Tangent plane and normal to an explicitly defined surface.

Tangent plane and normal to an explicitly defined surface.

The tangent plane to a surface at its point $M_0$ (point of tangency) is a plane containing all tangents to curves drawn on the surface through this point.

The normal to the surface is a line perpendicular to the tangent plane and passing through the point of tangency.

If the equation of the surface is given by $$F(x,y,z)=0,$$

then the equation of the tangent plane at the point $M_0(x_0, y_0, z_0)$ is given by $$F_x'(x_0, y_0, z_0)(x-x_0)+F_y'(x_0, y_0, z_0)(y-y_0)+F_z'(x_0, y_0, z_0)(z-z_0)=0.$$

The equation of the normal is given by $$\frac{x-x_0}{F_x'(x_0, y_0, z_0)}=\frac{y-y_0}{F_y'(x_0, y_0, z_0)}=\frac{z-z_0}{F_z'(x_0, y_0, z_0)}.$$

In the case of the surface given explicitly as $$z=f(x, y)$$ the equation of the tangent plane at the point $M_0(x_0, y_0, z_0)$ has the form $$z-z_0=f_x'(x_0, y_0)(x-x_0)+f_y'(x_0, y_0)(y-y_0),$$ takes the form $$\frac{x-x_0}{f_x'(x_0, y_0)}=\frac{y-y_0}{f_y'(x_0, y_0)}=\frac{z-z_0}{-1}.$$

Examples:

7.229. a) Find the equations of the tangent plane and normal to the surface $z=\sin x\cos y$ at the point $(\pi/4, \pi/4, \pi/4)$.

Solution.

For the surface

$$z=f(x, y)$$ the equation of the tangent plane at the point $M_0(x_0, y_0, z_0)$ is given by $$z-z_0=f_x'(x_0, y_0)(x-x_0)+f_y'(x_0, y_0)(y-y_0),$$ and the equation of the normal is given by $$\frac{x-x_0}{f_x'(x_0, y_0)}=\frac{y-y_0}{f_y'(x_0, y_0)}=\frac{z-z_0}{-1}.$$

Let's find the partial derivatives:

$z'_x=(\sin x\cos y)'_x=\cos x\cos y;$

$z'_x(\pi/4, \pi/4)=\cos \frac{\pi}{4}\cos \frac{\pi}{4}=\frac{1}{\sqrt 2}\cdot\frac{1}{\sqrt 2}=\frac{1}{2};$

$z'_y=(\sin x\cos y)'_y=-\sin x\sin y;$

$z'_y(\pi/4, \pi/4)=-\sin \frac{\pi}{4}\sin \frac{\pi}{4}=-\frac{1}{\sqrt 2}\cdot\frac{1}{\sqrt 2}=-\frac{1}{2};$

Thus, the equation of the tangent plane: $$z-\frac{\pi}{4}=\frac{1}{2}(x-\frac{\pi}{4})-\frac{1}{2}(y-\frac{\pi}{4})\Rightarrow$$ $$\frac{1}{2}x -\frac{1}{2}y-z+\frac{\pi}{4}=0.$$

The equation of the normal: $$\frac{x-\frac{\pi}{4}}{\frac{1}{2}}=\frac{y-\frac{\pi}{4}}{-\frac{1}{2}}=\frac{z-\frac{\pi}{4}}{-1}.$$

Answer: equation of the tangent plane: $\frac{1}{2}x -\frac{1}{2}y-z+\frac{\pi}{4}=0;$ normal equation: $\frac{x-\frac{\pi}{4}}{\frac{1}{2}}=\frac{y-\frac{\pi}{4}}{-\frac{1}{2}}=\frac{z-\frac{\pi}{4}}{-1}.$

7.232. For the surface $z=4x-xy+y^2$, find the equation of the tangent plane parallel to the plane $4x+y+2z+9=0$.

Solution.

For the surface $$z=f(x, y)$$ The equation of the tangent plane at the point $M_0(x_0, y_0, z_0)$ is given by$$z-z_0=f_x'(x_0, y_0)(x-x_0)+f_y'(x_0, y_0)(y-y_0).$$

We find the partial derivatives:

$z'_x=(4x-xy+y^2)'_x=4-y;$

$z'_y=(4x-xy+y^2)'_y=-x+2y;$

From this, we find the equation of the tangent plane: $$z-z_0=(4-y_0)(x-x_0)+(-x_0+2y_0)(y-y_0)\Rightarrow$$ $$(4-y_0)(x-x_0)+(-x_0+2y_0)(y-y_0)-z+z_0=0.$$

Let's find the point on the surface $M(x_0, y_0, x_0)$ at which the tangent plane is parallel to the plane $4x + y + 2z + 9 = 0$:

$$\frac{4-y_0}{4}=\frac{-x_0+2y_0}{1}=\frac{-1}{2}\Rightarrow 4-y_0=-2\Rightarrow y_0=6;$$ $$-x_0+12=-\frac{1}{2}\Rightarrow x_0=\frac{25}{2};$$ $$z_0=4\cdot\frac{25}{2}-\frac{25}{2}\cdot 6+6^2=50-75+36=11.$$

Thus, the equation of the tangent plane is: $$z-11=(4-6)(x-\frac{25}{2})+(-\frac{25}{2}+2\cdot 6)(y-6)\Rightarrow$$ $$z-11=-2(x-\frac{25}{2})-\frac{1}{2}(y-6)\Rightarrow 2x+\frac{1}{2}y+z-11-25-3=0\Rightarrow$$ $$\Rightarrow4x+y+2z-78=0.$$

Answer: $4x+y+2z-78=0.$

7.233. a) Find the equations of the tangent plane and the normal to the surface $x(y+z)(xy-z)+8=0$ at the point $(2, 1, 3)$.

Solution.

For the surface $$F(x,y,z)=0,$$ the equation of the tangent plane at the point $M_0(x_0, y_0, z_0)$ is$$F_x'(x_0, y_0, z_0)(x-x_0)+F_y'(x_0, y_0, z_0)(y-y_0)+F_z'(x_0, y_0, z_0)(z-z_0)=0.$$

$F(x, y, z)=x(y+z)(xy-z)+8=x^2y^2-xyz+x^2yz-xz^2+8=0$

We find the partial derivatives:

$F'_x=(x^2y^2-xyz+x^2yz-xz^2+8)'_x=2xy^2-yz+2xyz-z^2;$

$F'_x(2, 1, 3)=4-3+12-9=4;$

$F'_y=(x^2y^2-xyz+x^2yz-xz^2+8)'_y=2x^2y-xz+x^2z;$

$F'_y(2, 1, 3)=8-6+12=14;$

$F'_z=(x^2y^2-xyz+x^2yz-xz^2+8)'_z=-xy+x^2y-2xz;$

$F'_z(2, 1, 3)=-2+4-12=-10.$

From here, we derive the equation of the tangent plane: $$4(x-2)+14(y-1)-10(z-3)=0\Rightarrow 4x+14y-10z+8.$$

Answer: $2x+7y-5z+4=0.$

Homework:

7.229. b) Find the equations of the tangent plane and the normal to the surface $z = e^{x \cos y}$ at the point $(1, \pi, 1/e)$.

7.230. Find the distance from the origin to the tangent plane of the surface $z = y \tan\frac{x}{4}$ at the point $\left(\frac{\pi a}{4}, a, a\right)$.

7.233. b) Find the equations of the tangent plane and the normal to the surface $2^{x/z} + 2^{y/z} = 8$ at the point $(2, 2, 1)$.

7.234. For the surface $x^2 - z^2 - 2x + 6y = 4$, find the equation of the normal line that is parallel to the line $\frac{x+2}{1} = \frac{y}{3} = \frac{z+1}{4}$.

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