# The equation of a normal plane to a curve - confusion

• Ian_Brooks
In summary, the concept of tangent and normal lines is different in two dimensions versus three dimensions. In two dimensions, a smooth curve has a single tangent line and a single normal line, while in three dimensions, a smooth curve has a single tangent line and a single normal plane, resulting in an infinite number of tangent and normal lines at a point.

#### 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 (xo,yo,zo) is given by

$$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(x-xo) + $$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(y-yo) + $$\frac{{\partial d}}{{\partial z}}$$(xo,yo,zo)(z-zo) = 0

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

$$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial z}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(z-zo) = 0

$$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(z-zo) = 0

$$\frac{{\partial d}}{{\partial z}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(z-zo) = 0

$$\frac{{\partial d}}{{\partial z}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(z-zo) = 0

Ian_Brooks said:
Not a homework question

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

and the tangent plane to the curve at a point (xo,yo,zo) is given by

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

$$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial z}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(z-zo) = 0

$$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(z-zo) = 0

$$\frac{{\partial d}}{{\partial z}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(z-zo) = 0

$$\frac{{\partial d}}{{\partial z}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial d}}{{\partial y}}$$(xo,yo,zo)(y-yo) - $$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)(z-zo) = 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
$$\frac{{\partial d}}{{\partial x}}$$(xo,yo,zo)
You don't differentiate a point!
My guess would be that you mean
$$\frac{{\partial f}}{{\partial x}}$$(xo,yo,zo)
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 [itex]f'(t_0)(x- f(t_0))+ g'(t_0)(y- g(t_0))+ h'(t_0)(z- h(t_0))= 0.

Last edited by a moderator:
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 (xo,yo,zo)

we use the following equation

$$\frac{{\partial f}}{{\partial x}}$$(xo,yo,zo)(x-xo) + $$\frac{{\partial f}}{{\partial y}}$$(xo,yo,zo)(y-yo) + $$\frac{{\partial f}}{{\partial z}}$$(xo,yo,zo)(z-zo) = 0

which means I find the partial derivatives of f with respect to x,y and z then I substitute the points (xo,yo,zo). 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
$$\frac{{\partial f}}{{\partial x}}$$(xo,yo,zo)(x-xo) + $$\frac{{\partial f}}{{\partial y}}$$(xo,yo,zo)(y-yo) = 0

and the normal line is
$$\frac{{\partial f}}{{\partial y}}$$(xo,yo,zo)(x-xo) - $$\frac{{\partial df}}{{\partial x}}$$(xo,yo,zo)(y-yo) = 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.

Last edited:
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.

ahh that makes more sense,

thanks for bearing with me.

## 1. What is the equation of a normal plane to a curve?

The equation of a normal plane to a curve is a mathematical expression that describes the relationship between a curve and a plane that is perpendicular (or normal) to the curve at a given point. It can be represented in various forms, such as parametric or implicit equations.

## 2. How is the equation of a normal plane to a curve different from the equation of the curve itself?

The equation of a normal plane to a curve and the equation of the curve itself are two different equations that describe different aspects of the same curve. The equation of the curve represents the relationship between the independent and dependent variables, while the equation of the normal plane describes the relationship between the curve and a plane that is perpendicular to it at a specific point.

## 3. What does it mean for a plane to be normal to a curve?

A plane is considered normal to a curve if it is perpendicular to the curve at a particular point. This means that the plane intersects the curve at a 90-degree angle at that point.

## 4. How is the equation of a normal plane to a curve used in real-world applications?

The equation of a normal plane to a curve is used in various fields, such as engineering, physics, and computer graphics, to help understand and analyze the behavior of curves and surfaces. It is also useful for calculating the slope or gradient of a curve at a specific point.

## 5. What are some common sources of confusion when dealing with the equation of a normal plane to a curve?

Some common sources of confusion when dealing with the equation of a normal plane to a curve include understanding the concept of perpendicularity, the different forms of the equation, and the relationship between the normal plane and the curve itself. It is important to have a solid understanding of basic geometry and calculus principles to fully comprehend the equation of a normal plane to a curve.