- #1

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## Main Question or Discussion Point

Not a homework question

say we have a curve f(x,y,z)

and the tangent plane to the curve at a point (x

[tex]\frac{{\partial d}}{{\partial x}}[/tex](x

[tex]\frac{{\partial d}}{{\partial y}}[/tex](x

[tex]\frac{{\partial d}}{{\partial y}}[/tex](x

[tex]\frac{{\partial d}}{{\partial z}}[/tex](x

[tex]\frac{{\partial d}}{{\partial z}}[/tex](x

say we have a curve f(x,y,z)

and the tangent plane to the curve at a point (x

_{o},y_{o},z_{o}) is given by[tex]\frac{{\partial d}}{{\partial x}}[/tex](x

_{o},y_{o},z_{o})(x-x_{o}) + [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) + [tex]\frac{{\partial d}}{{\partial z}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0**then would the normal plane to the curve be either of the following?**[tex]\frac{{\partial d}}{{\partial y}}[/tex](x

_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial z}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0[tex]\frac{{\partial d}}{{\partial y}}[/tex](x

_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0[tex]\frac{{\partial d}}{{\partial z}}[/tex](x

_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0[tex]\frac{{\partial d}}{{\partial z}}[/tex](x

_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0