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November 23, 2014

November 23, 2014

Posted by **Alicia** on Thursday, October 4, 2012 at 1:10pm.

1)x^2+y^2-z=1 at the point (1,3,9)

2)x^2*y^2+z-40=0 at x=2, y=3

3) x^2+ln(xy)+z=6 at point (4,0.25,2)

For the first one I got the partial fraction in terms of x, y and z and my final answer was -2x+6y-z+2. I don't know if that is right.

For the second one, the back of the book says the answer is -36x-24y+148.

I didn't get the partial fraction with respect to z, I just plugged in x=2 and y=3 for z=-x+2y^2+40 and got 4. Then, my final answer was 36x+24y-140. I don't know how to get negative 4.

I haven't done prob. 3 because I'm not sure what is the right procedure.

- Calculus -
**Steve**, Thursday, October 4, 2012 at 2:27pm#1:

∇f = 2x,2y,-1

∇f(1,3,9) = 2,6,-1

plane: 2(x-1)+6(y-3)-1(z-9) = 0

2x+6y-z = 11

#2:

at x=2,y=3, z=4

∇f = 2xy^2,2x^2y,1

∇f(2,3,4) = 36,24,1

plane: 36(x-2)+24(y-2)+1(z-4) = 0

36x+24y+z = 124

#3:

∇f = 2x+1/x,1/y,1

∇f(4,1/4,2) = 33/4,4,1

plane: 33/4 (x-4) + 4(y-1/4) + 1(z-2) = 0

33x+16y+4z = 144

O' course, check my arithmetic

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