# Normal Lines and Tangent Planes • By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## Tangent Planes

Suppose $S$ is a surface with the equation $z=f(x,y)$ where $f$ has continuous first partial derivatives $f_x$ and $f_y.$ Let $P(a,b,c)$ be a point on $S$ and let $C_1$be the curve of intersection of $S$ with the plane $x=a$ and $C_2$ the intersection of $S$ with the plane $y=b.$ The tangent lines $T_1$ and $T_2$ to $C_1$ and $C_2$, respectively determine a unique plane and this plane actually contains the tangent to every smooth curve $C$ that passes through $P.$ We call this plane the tangent plane to $S$ at $P.$

Theorem. (Equation of Tangent Plane) If $z=f(x,y)$ is differentiable at $(a,b)$ then an equation of the tangent plane to the graph of $f$ at $(a,b)$ is \begin{equation} z-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b). \end{equation} The surface $z=2x^2+y^2$ and its tangent plane at $(1,1,3).$

Example. Find an equation of the tangent plane to the surface $$z=2x^2+y^2$$ at the point $(1,1,3).$

Solution. Let $f(x,y)=2x^2+y^2.$ Then $$f_x(x,y)=4x, \quad f_x(1,1)=4, \quad f_y(x,y)=2y, \quad f_y(1,1)=2.$$ Then an equation of the tangent plane is $z-3=4(x-1)+2(y-1)$ or \begin{equation} z=4x+2y-3. \end{equation} The surface $z=4-x^2-y^2$ and its tangent plane at $(1,1,2).$

Example. Find an equation of the tangent plane to the surface $$z=4-x^2-y^2$$ at the point $(1,1,2).$

Solution. Let $f(x,y)=4-x^2-y^2.$ Then $$f_x(x,y)=-2x, \quad f_x(1,1)=-2, \quad f_y(x,y)=-2y, \quad f_y(1,1)=-2.$$ Then an equation of the tangent plane is $z-2=-2(x-1)-2(y-1)$ or \begin{equation} z=-2x-2y+6 \end{equation} The surface $z=\sin x+\sin y$ and its tangent plane at $\left(\frac{\pi}{2},\frac{\pi}{3},1+\frac{\sqrt{3}}{2}\right).$

Example. Find an equation of the tangent plane to the surface $$z=\sin x+\sin y$$ at the point $\left(\frac{\pi}{2},\frac{\pi}{3},1+\frac{\sqrt{3}}{2}\right).$

Solution. Let $f(x,y)=\sin x+\sin y.$ Then $$f_x(x,y)=\cos x, \quad f_x\left(\frac{\pi}{2},\frac{\pi}{3}\right)=0, \quad f_y(x,y)=\cos y, \quad f_y\left(\frac{\pi}{2},\frac{\pi}{3}\right)=\frac{1}{2}.$$ Then an equation of the tangent plane is $$z-\left(1+\frac{\sqrt{3}}{2}\right)=0\left(x-\frac{\pi}{2}\right)+\frac{1}{2}\left(y-\frac{\pi}{3}\right)$$ or \begin{equation}
z=\frac{1}{2}y-\frac{\pi}{6}+1+\frac{\sqrt{3}}{2} \end{equation}

Theorem. (Tangent Plane To A Surface) Suppose $S$ is a surface with the equation $F(x,y,z)=C$ and let $P(a,b,c)$ be a point on $S$ where $F$ is differentiable with $\nabla F_P\neq 0.$ Then an equation of the tangent plane to $S$ at $P$ is \begin{equation} F_x(a,b,c)(x-a)+F_y(a,b,c)(y-b)+F_z(a,b,c)(z-c)=0 \end{equation} and the normal line to $S$ at $P$ has parametric equations \begin{equation} x=a+F_x(a,b,c) t, \qquad y=b+F_y(a,b,c) t, \qquad z=c+F_z(a,b,c) t.\end{equation}

Proof. Any plane that passes through $P(a,b,c)$ has an equation of the form $$A(x-a)+B(y-b)+C(z-c)=0.$$ By dividing this equation by $C$ and letting $a_1=-A/C$ and $b_1=-B/C$, we can write it in the form $$z-c=a_1(x-a)+b_1(y-b).$$ If this equation represents the tangent plane at $P$, then its intersection with the plane $y=b$ must be the tangent line with slope $f_x(a,b).$ Therefore, $a_1=f_x(a,b).$ Similarly, putting $x=a$, we get $z-c=b_2(y-b),$ which must represent the tangent line with slope $f_y(a,b)$ and so $b_2=f_y(a,b).$ Thus, the equation of the tangent plane is $$z-c=f_x(a,b)(x-a)+f_y(a,b)(y-b).$$

Example. Find an equation of the tangent plane to the surface \begin{equation} \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 \end{equation} at the point $(x_0,y_0,z_0).$

Solution. Since $F_x=2x/a^2,$ $F_y=2y/b^2,$ and $F_z=2z/c^2,$ the tangent plane at $P_0\left(x_0,y_0,z_0\right)$ has equation \begin{equation} \frac{2x_0}{a^2}\left(x-x_0\right)+\frac{2y_0}{b^2}\left(y-y_0 \right) + \frac{2z_0}{c^2}\left(z-z_0\right)=0\end{equation} \begin{equation} \frac{x_0 x}{a^2}-\frac{x_0{}^2}{a^2}+\frac{y_0y}{b^2}-\frac{y_0^2}{b^2}+\frac{z_0z}{c^2}-\frac{z_0{}^2}{c^2}=0\end{equation} \begin{equation} \frac{x_0 x}{a^2}+\frac{y_0y}{b^2}+\frac{z_0z}{c^2}=1. \end{equation}

Example. Find an equation of the tangent plane to the surface \begin{equation} \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1 \end{equation} at the point $(x_0,y_0,z_0).$

Solution. Since $F_x=2x/a^2,$ $F_y=2y/b^2,$ and $F_z=-2z/c^2,$ the tangent plane at $P_0\left(x_0,y_0,z_0\right)$ has equation \begin{equation} \frac{2x_0}{a^2}\left(x-x_0\right)+\frac{2y_0}{b^2}\left(y-y_0\right)-\frac{2z_0}{c^2}\left(z-z_0\right)=0\end{equation} \begin{equation} \frac{x_0 x}{a^2}-\frac{x_0^2}{a^2}+\frac{y_0y}{b^2}-\frac{y_0^2}{b^2}+\frac{z_0z}{c^2}+\frac{z_0^2}{c^2}=0\end{equation} \begin{equation} \frac{x_0 x}{a^2}+\frac{y_0y}{b^2}-\frac{z_0z}{c^2}=1. \end{equation}

Example. Find an equation of the tangent plane to the surface \begin{equation} \frac{z}{c}=\frac{x^2}{a^2}+\frac{y^2}{b^2} \end{equation} at the point $(x_0,y_0,z_0).$

Solution. Since $F_x=2x/a^2,$ $F_y=2y/b^2,$ and $F_z=-1/c,$ the tangent plane at $P_0\left(x_0,y_0,z_0\right)$ has equation \begin{equation}
\frac{2x_0}{a^2}\left(x-x_0\right)+\frac{2y_0}{b^2}\left(y-y_0\right)-\frac{1}{c}\left(z-z_0\right)=0\end{equation} \begin{equation} \frac{x_0 x}{a^2}-\frac{x_0^2}{a^2}+\frac{y_0y}{b^2}-\frac{y_0{}^2}{b^2}+\frac{z}{c}+\frac{z_0}{c}=0 \end{equation} \begin{equation} \frac{x_0 x}{a^2}+\frac{y_0y}{b^2}-\frac{z}{c}=\frac{z_0}{c}. \end{equation}

## Normal Lines

Example. Find the equations for the tangent plane and the normal line to the cone $$z^2=x^2+y^2$$ at the point where $x=3, y=4,$ and $z>0.$

Solution. If $P(a,b,c)$ is the point of tangency and $a=3, y=4,$ and $c>0,$ then $c=\sqrt{a^2+b^2}=5.$ If we consider $F(x,y,z)=x^2+y^2-z^2,$ then the cone can be regarded as the level surface $F(x,y,z)=0.$ The partial derivatives of $F$ are $F_x=2x$, $F_y=2y$, and $F_z=-2z$ so at $P(3,4,5)$ we find $F_x(3,4,5)=6$, $F_y(3,4,5)=8,$ and $F_z(3,4,5)=-10.$ Thus the tangent plane has an equation \begin{equation}6(x-3)+8(y-4)-10(z-5)=0\end{equation} or $3x+4y-5z=0$ and the normal line is given parametrically by the equations $x=3+6t ,$ $y=4+8t,$ and $z=5-10t.$

Example. Find the equations of the tangent plane and the normal line at the point $(-2,1,-3)$ to the ellipsoid $$\frac{x^2}{4}+y^2+\frac{z^2}{9}=3.$$

Solution. The ellipsoid is a level surface of the function $$F(x,y,z)=\frac{x^2}{4}+y^2+\frac{z^2}{9}.$$ Therefore we have \begin{equation} F_x(x,y,z)=\frac{x}{2}, \quad F_y(x,y,z)=2y, \quad F_z(x,y,z)=\frac{2z}{9} \end{equation} Thus, $$F_x(-2,1,3)=-1, \quad F_y(-2,1,3)=2, \quad F_z(-2,1,3)=-2/3.$$ Then the tangent plane at $(-2,1,-3)$ is $$-1(x+2)+2(y-1)-(2/3)(z+3)=0$$ which simplifies to $$3x-6y+2z+18=0$$ and parametric equations for the normal line are $$x=-2- t, \quad y=1+2t, \quad z=-3-(2/3)t.$$

Example. Find an equation of the tangent plane to the surface $z=4-x^2-y^2$ at the point $(1,1,2).$

Solution. Let $f(x,y)=4-x^2-y^2.$ Then $$f_x(x,y)=-2x, \quad f_x(1,1)=-2, \quad f_y(x,y)=-2y, \quad f_y(1,1)=-2.$$ Then an equation of the tangent plane is $z-2=-2(x-1)-2(y-1)$ or \begin{equation} z=-2x-2y+6 \end{equation}

Example. Find an equation of the tangent plane to the surface $$x^3-xy^2+z^3-2x+6=0$$ at the point $P(1,2,-1).$

Solution. Let $F(x,y,z)=x^3-xy^2+z^3-2x+6.$ The normal vector of the tangent plane at the point $P(1,2,-1)$ is $\nabla F(1,2,-1)$ which is the directional vector of the the normal line at the same point. We have \begin{align*}\nabla F(x,y,z) & = F_x\mathbf{i}+F_y\mathbf{j}+F_z\mathbf{k} \\ & =(3x^2-y^2-2)\mathbf{i} -2xy\mathbf{j}+3z^2\mathbf{k}\end{align*} and so $\nabla F(1,2,-1) = -3\mathbf{i} +4\mathbf{j}+3\mathbf{k}.$ Therefore the equation for the tangent plane is $$-3(x-1)+4(y-2)+3(z+1)=0,$$ and the equation for the normal line is $$\frac {x-1}{-3}=\frac {y-2}4=\frac {z+1}3.$$

## Exercises on Normal Lines and Tangent Planes

Exercise. Find the maximum rate of change of the function at the given point and the direction in which it occurs $f(x,y)=\sqrt{x^2+2y}$, $(4,10)$

Exercise. Show that if $x$ and $y$ are sufficiently close to zero and $f$ is differentiable at $(0,0),$ then \begin{equation} f(x,y)\approx f(0,0)+x f_x(0,0)+y f_y(0,0) \end{equation} Use this approximation for the expressions \begin{equation} \frac{1}{1+x-y} \qquad \text{and} \qquad \frac{1}{(x+1)^2+(y+1)^2} \end{equation} around $(0,0).$

Exercise. Find the total differential of the function $z=x^2-2x y +3y^2.$

Exercise. Find the tangent plane to the surface $z=x \sin y-x \cos x$ at $(\pi,0).$

Exercise. Find an equation for each horizontal tangent plane to the surface $z=5-x^2-y^2+4y.$

Exercise. Determine the total differential.

$(1) \quad z=5x^2y^3$

$(2) \quad z=\cos x^2y$

$(3) \quad z=y e^x$

$(4) \quad w=\sin x+\sin y+\cos z$

$(5) \quad w=z^2\sin (2x-3y)$

$(6) \quad w=3y^2z \cos x$

Exercise. Show that each functions is differentiable on $\mathbb{R}^2.$

$(1) \quad f(x,y)=x y^3+3 x y^2$

$(2) \quad f(x,y)=\sin \left(x^2+3y\right)$

$(3) \quad f(x,y)=x^2+4x-y^2$

$(4) \quad f(x,y)=e^{2x+y^2}$

Exercise. Determine the standard-form equation for the tangent plane to the surface at the specified point.

$(1) \quad f(x,y)=\sqrt{x^2+y^2}$ at $P_0\left(3,1,\sqrt{10}\right)$

$(2) \quad f(x,y)=x^2+y^2+ \sin x y$ at $P_0(0,2,4)$

$(3) \quad f(x,y)=e^{-x} \sin y$ at $P_0\left(0,\frac{\pi}{2},1\right)$

$(4) \quad z=10-x^2-y^2$ at $P(2,2,2)$

$(5) \quad z=\ln \left|x+y^2\right|$ at $P(-3,-2,0)$

Exercise. Use linear approximation (differentials) to find approximate value.

$(1) \quad \left(\sqrt{\frac{\pi }{2}}-0.01\right)$

$(2) \quad \sin \left(\sqrt{\frac{\pi }{2}}+0.01\right)$

$(3) \quad e^{1.01^20.98^2}$

Exercise. Show that if $x$ and $y$ are sufficiently close to zero and $f$ is differentiable at $(0,0),$ then \begin{equation} f(x,y)\approx f(0,0)+x f_x(0,0)+y f_y(0,0) \end{equation} Use this approximation for the expressions \begin{equation} \frac{1}{1+x-y} \qquad \text{and} \qquad \frac{1}{(x+1)^2+(y+1)^2} \end{equation} around $(0,0).$

Exercise. Find a unit vector that is normal to the given graph at the point $P_0\left(x_0,y_0\right)$ on the graph.

(1) the circle $x^2+y^2=a^2$

(2) the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

Exercise. Find a unit vector that is normal to each surface given at the prescribed point, and the standard form of the equation of the tangent plane at that point.

$(1) \quad \ln \left(\frac{x}{y-z}\right)=0$ at $(2,5,3)$

$(2) \quad z e^{x^2-y^2}=3$ at $(1,1,3)$