Math 241 - Exam #3

Covers 11.6, 11.7 and 11.8

Solutions

1. f_x = 2z + 2xz + z^2
f_y = 0
f_z = 2x + x^2 + 2xz

2. grad f = < 1/y, -x/y^2 >
grad f(3,-2) = < -1/2, -3/4 >
u = v/|v| = <1,1>/sqrt(2)
D_u f = grad f(3,-2) . u = -5/(4 sqrt(2))

3. grad f = /sqrt(x^2+y^2)
grad f(3,-4) = < 3/5, -4/5 >
line: (x,y) = (3,-4) + t(3/5,-4/5)
or (x-3)/(3/5) = (y+4)/(4/5)
which simplifies to y = -4/3 x
thus it does go through the orign

4. grad h = <-0.02x, -0.04y>
grad h(60,100) = <-1.2, -4>

4a. direction of <-1.2, -4>

4b. rate of change |grad h| = sqrt(1.2^2 +4^2) = 4.176

4c. angle = arctan(4.176) = 76.5 degrees

5. grad f = < 6x^2+y^2+10x, 2xy+2y >
critical points are (0,0), (-5/3,0), (-1,2), (-1,-2)
f_xx = 12x + 10, f_xy = 2y, f_yy = 2x+2
(0,0): D>0, f_xx >0, local min
(-5/3,0): D>0, f_xx<0, local max
(-1,2): D<0, saddle point
(-1,-2): D<0, saddle point

6. the absolute extreme values occur either in the interior of D
or on the boundary of D, i.e. where x^2 + y^2 = 9.

Interior critical points are (0,0) and (1,2) with values 0 and -4
The boundary gives a max of 4 and a min of -2
so the absolute maximum value is 4 and absolute minimum value is -4.

7. grad f = l grad g
1 = 2x l, 3 = 2y l, 5 = 2z l, x^2 + y^2 + z^2 = 1
express x, y and z in terms of l, and plug into last
equation to find l. Get l = +/- sqrt(35/4) and this gives
f = +/-sqrt(35)
So the absolute maximum is sqrt(35), minimum -sqrt(35).

ccollins@math.utk.edu
Last Modified: Mar. 22, 1999