Tangent Plane and Normal Vector

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

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

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

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

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 $z = f(x,y)$

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 $(1,1,6)$

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

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.

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

Now let

be an arbitray point on the normal line. Then the vector connecting $(x,y,z)$

and

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

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

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

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

SOLUTION Since $\displaystyle{f_x(x,y) = \frac{x}{2}, f_y(x,y)= \frac{2y}{9}}$ , the tangent plane of $f(x,y)$

is given by

$\displaystyle z$	$\displaystyle =$	$\displaystyle f(-2,3) + f_x(-2,3)(x+2) + f_y(-2,3)(y-3)$
	$\displaystyle =$	$\displaystyle 2 -(x+2) + \frac{2}{3}(y-3)$
	$\displaystyle =$	$\displaystyle -x + \frac{2y}{3} -2.$

Now let

be an arbitray point on the normal line. Then the vector connecting $(x,y,z)$

and

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

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

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

Exercise A

1.

Find the gradient and total differential of the following functions.

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

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 .

(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 .

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 . Find the equation of the normal line at the point corresponds to .
(a) $\displaystyle{f(x,y) = x^3 y^4}$ (b) $\displaystyle{f(x,y) = x^3 y + x^2 y^4}$ (c) $\displaystyle{z = x^2 y e^{2x}}$ (d) $\displaystyle{z = \cos{xy}}$
2.: Approximate the following value by using total diferential.．
(a) $\displaystyle{\sqrt{125}\sqrt[4]{17}}$ (b) $\displaystyle{\sin{(\frac{6\pi}{7})}\cos{(\frac{\pi}{3})}}$