<p> Multivariable Calculus Self Assessment B - answers Name:______</p><p> 2 Z 1. If z=f(x, y) and xy + yz + z2= 0, find at (-2,1,1) X Y Solution: F z y x  z 1.      w  f (x, y, z) y F y  2z z 2 z   z  w f f y f z 2.          xy x  y  x x y x z x 1 (y  2z)(1)  (x  z)(2)  y     0  .  y  2z (y  2Z)2  y  2z  1 y  2x  y  1 y2  2xy    .     y  2z (y  2Z)2  y  2z  y  2z (y  2z)3  y2  4yz  4z 2  y2  2xy  4yz  2xy  4z 2   (y  2z)3 (y  2z)3  4  4  4 4    (1 2)3 27 another way:  z   z  2 y  2z1   x  z2   z   x  z   x   x       xy x  y  2z  y  2z2  y   2y  y  2z1   x  z   y  2z   y  2z  y  2zy  2z  y x  z2y         y  2z2 y  2z3 y  2z2z x  z2y 32 12 4       y  2z3 1 23 27 4 answer: ______ ______27</p><p>Self-Assessment page 1 Form B dy 2. Use the Jacobian to find at (1, -1, 1) if -2xy + 3z + x2= 6 and 2x - y2 + yz = 2 dx Solution: (F,G)  2y  2x 3 4 3 dy (x, z) 2 y 2 1 10 10 10           answer:  dx (F,G)  2x 3  2 3  7 7 7 (y, z)  2y  z y 3 1</p><p>3. Find the directional derivative of w= 2x2yz in the direction of v=3i-4j at he point (-1, 1, 3, 6) Solution : The gradient is given by : w  4xyzi  2x2 zj  2x2 yk  12i  6 j  2k   3 4 The unit vector in the direction of v is u  i  j  0k 5 5   36 24 60 Dw  w  u    0    12 answer : 12 v 5 5 5</p><p>4. Use the Lagrangean coefficients technique to find the minimum value of w  y  z  2x2 subject to the condition x  y2  z 2. We want to find the minimum value of w  y  z  2x2 for point located in the paraboloid x  y2  z 2. Solution: using Lagrangean coefficients. The lagrangean equation is given by F(x, y, z,)  y  z  2x2  (x  y2  z 2 )  F   4x    0  x    x 4  F 1  1 2y  0  y   y 2  F 1   1 2z  0  z   z 2  F 2 2  1 1 3 3   x  y  z  0      0    2     2   4 42 42 Therefore, 2 1 1   3 2  1 3 4  3 4 3 4 33 4 w  y  z  2x2    2        3 3   3 2 2 2 2  4  2 8 2 8 8 33 4 answer: ___  ______8</p><p>Self-Assessment page 2 Form B 5. Find all critical points of f(x, y) = y3 + xy2 -2xy . Determine whether each critical point yields a relative maximum value, a relative minimum, or a saddle point. Step 1. f  y2  2y  0 x y(y  2)  0 y  0 and x  0 or  f   2     3y2  2xy  2x  0 3y  2xy  2x  0  y  2 and 2x 12  0  x  6  y Critical Points: The critical points are (0, 0, 0) and (-6, 2, 8)</p><p>Step 2.  2 f   0  2   x  2 f at x  0 and y  0  0    4, saddle point 2   2  f   y   0 2 x2    f     2 2 f  yx  6y  2x    2  2  y   f 2   0   f  x2   2y  2   yx 2 f   at x  6 and y  2  2    4, saddle point   2  y  2    f    2   yx Answer: both (0, 0, 0) and (-6, 2, 8) are saddle points </p><p>6) An ice cone is melting in such a way that the radius is decreasing at the rate of 2 inches per minute and the height is decreasing at the rate of 3 inches per minute. At what rate is the volume changing when r=3 feet and h=5 feet.</p><p>. 1 V V dr V dh V 2 1 V   r 2h         rh 2   r 2 3 3 t r dt h dt t 3 3 V 2 1    (36)(60) 2   (36)2  3  2880 1296 t 3 3 V  4176 sq  inches per min  29 sq  ft per min  91.106 ft  min t</p><p> answer: ______ 91.106 ft  min ______</p><p>Self-Assessment page 3 Form B  2 z 7. If z = x2 +y2 , x=r+2s and y = 3r +s, find  r s Solution: z z x z y      2x  2  2y 1 4x  2y  w s x s y s 2 z  w x w y  (w)      41  2(3) 10 rs r x r y r answer: __ 10______</p><p> 2 z 8. If f(x, y) = x2 -xy + y2 find  x y Solution:  2 z   z        x  2y  1  x y x  y  x answer:____-1 ______</p><p>x y2 9. If x=f(y, z), 4 = x3 -y3 -z3 , find answer: : ______y x2</p><p>Solution: F x y  3y2 y2      y F 3x2 x2 x</p><p>10. Use differentials to approximate 3 6 . 3 3 7 . 6 Solution: Let f (x, y)   x3 y , x  36, x  0.3, y  8, y  0.4 f (36,8)   363 812 f f 3 y x f  df  x  y  x  y  x y 2 x 33 x2 2 6 6 24 18  3  0.3   0.4       0.15 12 12 120 120 120 20 Therefore, 36.33 7.6 12  0.15 11.85 answer :11.85</p><p>Self-Assessment page 4 Form B 11. Find the maximum value of w= x2 +3y2 +2z2 subject to the condition 2x-3y+5z=1. Solution: using Lagrangean coefficients. The lagrangean equation is given by F(x, y, z,)  x2  3y2  2z 2  (2x  3y  5z 1)  F 39 4  2x  2  0  x     1     x  4 39  F   4   6y  3  0  y   x   y 2  39    F 5 2   4z  5  0  z     y    z 4  39  F 3 25  5  2x  3y  5z 1  0  2     1  0 z    2 4  39 2 2 2 16 12 50 78 2 Therefore, w  x  3y  2z      answer 392 392 392 392 39</p><p>12. Find the point on the plane 2x-y +3z -5 =0 that is closest to the origin. Solution: We want to minimize d  x2  y2  z 2 or D  d 2  x2  y2  z 2 subject to the constraint that 2x - y + 3z -5 =0. The lagrangean equation is given by F(x, y, z,)  x2  y2  z 2  (2x  y  3z  5)  F  5  2x  2  0  x   14  10      x  7  F   5   2y    0  y   x   y 2  7    F 3 5   2z  3  0  z     y    z 2  14  F 1 9  15  2x  y  3z  5  0  2      5  0 z    2 2  14  5 5 15  Therefore, the closest point is  ,  ,   7 14 14 </p><p>Self-Assessment page 5 Form B 13. The gradient is normal to the level surface F(x, y, z) =0. It is because of this normality of the gradient that the maximum directional derivative ( which is ) at a point is often called the normal d f derivative at Normal Derivative at that point. We use the notation for the normal derivative of a d n function f.</p><p>Find the normal derivative of z = x3 y at the point (-1, 2, 2) and the corresponding unit vector. Solution: df z  3x2 yi  x3 j  6i  j   z  62  (1)2  37 dn  6 1 u  i  j 37 37</p><p> x 2 y 14. Show that l i m 4 2  0 along any line y=mx, for any m. What is x 0 x  y y  0 x 2 y x 2 y l i m 4 2 along y=x2 . Does l i m 4 2 exist?. Explain. x 0 x  y x 0 x  y y  0 y  0</p><p>Solution: x2 y mx3 mx 0 y  mx, m  0  lim  lim  lim   0 x0 x4  y2 x0 x4  m2 x2 x0 x2  m2 m2 y0 y0 y0 x2 y 0 m  0 y  0 and lim  lim  0 x0 x4  y2 x0 x4 y0 y0 x2 y x4 1 y  x2  lim  lim  x0 x4  y2 x0 x4  x4 2 y0 y0 Therefore the limit does not exist because it is not unique</p><p>Self-Assessment page 6 Form B</p>