# Calculus

posted by on .

Find the equation of the tangent plane at the given point.

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 - ,

#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