Section 12.7 Tangent Lines, Normal Lines, and Tangent Planes
Subsection 12.7.1 Tangent Lines
Derivatives and tangent lines go hand-in-hand. Given y=f(x), the line tangent to the graph of f at x=x0 is the line through (x0,f(x0)) with slope fβ²(x0); that is, the slope of the tangent line is the instantaneous rate of change of f at x0. When dealing with functions of two variables, the graph is no longer a curve but a surface. At a given point on the surface, it seems there are many lines that fit our intuition of being βtangentβ to the surface. In Section 12.3.2 we introduced the concept of the tangent plane, which could be thought of as consisting of all possible lines tangent to the surface at a given point. In this section, we explore this idea in more detail.Definition 12.7.2. Directional Tangent Line.
Let z=f(x,y) be differentiable on a set S containing (x0,y0) and let βu=β¨u1,u2β© be a unit vector.
The line βx through (x0,y0,f(x0,y0)) parallel to β¨1,0,fx(x0,y0)β© is the tangent line to f in the direction of x at (x0,y0).
The line βy through (x0,y0,f(x0,y0)) parallel to β¨0,1,fy(x0,y0)β© is the tangent line to f in the direction of y at (x0,y0).
The line ββu through (x0,y0,f(x0,y0)) parallel to β¨u1,u2,Dβuf(x0,y0)β© is the tangent line to f in the direction of βu at (x0,y0).
Example 12.7.3. Finding directional tangent lines.
Find the lines tangent to the surface z=sin(x)cos(y) at (Ο/2,Ο/2) in the x and y directions and also in the direction of βv=β¨β1,1β©.
The partial derivatives with respect to \(x\) and \(y\) are:
At \((\pi/2,\pi/2)\text{,}\) the \(z\)-value is 0.
Thus the parametric equations of the line tangent to \(f\) at \((\pi/2,\pi/2)\) in the directions of \(x\) and \(y\) are:
The two lines are shown with the surface in Figure 12.7.4.(a).
To find the equation of the tangent line in the direction of \(\vec v\text{,}\) we first find the unit vector in the direction of \(\vec v\text{:}\) \(\vec u = \la -1/\sqrt{2},1/\sqrt{2}\ra\text{.}\) The directional derivative at \((\pi/2,\pi,2)\) in the direction of \(\vec u\) is
Thus the directional tangent line is
The curve through \((\pi/2,\pi/2,0)\) in the direction of \(\vec v\) is shown in Figure 12.7.4.(b) along with \(\ell_{\vec u}(t)\text{.}\)
Example 12.7.5. Finding directional tangent lines.
Let f(x,y)=4xyβx4βy4. Find the equations of all directional tangent lines to f at (1,1).
First note that \(f(1,1) = 2\text{.}\) We need to compute directional derivatives, so we need \(\nabla f\text{.}\) We begin by computing partial derivatives.
Thus \(\nabla f(1,1) = \la 0,0\ra\text{.}\) Let \(\vec u = \la u_1,u_2\ra\) be any unit vector. The directional derivative of \(f\) at \((1,1)\) will be \(D_{\vec u\,}f(1,1) = \la 0,0\ra\cdot \la u_1,u_2\ra = 0\text{.}\) It does not matter what direction we choose; the directional derivative is always 0. Therefore
Figure 12.7.6 shows a graph of \(f\) and the point \((1,1,2)\text{.}\) Note that this point comes at the top of a βhill,β and therefore every tangent line through this point will have a βslopeβ of 0.
That is, consider any curve on the surface that goes through this point. Each curve will have a relative maximum at this point, hence its tangent line will have a slope of 0. The following section investigates the points on surfaces where all tangent lines have a slope of 0.
Subsection 12.7.2 Normal Lines
When dealing with a function y=f(x) of one variable, we stated that a line through (c,f(c)) was tangent to f if the line had a slope of fβ²(c) and was normal (or, perpendicular, orthogonal) to f if it had a slope of β1/fβ²(c). We extend the concept of normal, or orthogonal, to functions of two variables. Let z=f(x,y) be a differentiable function of two variables. By Definition 12.7.2, at (x0,y0), βx(t) is a line parallel to the vector βdx=β¨1,0,fx(x0,y0)β© and βy(t) is a line parallel to βdy=β¨0,1,fy(x0,y0)β©. Since lines in these directions through (x0,y0,f(x0,y0)) are tangent to the surface, a line through this point and orthogonal to these directions would be orthogonal, or normal, to the surface. We can use this direction to create a normal line. The direction of the normal line is orthogonal to βdx and βdy, hence the direction is parallel to βdn=βdxΓβdy. It turns out this cross product has a very simple form:Definition 12.7.7. Normal Line.
Let z=f(x,y) be differentiable on a set S containing (x0,y0) where
are defined.
A nonzero vector parallel to βn=β¨a,b,β1β© is orthogonal to f at P=(x0,y0,f(x0,y0)).
The line βn through P with direction parallel to βn is the normal line to f at P.
Example 12.7.8. Finding a normal line.
Find the equation of the normal line to z=βx2βy2+2 at (0,1).
We find \(z_x(x,y) = -2x\) and \(z_y(x,y) = -2y\text{;}\) at \((0,1)\text{,}\) we have \(z_x = 0\) and \(z_y = -2\text{.}\) We take the direction of the normal line, following Definition 12.7.7, to be \(\vec n=\la 0,-2,-1\ra\text{.}\) The line with this direction going through the point \((0,1,1)\) is
The surface \(z=-x^2-y^2+2\text{,}\) along with the found normal line, is graphed in Figure 12.7.9.
Example 12.7.10. Finding the distance from a point to a surface.
Let f(x,y)=2βx2βy2 and let Q=(2,2,2). Find the distance from Q to the surface defined by f.
This surface is used in Example 12.7.5, so we know that at \((x,y)\text{,}\) the direction of the normal line will be \(\vec d_n = \la -2x,-2y,-1\ra\text{.}\) A point \(P\) on the surface will have coordinates \((x,y,2-x^2-y^2)\text{,}\) so \(\overrightarrow{PQ} = \la 2-x,2-y,x^2+y^2\ra\text{.}\) To find where \(\overrightarrow{PQ}\) is parallel to \(\vec d_n\text{,}\) we need to find \(x\text{,}\) \(y\) and \(c\) such that \(c\overrightarrow{PQ} = \vec d_n\text{.}\)
This implies
\begin{align*} c(2-x) \amp = -2x\\ c(2-y) \amp = -2y\\ c(x^2+y^2) \amp = -1 \end{align*}In each equation, we can solve for \(c\text{:}\)
The first two fractions imply \(x=y\text{,}\) and so the last fraction can be rewritten as \(c=-1/(2x^2)\text{.}\) Then
This last equation is a cubic, which is not difficult to solve with a numeric solver. We find that \(x= 0.689\text{,}\) hence \(P = (0.689,0.689, 1.051)\text{.}\) We find the distance from \(Q\) to the surface of \(f\) is
Example 12.7.11. Finding a point a set distance from a surface.
Let f(x,y)=xβy2+3. Let P=(2,1,f(2,1))=(2,1,4). Find points Q in space that are 4 units from the surface of f at P. That is, find Q such that β and \overrightarrow{PQ} is orthogonal to f at P\text{.}
We begin by finding partial derivatives:
The vector \(\vec n=\la 1,-2,-1\ra\) is orthogonal to \(f\) at \(P\text{.}\) For reasons that will become more clear in a moment, we find the unit vector in the direction of \(\vec n\text{:}\)
Thus a the normal line to \(f\) at \(P\) can be written as
An advantage of this parametrization of the line is that letting \(t=t_0\) gives a point on the line that is \(\abs{t_0}\) units from \(P\text{.}\) (This is because the direction of the line is given in terms of a unit vector.) There are thus two points in space 4 units from \(P\text{:}\)
The surface is graphed along with points \(P\text{,}\) \(Q_1\text{,}\) \(Q_2\) and a portion of the normal line to \(f\) at \(P\text{.}\)
Subsection 12.7.3 Tangent Planes
We can use the direction of the normal line to define a plane. With a=f_x(x_0,y_0)\text{,} b=f_y(x_0,y_0) and P = \big(x_0,y_0,f(x_0,y_0)\big)\text{,} the vector \vec n=\la a,b,-1\ra is orthogonal to f at P\text{.} (See Definition 12.3.10.) The plane through P with normal vector \vec n is therefore tangent to f at P\text{.}Definition 12.7.13. Tangent Plane.
Let z=f(x,y) be differentiable on a set S containing (x_0,y_0)\text{,} where a = f_x(x_0,y_0)\text{,} b=f_y(x_0,y_0)\text{,} \vec n= \la a,b,-1\ra and P=\big(x_0,y_0,f(x_0,y_0)\big)\text{.}
The plane through P with normal vector \vec n is the tangent plane to f at P\text{.} The standard form of this plane is
Example 12.7.14. Finding tangent planes.
Find the equation of the tangent plane to z=-x^2-y^2+2 at (0,1)\text{.}
Note that this is the same surface and point used in Example 12.7.8. There we found \(\vec n = \la 0,-2,-1\ra\) and \(P = (0,1,1)\text{.}\) Therefore the equation of the tangent plane is
The surface \(z=-x^2-y^2+2\) and tangent plane are graphed in Figure 12.7.15.
Example 12.7.16. Using the tangent plane to approximate function values.
The point (3,-1,4) lies on the surface of an unknown differentiable function f where f_x(3,-1) = 2 and f_y(3,-1) = -1/2\text{.} Find the equation of the tangent plane to f at P\text{,} and use this to approximate the value of f(2.9,-0.8)\text{.}
Knowing the partial derivatives at \((3,-1)\) allows us to form the normal vector to the tangent plane, \(\vec n = \la 2,-1/2,-1\ra\text{.}\) Thus the equation of the tangent line to \(f\) at \(P\) is:
Just as tangent lines provide excellent approximations of curves near their point of intersection, tangent planes provide excellent approximations of surfaces near their point of intersection. So \(f(2.9,-0.8) \approx z(2.9,-0.8) = 3.7\text{.}\)
This is not a new method of approximation. Compare the right hand expression for \(z\) in Equation (12.7.1) to the total differential:
Thus the βnew \(z\)-valueβ is the sum of the change in \(z\) (i.e., \(dz\)) and the old \(z\)-value (4). As mentioned when studying the total differential, it is not uncommon to know partial derivative information about a unknown function, and tangent planes are used to give accurate approximations of the function.
Subsection 12.7.4 The Gradient and Normal Lines, Tangent Planes
The methods developed in this section so far give a straightforward method of finding equations of normal lines and tangent planes for surfaces with explicit equations of the form z=f(x,y)\text{.} However, they do not handle implicit equations well, such as x^2+y^2+z^2=1\text{.} There is a technique that allows us to find vectors orthogonal to these surfaces based on the gradient.Definition 12.7.17. Gradient.
Let w=F(x,y,z) be differentiable on a set D that contains the point (x_0,y_0,z_0)\text{.}
The gradient of F is \nabla F(x,y,z) = \la f_x(x,y,z),f_y(x,y,z),f_z(x,y,z)\ra\text{.}
-
The gradient of F at (x_0,y_0,z_0) is
\begin{equation*} \nabla F(x_0,y_0,z_0) = \la f_x(x_0,y_0,z_0),f_y(x_0,y_0,z_0),f_z(x_0,y_0,z_0)\ra\text{.} \end{equation*}
Theorem 12.7.18. The Gradient and Level Surfaces.
Let w=F(x,y,z) be differentiable on a set D containing (x_0,y_0,z_0) with gradient \nabla F\text{,} where F(x_0,y_0,z_0) = c\text{.}
The vector \nabla F(x_0,y_0,z_0) is orthogonal to the level surface F(x,y,z)=c at (x_0,y_0,z_0)\text{.}
Example 12.7.19. Using the gradient to find a tangent plane.
Find the equation of the plane tangent to the ellipsoid \frac{x^2}{12} +\frac{y^2}{6}+\frac{z^2}{4}=1 at P = (1,2,1)\text{.}
We consider the equation of the ellipsoid as a level surface of a function \(F\) of three variables, where \(F(x,y,z) = \frac{x^2}{12} +\frac{y^2}{6}+\frac{z^2}{4}\text{.}\) The gradient is:
At \(P\text{,}\) the gradient is \(\nabla F(1,2,1) = \la 1/6, 2/3, 1/2\ra\text{.}\) Thus the equation of the plane tangent to the ellipsoid at \(P\) is
The ellipsoid and tangent plane are graphed in Figure 12.7.20.
Exercises 12.7.5 Exercises
Terms and Concepts
1.
Explain how the vector \vec v=\la 1,0,3\ra can be thought of as having a βslopeβ of 3.
2.
Explain how the vector \vec v=\la 0.6,0.8, -2\ra can be thought of as having a βslopeβ of -2\text{.}
3.
True or False? Let z=f(x,y) be differentiable at P\text{.} If \vec n is a normal vector to the tangent plane of f at P\text{,} then \vec n is orthogonal to \ell_x and \ell_y at P\text{.}
True
False
4.
Explain in your own words why we do not refer to the tangent line to a surface at a point, but rather to directional tangent lines to a surface at a point.
In the following exercises, a function f(x,y)\text{,} a vector \vec v and a point P are given. Give the parametric equations of the following directional tangent lines to z=f(x,y) at P\text{:}
\displaystyle \ell_x(t)
\displaystyle \ell_y(t)
\ell_{\vec u\,}(t)\text{,} where \vec u is the unit vector in the direction of \vec v\text{.}
5.
f(x,y) = 2x^2y-4xy^2\text{,} \vec v = \la 1,3\ra\text{,} P=(2,3)\text{.}
6.
Let f(x,y) = 3\cos(x) \sin(y) and P=(\pi/3, \pi/6)\text{.}
Find parametric equations for the directional tangent line \ell_{x} at P\text{.}
Find parametric equations for the directional tangent line \ell_{y} at P\text{.}
Find parametric equations for the directional tangent line \ell_{\vec{v}} at P\text{,} where \vec v = \la 1,2\ra\text{.}
7.
f(x,y) = 3x-5y\text{,} \vec v = \la 1,1\ra\text{,} P=(4,2)\text{.}
8.
Let f(x,y) = x^2-2x-y^2+4y and P=(1, 2)\text{.}
Find parametric equations for the directional tangent line \ell_{x} at P\text{.}
Find parametric equations for the directional tangent line \ell_{y} at P\text{.}
Find parametric equations for the directional tangent line \ell_{\vec{v}} at P\text{,} where \vec v = \la 1,1\ra\text{.}
In the following exercises, a function f(x,y) and a point P are given. Find the equation of the normal line to z=f(x,y) at P\text{.} Note: these are the same functions as in Exercises 12.7.5.5 β Exercise 12.7.5.8.
9.
f(x,y) = 2x^2y-4xy^2\text{,} P=(2,3)\text{.}
10.
Let f(x,y) = 3\cos(x) \sin(y) and P=(\pi/3, \pi/6)\text{.} Find parametric equations for the normal line at P\text{.}
11.
f(x,y) = 3x-5y\text{,} P=(4,2)\text{.}
12.
Let f(x,y) = x^2-2x-y^2+4y and P=(1,2)\text{.} Find parametric equations for the normal line at P\text{.}
In the following exercises, a function f(x,y) and a point P are given. Find the two points that are 2 units from the surface z=f(x,y) at P\text{.} Note: these are the same functions as in Exercises 12.7.5.5β12.7.5.8.
In the following exercises, a function f(x,y) and a point P are given. Find an equation of the tangent plane to z=f(x,y) at P\text{.} Note: these are the same functions as in Exercises 12.7.5.5β12.7.5.8.
In the following exercises, an implicitly defined function of x\text{,} y and z is given along with a point P that lies on the surface. Use the gradient \nabla F to:
find the equation of the normal line to the surface at P\text{,} and
find the equation of the plane tangent to the surface at P\text{.}
21.
\ds \frac{x^2}{8}+\frac{y^2}4+\frac{z^2}{16}=1\text{,} at P = (1,\sqrt{2},\sqrt{6})
22.
Given z^2-\frac{x^2}{4} - \frac{y^2}9=0\text{,} at P = (4,-3,\sqrt{5})\text{:}
Find parametric equations for the normal line.
Find an equation of the tangent plane.
23.
\ds xy^2-xz^2=0\text{,} at P = (2,1,-1)
24.
Given \sin(xy)+\cos(yz)=0\text{,} at P = (2, \pi/12, 4)\text{:}
Find parametric equations for the normal line.
Find an equation of the tangent plane.