Section 13.7 Tangent Lines, Normal Lines, and Tangent Planes
Subsection 13.7.1 Tangent Lines
Derivatives and tangent lines go hand-in-hand. Given the line tangent to the graph of at is the line through with slope that is, the slope of the tangent line is the instantaneous rate of change of at
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 “tangent” to the surface.
In Subsection 13.3.2 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.
In Figure 13.7.1 we see lines that are tangent to curves in space. Since each curve lies on a surface, it makes sense to say that the lines are also tangent to the surface. The next definition formally defines what it means to be “tangent to a surface.”
It is instructive to consider each of three directions given in the definition in terms of “slope.” The direction of is that is, the “run” is one unit in the -direction and the “rise” is units in the -direction. Note how the slope is just the partial derivative with respect to A similar statement can be made for The direction of is the “run” is one unit in the direction (where is a unit vector) and the “rise” is the directional derivative of in that direction.
Definition 13.7.2 leads to the following parametric equations of directional tangent lines:
Example 13.7.3. Finding directional tangent lines.
Solution.
The two lines are shown with the surface in Figure 13.7.4.(a).
To find the equation of the tangent line in the direction of we first find the unit vector in the direction of The directional derivative at in the direction of is
Example 13.7.5. Finding directional tangent lines.
Solution.
First note that We need to compute directional derivatives, so we need We begin by computing partial derivatives.
Thus Let be any unit vector. The directional derivative of at will be It does not matter what direction we choose; the directional derivative is always 0. Therefore
Figure 13.7.6 shows a graph of and the point Note that this point comes at the top of a “hill,” and therefore every tangent line through this point will have a “slope” 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 13.7.2 Normal Lines
When dealing with a function of one variable, we stated that a line through was tangent to if the line had a slope of and was normal (or, perpendicular, orthogonal) to if it had a slope of We extend the concept of normal, or orthogonal, to functions of two variables.
Let be a differentiable function of two variables. By Definition 13.7.2, at is a line parallel to the vector and is a line parallel to Since lines in these directions through 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 and hence the direction is parallel to It turns out this cross product has a very simple form:
It is often more convenient to refer to the opposite of this direction, namely This leads to a definition.
Definition 13.7.7. Normal Line.
Example 13.7.8. Finding a normal line.
Solution.
We find and at we have and We take the direction of the normal line, following Definition 13.7.7, to be The line with this direction going through the point is
The direction of the normal line has many uses, one of which is the definition of the tangent plane which we define shortly. Another use is in measuring distances from the surface to a point. Given a point in space, it is a general geometric concept to define the distance from to the surface as being the length of the shortest line segment over all points on the surface. This, in turn, implies that will be orthogonal to the surface at Therefore we can measure the distance from to the surface by finding a point on the surface such that is parallel to the normal line to at
Example 13.7.10. Finding the distance from a point to a surface.
Solution.
This surface is used in Example 13.7.5, so we know that at the direction of the normal line will be A point on the surface will have coordinates so To find where is parallel to we need to find and such that
This implies
This last equation is a cubic, which is not difficult to solve with a numeric solver. We find that hence We find the distance from to the graph of is
We can take the concept of measuring the distance from a point to a surface to find a point a particular distance from a surface at a given point on the surface.
Example 13.7.11. Finding a point a set distance from a surface.
Let Let Find points in space that are 4 units from the graph of at That is, find such that and is orthogonal to at
Solution.
The vector is orthogonal to at For reasons that will become more clear in a moment, we find the unit vector in the direction of
An advantage of this parametrization of the line is that letting gives a point on the line that is units from (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
Subsection 13.7.3 Tangent Planes
We can use the direction of the normal line to define a plane. With and the vector is orthogonal to at (See Definition 13.3.10.) The plane through with normal vector is therefore tangent to at
Definition 13.7.13. Tangent Plane.
The plane through with normal vector is the tangent plane to at The standard form of this plane is
Example 13.7.14. Finding tangent planes.
Solution.
Note that this is the same surface and point used in Example 13.7.8. There we found and Therefore the equation of the tangent plane is
Example 13.7.16. Using the tangent plane to approximate function values.
The point lies on the graph of an unknown differentiable function where and Find the equation of the tangent plane to at and use this to approximate the value of
Solution.
Knowing the partial derivatives at allows us to form the normal vector to the tangent plane, Thus the equation of the tangent line to at 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
This is not a new method of approximation. Compare the right hand expression for in Equation (13.7.1) to the total differential:
Thus the “new -value” is the sum of the change in (i.e., ) and the old -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 13.7.4 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 However, they do not handle implicit equations well, such as There is a technique that allows us to find vectors orthogonal to these surfaces based on the gradient.
Definition 13.7.17. Gradient.
Recall that when the gradient is orthogonal to level curves of An analogous statement can be made about the gradient where Given a point let Then is a level surface that contains the point The following theorem states that is orthogonal to this level surface.
Theorem 13.7.18. The Gradient and Level Surfaces.
The gradient at a point gives a vector orthogonal to the surface at that point. This direction can be used to find tangent planes and normal lines.
Example 13.7.19. Using the gradient to find a tangent plane.
Solution.
We consider the equation of the ellipsoid as a level surface of a function of three variables, where The gradient is:
The ellipsoid and tangent plane are graphed in Figure 13.7.20.
To understand why Theorem 13.7.18 is true, recall the method of implicit differentiation given in Subsection 13.5.2. A level surface can be viewed as defining implicitly. We found that the partial derivatives of with respect to and are then given by
If we plug these values into the tangent plane equation
we get, with
If we move everything to the left-hand side of the equation and multiply through by we get
which is the equation of a plane with normal vector
Example 13.7.21. Finding the tangent plane of a level surface.
Determine the equation of the tangent plane to the level surface at the point (Note that this is the same problem as Example 13.5.15.)
Solution.
The equation of the tangent plane is therefore
Tangent lines and planes to surfaces have many uses, including the study of instantaneous rates of changes and making approximations. Normal lines also have many uses. In this section we focused on using them to measure distances from a surface. Another interesting application is in computer graphics, where the effects of light on a surface are determined using normal vectors.
The next section investigates another use of partial derivatives: determining relative extrema. When dealing with functions of the form we found relative extrema by finding where We can start finding relative extrema of by setting and to 0, but it turns out that there is more to consider.
Exercises 13.7.5 Exercises
Terms and Concepts
1.
Explain how the vector can be thought of as having a “slope” of 3.
2.
3.
True.
False.
Let be differentiable at If is a normal vector to the tangent plane of at then is orthogonal to and at
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.
Problems
Exercise Group.
A function a vector and a point are given. Give the parametric equations of the following directional tangent lines to at
Exercise Group.
A function and a point are given. Find the equation of the normal line to at Note: these are the same functions as in Exercises 5–8.
Exercise Group.
A function and a point are given. Find the two points that are units from the surface at Note: these are the same functions as in Exercises 5–8.
Exercise Group.
A function and a point are given. Find an equation of the tangent plane to at Note: these are the same functions as in Exercises 5–8.
Exercise Group.
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