Dated: 30-10-2024

Tangent Planes to the Surfaces

If \(C\) is a parametric curve on 3D then the tangent line¹ to the \(C\) at a point \(P_0\) is the line² going through \(P_0\) and the unit tangent vector.³

Tangent Plane

If \(P_0(x_0, y_0, z_0)\) is any point on the surface \(S\) then the plane⁴ which contains all the tangents¹ to the point \(P_0\) is called tangent plane.

Surface Normal

The vector³ which is perpendicular to tangent plane is called surface normal.

Parametric Equations of a line²

The parametric equations of a line² in 2D, going through \(P(x_0, y_0)\), parallel to the vector³ \(a \hat i + b \hat j\) is given by

\[x = x_0 + at \text{ and } y = y_0 + bt\]

Eliminating \(t\), we get

\[\frac{x-x_{0}}{a}=\frac{y-y_0}{b}\]

\[y-y_{0}=\frac{b}{a}(x-x_{0})\]

Similarly for 3D,

\[\frac{x-x_{0}}{a}=\frac{y-y_0}{b} = \frac{z - z_0}{c}\]

Parametric Vector Form

\[\vec{r}(t) = (x_0 + at) \hat i + (y_0 + bt) \hat j\]

The idea is to make the \(x\) and \(y\) variables dependent on some other variable which is \(t\) in this case.

Equation of a Tangent Plane

We can construct a plane⁴ if we know

1. A point on the plane⁴
2. A normal vector to it.

Let there be a point \(P_0(x_0, y_0, z_0)\) on the plane⁴ and normal being in the direction of \(\vec n = a \hat i + b \hat j + c \hat k\)
Then let \(P(x, y, z)\) be any arbitrary point on this plane.⁴
Then we can construct number of vectors³ on this plane⁴ by

\[\vec{P_0P} = (x - x_0) \hat i + (y - y_0) \hat j + (z - z_0) \hat k\]

Also

\[\vec n \cdot \vec{P_0P} = 0\]

Then the equation of the tangent plane is

\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]

Example

We know that the general equation for the plane⁴ is

\[ax + by + cz + d = 0\]

Let their be 2 points on it

\[(x_1, y_1, z_1)\]

\[(x_2, y_2, z_2)\]

Putting these in our equation of the plane,⁴ we get

\[ax_1 + by_1 + cz_1 + d = 0\]

\[ax_2 + by_2 + cz_2 + d = 0\]

Subtracting these, we get

\[a(x_2 - x_1) + b(y_2 - y_1) + c(z_2 - z_1) = 0\]

From the definition of dot product,³ we can reverse it as

\[( a \hat i + b \hat j + c \hat k ) \cdot \left((x_2 - x_1) \hat i + (y_2 - y_1) \hat j + (z_2 - z_1) \hat k\right) = 0\]

Here

\[\vec V = (x_2 - x_1) \hat i + (y_2 - y_1) \hat j + (z_2 - z_1) \hat k\]

If

\[\Phi = ax + by + cz\]

then

\[\Phi_x = a \text{ and } \Phi_y = b \text{ and } \Phi_z = c\]

\[\therefore \nabla \Phi = a \hat i + b \hat j + c \hat k\]

Plugging it into the equation we got from the dot product³ definition, we get

\[\nabla \Phi \cdot \vec V = 0\]

This shows that \(\nabla \Phi\) is always normal to the surface.

Gradients and Tangents to the Surface

\[z = f(x, y) = c\]

If this differentiable function⁵ has a constant value \(c\) along some smooth curve having parametric equations

\[x = g(t)\]

\[y = h(t)\]

\[\vec r = g(t) \hat i + h(t) \hat j\]

Then

\[\frac{d}{dt} \left(f\left(g(t), h(t)\right)\right) = \frac{d}{dt} c\]

Applying chain rule,⁶ we get

\[\frac{\partial f}{\partial x}\frac{dg}{dt}+\frac{\partial f}{\partial y}\frac{dh}{dt}=0\]

From the dot product³ definition

\[\left(\frac{\partial f}{\partial x}\hat i+\frac{\partial f}{\partial y} \hat j\right)\cdot\left(\frac{dg}{dt} \hat i+\frac{dh}{dt}\hat j\right)=0\]

Some additional reversing, we get

\[\nabla f \cdot \frac{d \vec{r}}{dt} = 0\]

Example

Find tangent plane to the surface

\[f(x, y, z) = 9 x^2 + 4y^2 - z^2 - 36 \text{ at } P(2, 3, 6)\]

Solution

\[\frac{\partial}{\partial x} f(x, y, z) = 18x\]

\[\frac{\partial}{\partial y} f(x, y, z) = 8y\]

\[\frac{\partial}{\partial z} f(x, y, z) = -2z\]

\[f_x(P) = 18(2) = 36\]

\[f_y(P) = 8(3) = 24\]

\[f_z(P) = -2(6) = -12\]

So the equation for tangent plane will be

\[36(x-2)+24(y-3)-12(z-6)=0\]

\[3x+2y-z-6=0\]

References

Read more about notations and symbols.

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1. Read more about tangents. ↩↩

2. Read more about lines. ↩↩↩

3. Read more about vectors. ↩↩↩↩↩↩↩

4. Read more about planes. ↩↩↩↩↩↩↩↩

5. Read more about functions. ↩

6. Read more about the chain rule. ↩