Tangent Planes

Consider the surface F(x, y, z) = c, perhaps as z = f (x, y), and suppose that f and F have continuous partial derivatives. Suppose now we have a smooth curve on the surface, say $\phi$(t) = (x(t), y(t), z(t)). Then since the curve lies in the surface, we have

F(x(t), y(t), z(t)) = c,

and so, applying the chain rule, we have

$${\frac{{dF}}{{dt}}}$$ = $${\frac{{\partial F}}{{\partial x}}}$$$${\frac{{dx}}{{dt}}}$$ + $${\frac{{\partial F}}{{\partial y}}}$$$${\frac{{dy}}{{dt}}}$$ + $${\frac{{\partial F}}{{\partial z}}}$$$${\frac{{dz}}{{dt}}}$$ = 0.

or, writing this in terms of vectors, we have

$$\nabla$$F.v(t) = $$\left(\vphantom{\frac{\partial F}{\partial x}, \frac{\partial F}{\partial z}, \frac{\partial F}{\partial z} }\right.$$$${\frac{{\partial F}}{{\partial x}}}$$,$${\frac{{\partial F}}{{\partial z}}}$$,$${\frac{{\partial F}}{{\partial z}}}$$$$\left.\vphantom{\frac{\partial F}{\partial x}, \frac{\partial F}{\partial z}, \frac{\partial F}{\partial z} }\right)$$.$$\left(\vphantom{\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}}\right.$$$${\frac{{dx}}{{dt}}}$$,$${\frac{{dy}}{{dt}}}$$,$${\frac{{dz}}{{dt}}}$$$$\left.\vphantom{\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}}\right)$$ = 0.

Since the RH vector is the velocity of a point on the curve, which lies on the surface, we see that the left hand vector must be the normal to the curve.

Note that we have defined the gradient vector $\nabla$F associated with the function F by

$$\nabla$$F = $$\left(\vphantom{\frac{\partial F}{\partial x}, \frac{\partial F}{\partial z}, \frac{\partial F}{\partial z} }\right.$$$${\frac{{\partial F}}{{\partial x}}}$$,$${\frac{{\partial F}}{{\partial z}}}$$,$${\frac{{\partial F}}{{\partial z}}}$$$$\left.\vphantom{\frac{\partial F}{\partial x}, \frac{\partial F}{\partial z}, \frac{\partial F}{\partial z} }\right)$$.

Theorem 8.25   The tangent to the surface F(x, y, z) = c at the point (x₀, y₀, z₀) is given by

$${\frac{{\partial F}}{{\partial x}}}$$(x - x₀) + $${\frac{{\partial F}}{{\partial y}}}$$(y - y₀) + $${\frac{{\partial F}}{{\partial z}}}$$(z - z₀) = 0.

Proof. This is a simple example of the use of vector geometry. Given that (x₀, y₀, z₀) lies on the surface, and so in the tangent, then for any other point (x, y, z) in the tangent plane, the vector (x - x₀, y - y₀, z - z₀) must lie in the tangent plane, and so must be normal to the normal to the curve (i.e. to $\nabla$F). Thus (x - x₀, y - y₀, z - z₀) and $\nabla$F are perpendicular, and that requirement is the equation which gives the tangent plane. $\qedsymbol$

Example 8.26   Find the equation of the tangent plane to the surface

F(x, y, z) = x² + y² + z - 9 = 0

at the point P = (1, 2, 4).

Solution. We have $\nabla$F|_{(1, 2, 4)} = (2, 4, 1), and the equation of the tangent plane is

2(x - 1) + 4(y - 2) + (z - 4) = 0.

Exercise 8.27   Show that the tangent plane to the surface z = 3xy - x³ - y³ is horizontal only at (0, 0, 0) and (1, 1, 1).