Tangent Plane and Normal Vector

If $z = f(x,y)$ is totally differentiable at $(x_0, y_0)$, then

$$f(x,y) = f(x_0,y_0) + f_{x}(x_0,y_0)\Delta x + f_{y}(x_0,y_0)\Delta y + \varepsilon(\Delta x, \Delta y)$$

where $(\Delta x, \Delta y) = (x-x_0, y-y_0)$. Note that

$$z = f(x_0,y_0) + f_{x}(x_0,y_0)\Delta x + f_{y}(x_0,y_0)\Delta y$$

represents the equation of the plane goes through a point $(x_0, y_0, f(x_0, y_0))$. Note also that these two equations behave almost the same when $(x,y)$ is close to $(x_0, y_0)$. Thus the equation above is called tangent plane of the function $z = f(x,y)$ at $(x_0, y_0, f(x_0, y_0))$.

NOTE Given the point $(x_0, y_0, f(x_0, y_0))$ and take a point $(x,y,z)$. Then we can form a vector $(x-x_0, y-y_0, z-f(x_0,y_0)$. Now $((x-x_0, y-y_0, z-f(x_0,y_0)\cdot (f_{x}(x_0,y_0), f_y(x_0,y_0),-1) = f_{x}(x-x_0) + f_y(x_0,y_0) - (z - f(x_0,y_0)) = 0$ Thus $(f_{x}(x_0,y_0), f_y(x_0,y_0),-1)$ is orthogonal to the tangent plane.

Example 4..11   Find the tangent plane and the normal line of the following function at $(1,1,6)$.

$$z = f(x,y) = 3x + xy + 2y$$

SOLUTION Since $f_x(x,y) = 3+y, f_y(x,y)=x+2$, the tangent plane of $f(x,y)$ at $(1,1,6)$ is given by the following.

$$z = f(1,1) + f_x(1,1)(x-1) + f_y(1,1)(y-1) = 6 + 4(x-1) + 3(y-1).$$

Now let $(x,y,z)$ be an arbitray point on the normal line. Then the vector connecting $(x,y,z)$ and $(1,1,6)$ is given by $(x,y,z) - (1,1,6)$ and this vector $(x-1,y-1,z-6)$ is on the normal line with the normal vector $(4,3,-1)$, Thus the normal line is

$$(x-1,y-1,z-6) = t(4,3,-1)$$   or$$t = \frac{x-1}{4} = \frac{y-1}{3} = \frac{z-6}{-1} \ensuremath{ \blacksquare}$$

Exercise 4..11   Find the tangent plane and the normal line of the following function at $(-2,3,2)$.

$$z = \frac{x^2}{4} + \frac{y^2}{9}$$

SOLUTION Since $${f_x(x,y) = \frac{x}{2}, f_y(x,y)= \frac{2y}{9}}$$, the tangent plane of $f(x,y)$ at $(-2,3,2)$ is given by

$$z = f(-2,3) + f_x(-2,3)(x+2) + f_y(-2,3)(y-3)$$

$$= 2 -(x+2) + \frac{2}{3}(y-3)$$

$$= -x + \frac{2y}{3} -2.$$

Now let $(x,y,z)$ be an arbitray point on the normal line. Then the vector connecting $(x,y,z)$ and $(-2,3,2)$ is given by $(x,y,z) - (-2,3,2)$ and this vector $(x+2,y-3,z-2)$ is on the normal line with the normal vector $(-1,\frac{2}{3},-1)$, Thus the normal line is

$$(x+2,y-3,z-2) = t(-1,\frac{2}{3},-1)$$

or

$$\frac{x+2}{-1} = \frac{3(y-3)}{2} = \frac{z-2}{-1} \ensuremath{ \blacksquare}$$

Exercise A

1.

Find the gradient and total differential of the following functions.

(a) $${f(x,y) = x^3 + y^{2}}$$ (b) $${f(x,y) = 3x^{2} - xy + y}$$ (c) $${z = x^2 y^{-2}}$$ (d) $${z = x^2 y}$$ (e) $${z = e^{x}\cos{y}}$$

2.

Answer the following questions．

(a) Find the equation of the tangent plane to the surface whose normal vector is $(3,2,-1)$. Find the equation of the normal line through the point $(1,1,1)$.

(b) Find the equation of the tangent plane to the surface $z = xy$ at the point $(2,1,2)$. Find the equation of the normal line through the point $(2,1,2)$.

(c) Find the equation of the tangent plane to the surface $z = x^{2} + xy + 2y^{2}$ at the point $(1,1,4)$. Find the equation of the normal line through the point $(1,1,4)$.

Exercise B

1.

Find the gradient and total differential of the following functions．Find the equation of the tangent plane to the surface at the point corresponds to $(1,1)$. Find the equation of the normal line at the point corresponds to $(1,1)$.

(a) $${f(x,y) = x^3 y^4}$$ (b) $${f(x,y) = x^3 y + x^2 y^4}$$ (c) $${z = x^2 y e^{2x}}$$ (d) $${z = \cos{xy}}$$

2.

Approximate the following value by using total diferential.．

(a) $${\sqrt{125}\sqrt[4]{17}}$$ (b) $${\sin{(\frac{6\pi}{7})}\cos{(\frac{\pi}{3})}}$$