http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Thu, 10 Jul 2014 09:55:49 -0500http://calc3.askbot.com/question/79/tangent-planes-level-curves-level-surfaces/I understand how to plug things into the tangent plane lines, and to get the equations, however in class he said that the answers to a and b from part two should be the same, but mine ended up different. We'll start with part a. $$f(x,y)=x^3y^2$$ $$f_x = 3x^2y^2 \quad f_y = 2yx^3$$ $$p_0 = \langle 1,2\rangle$$ $$ f(x_0,x_0)= 1^3(2^2) = 4$$ $$f_x(x_0,y_0)= 3(1^2)(2^2) = 12$$ $$ f_y(x_0,y_0)= 2(2)(1^3) = 4$$ $$ L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$ $$L(x,y) = 4 + 12(x-1) + 4(y-2)$$ for part b: $$F(x,y,z)=z-x^3y^2$$ $$p_0=\langle1,2,4\rangle$$ $$F_x= -3x^2y^2 \quad F_y= -2yx^3 \quad F_z=1$$ $$ f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)=0$$ $$f_x(x_0,y_0,z_0)= -3(1^2)(2^2)=-12$$ $$f_y(x_0,y_0,z_0)= -2(2)(1^3)= -4$$ So: $$-12(x-1)-4(y-2)+(z-4) =0$$ I know that if you set z=0 and then reverse the side of the equations you can end up with the same equation. Is that what he meant by it should be the same equation? Or did I screw up somewhere along the way. Also I know he explained the level surface and level curve. From what I got, a level curve is a 2D, and a level surface is a 3D? and basically to find those curves and surfaces you make sure all of the variables are on one side of the equation and set it equal to a constant? If someone could clarify that for me it would be greatly appreciated! :) *Comment*: Your answers *are* the same! To this totally clearly, you've just got to set $L(x,y)=z$ in your first (since that's how function values are interpreted geometrically) and then isolate the $z$ on one side in your second answer.TiffanyThu, 10 Jul 2014 09:55:49 -0500http://calc3.askbot.com/question/79/