The equation of a normal plane to a curve - confusion

[Ian_Brooks]

Not a homework question

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

\frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(x-x_o) + \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(y-y_o) + \frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(z-z_o) = 0

then would the normal plane to the curve be either of the following?

\frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(z-z_o) = 0

\frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(z-z_o) = 0

\frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(z-z_o) = 0

\frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(z-z_o) = 0

 

[HallsofIvy]

Not a homework question

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

\frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(x-x_o) + \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(y-y_o) + \frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(z-z_o) = 0
then would the normal plane to the curve be either of the following?

\frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(z-z_o) = 0

\frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(z-z_o) = 0

\frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(z-z_o) = 0

\frac{{\partial d}}{{\partial z}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial d}}{{\partial y}}(x_o,y_o,z_o)(y-y_o) - \frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)(z-z_o) = 0

I don't know where to start! In the first place, "f(x,y,z)" is NOT a curve. If you mean an equation such as f(x,y,z)= 0 that describes a surface, not a curve.

Second, I don't know what you mean by
\frac{{\partial d}}{{\partial x}}(x_o,y_o,z_o)
You don't differentiate a point!
My guess would be that you mean
\frac{{\partial f}}{{\partial x}}(x_o,y_o,z_o)
but then that is the tangent plane to a surface, not to a curve.


If you are given a curve as x= f(t), y= g(t), z= h(t) for some parameter t, then the tangent vector at (x_0, y_0, z_0) is f'(t_0)\vec{i}+ g'(t_0)\vec{j}+ h'(t_0)\vec{k} where x_0= g(t_0), etc.

The tangent line to that curve is given by x= f'(t_0)t+ f(t_0), y= g'(t_0)t+ g(t_0), z= h'(t_0)t+ h(t_0). A curve does not have a "tangent plane"- there are an infinite number of planes tangent to a smooth curve at a given point.

and the normal plane by f'(t_0)(x- f(t_0))+ g'(t_0)(y- g(t_0))+ h'(t_0)(z- h(t_0))= 0.

 

[Ian_Brooks]

I worded the question wrong

Say we had a function of x, y, z that resulted in a surface in 3 dimensions.

f(x,y,z) = 3x³ -9xy² + 15x²y + 19

then to find the tangent line at an arbitrary point say (x_o,y_o,z_o)

we use the following equation

\frac{{\partial f}}{{\partial x}}(x_o,y_o,z_o)(x-x_o) + \frac{{\partial f}}{{\partial y}}(x_o,y_o,z_o)(y-y_o) + \frac{{\partial f}}{{\partial z}}(x_o,y_o,z_o)(z-z_o) = 0

which means I find the partial derivatives of f with respect to x,y and z then I substitute the points (x_o,y_o,z_o). I get a scaler value which i multiply with the corresponding form needed to find the equation of a line.then how would we modify the above for to find the equation of the normal line to the curve in 3D? I thought it would have several forms as shown above.

why?
since in 2 dimensions we have
\frac{{\partial f}}{{\partial x}}(x_o,y_o,z_o)(x-x_o) + \frac{{\partial f}}{{\partial y}}(x_o,y_o,z_o)(y-y_o) = 0

and the normal line is
\frac{{\partial f}}{{\partial y}}(x_o,y_o,z_o)(x-x_o) - \frac{{\partial df}}{{\partial x}}(x_o,y_o,z_o)(y-y_o) = 0

so in 3 dimensions I thought it would be different. Or we can take the tangent line and find the negative reciprocal for one partial derivative at the point selected.

 

[HallsofIvy]

Now you are asking for "the" tangent line to a surface at a point? Again, there is no such thing! A smooth surface has a tangent plane at any point. Any line lying in that tangent plane and passing through that point is a tangent line.

Yes, it is "different in three dimension" but more different than you seem to think! The geometry of three dimensions simply isn't the geometry of two dimensions.

In two dimensions, a smooth curve has a single tangent line and a single normal line.

In three dimensions, a smooth curve has a single tangent line and a single normal plane, so an infinite number of normal lines, at a point.

In three dimensions, a smooth surface has a single normal line and a single tangent plane, so an infinite number of tangent line at a point.

 

[Ian_Brooks]

ahh that makes more sense,

thanks for bearing with me.