or the y slopes, which are positive, get flatter as you move to the right. 6