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 $$F_x^{\prime }(x_0,y_0,z_0)(x-x_0)+F_y^{\prime }(x_0,y_0,z_0)(y-y_0)+F_z^{\prime }(x_0,y_0,z_0)(z-z_0)=0.$$

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

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 $$z-z_0=f_x^{\prime }(x_0,y_0)(x-x_0)+f_y^{\prime }(x_0,y_0)(y-y_0),$$ takes the form $$\frac{x-x_0}{f_x^{\prime }(x_0,y_0)}=\frac{y-y_0}{f_y^{\prime }(x_0,y_0)}=\frac{z-z_0}{-1}.$$

Examples:

1. 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 $$z-z_0=f_x^{\prime }(x_0,y_0)(x-x_0)+f_y^{\prime }(x_0,y_0)(y-y_0),$$ and the equation of the normal is given by $$\frac{x-x_0}{f_x^{\prime }(x_0,y_0)}=\frac{y-y_0}{f_y^{\prime }(x_0,y_0)}=\frac{z-z_0}{-1}.$$

Let's find the partial derivatives:

$z_x^{\prime }=(\sin x\cos y{)}_x^{\prime }=\cos x\cos y;$

$z_x^{\prime }(\pi /4,\pi /4)=\cos \frac{\pi }{4}\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2};$

$z_y^{\prime }=(\sin x\cos y{)}_y^{\prime }=-\sin x\sin y;$

$z_y^{\prime }(\pi /4,\pi /4)=-\sin \frac{\pi }{4}\sin \frac{\pi }{4}=-\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=-\frac{1}{2};$

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: $\frac{x-\frac{\pi }{4}}{\frac{1}{2}}=\frac{y-\frac{\pi }{4}}{-\frac{1}{2}}=\frac{z-\frac{\pi }{4}}{-1}.$

2. 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$$z-z_0=f_x^{\prime }(x_0,y_0)(x-x_0)+f_y^{\prime }(x_0,y_0)(y-y_0).$$

We find the partial derivatives:

$z_x^{\prime }=(4x-xy+y^2{)}_x^{\prime }=4-y;$

$z_y^{\prime }=(4x-xy+y^2{)}_y^{\prime }=-x+2y;$

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: $$z-11=(4-6)(x-\frac{25}{2})+(-\frac{25}{2}+2\cdot 6)(y-6)\Rightarrow$$ $$z-11=-2(x-\frac{25}{2})-\frac{1}{2}(y-6)\Rightarrow 2x+\frac{1}{2}y+z-11-25-3=0\Rightarrow$$ $$\Rightarrow 4x+y+2z-78=0.$$

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

3. 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$$F_x^{\prime }(x_0,y_0,z_0)(x-x_0)+F_y^{\prime }(x_0,y_0,z_0)(y-y_0)+F_z^{\prime }(x_0,y_0,z_0)(z-z_0)=0.$$

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

We find the partial derivatives:

$F_x^{\prime }=(x^2{y}^2-xyz+x^{2}yz-x{z}^2+8{)}_x^{\prime }=2x{y}^2-yz+2xyz-z^2;$

$F_x^{\prime }(2,1,3)=4-3+12-9=4;$

$F_y^{\prime }=(x^2{y}^2-xyz+x^{2}yz-x{z}^2+8{)}_y^{\prime }=2x^{2}y-xz+x^{2}z;$

$F_y^{\prime }(2,1,3)=8-6+12=14;$

$F_z^{\prime }=(x^2{y}^2-xyz+x^{2}yz-x{z}^2+8{)}_z^{\prime }=-xy+x^{2}y-2xz;$

$F_z^{\prime }(2,1,3)=-2+4-12=-10.$

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:

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

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

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

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

Tags: Tangent plane, calculus, derivative, mathematical analysis, normal