Chapter 7: Implicit Functions and Implicit Differentiation

Implicit Functions

We have seen that not all functions are given by formulas. Sometimes a function is determined by interpolating a table of experimentally determined data (drawing a smooth curve between the points). Another way a function can be determined is implicitly. That means that we have a formula which and together must satisfy. The equation may take the form , or perhaps .

(a) The circle can be written , which is actually the implicit equation of two functions at the same time: the upper semicircle function y = and the lower semicircle function .

(b) We can avoid fractional powers by raising both sides to the power and writing the implicit equation .

(c) We can avoid negative powers by writing in the form .

[distance from (x,y) to ( )] + [distance from (x,y) to ( )] =

which leads to the following implicit equation for the ellipse:

.

(e) In the case when the ellipse's axes of symmetry are parallel to the x- and y-axes, a simpler implicit equation can be given, namely

.

With many implicit equations, it is possible to solve for y in terms of x. For example, we can do this for the circle.

Implicit Differentiation

Suppose we want to find the derivative at a point (x,y) of an implicitly defined curve. What you do is: (1) take of both sides of the equation; (2) separate the terms involving to one side of the equation and the terms without to the other side; and (3) solve for . In step (1), it is important to remember that you are differentiating with respect to x (or whatever letter is playing the role of the x-variable), and so the chain rule must be used for expressions like . In other words, the letter y stands for a certain function of x, and so taking of is just like taking of : = .

Find at the point ( ) for each implicitly defined function in Example 7.1

(a) When we take of both sides of , remember that (1) the derivative of is not but rather , and (2) the derivative of is not but rather , since r' = 0 as r is a constant.

Thus, taking of both sides gives:

.

We next move the to the right, and divide both sides by t o solve for :

= .

This answer agrees with our earlier formula for the derivative of , namely, , since y and are the same.

Notice that when you differentiate an explicit function y = f(x) your answer is a formula in terms of x, whereas implicit differentiation generally leads to a formula involving (it both y and x).

Note that if we take the lower semicircle function , the expression for becomes which is a different formula. However, the formula is still the same! Thus the implicit differentiation not only avoids the solution process for y in terms of x, but also gives the right answer for all possible solutions at once.

Parts (b) and (c) of this example will show us how to derive the nx^(n-1) rule when the power of x is a fraction or , where l and m are positive integers. (In the section on the product rule, we already saw how to get the rule for any positive integer n.)

(b) We take of both sides of , and then solve for :
, so that .

The last ratio simplifies algebraically, if we use the fact that = . That is,

= = (here n = ).

(c) We take of both sides of (using the product rule), and then solve for :

, so that .

Canceling from the numerator and denominator in the last expression, we obtain:

= = , where .

(d) When we take of both sides, we must use the chain rule several times, first with the square root as the outside function and then when we find the derivative of , etc. The result is:

.

Solving for is not difficult, but the expression we end up with is rather messy, so we shall omit it. If we had specific values for the foci ( ) and ( ), and if we wanted to find the slope y' at a particular point (x,y), then we could substitute all of these numerical values in place of x, y, , ending up with an equation of the form y' = 0 (where comes from bringing together all of the terms with and comes from bringing together all the terms without ). Such an equation has solution simply . We will see such situations later in the story problems.

(e) Here we get: . Solving for , we obtain

.

If you know that the curve (where C is a constant) passes through the point ( ), find the y-coordinate of the point ( , ?) on the curve near ( ). Use the tangent line approximation.

( Note: ) The constant is not given. We could easily compute it from the fact that the point ( ) satisfies the equation. However, we will never need to know the value of C , so we won't bother to compute it.)

In this problem there is no way you can find an exact expression for the y-coordinate corresponding to , because when you substitute you get an equation for that cannot be solved algebraically. The equation for could be solved by an approximation method that we'll learn later Newton's method . In the meantime, the best we can do is to use the tangent line approximation to find the y-coordinate. In this problem we know that the implicitly defined has value 2 when (this is what it means to say that the curve passes through the point ( )). We want to know . According to the tangent line approximation , approx = , where denotes the derivative at the point ( ). To find that derivative we first take of both sides of our implicit equation . We use the chain rule and the product rule to find , and on the right we get (the derivative of any constant is ):

.

Once we've taken we can substitute the particular and values of the point where we want the derivative. After that our equation will look a lot simpler --- just some numbers together with the unknown :

,

(i.e., ) ,

(i.e., ) = .

Finally, substituting this value of in the tangent line approximation
gives the y-coodinate as .

Suppose that the ellipse in Example 7.1(d) has foci at the points ( ) and ( ), and passes through the point ( ). Find the -coordinate of the point ( ,?) through which it passes. Use the tangent line approximation.

The tangent line approximation tells us that = approx ) = , where is the derivative at the point ( ). The implicit differentiation of the equation of this ellipse has already been carried out, so it remains just to substitute the particular values we're given: , , , , x = 2, y = 3:

, so that ,

i.e., . Thus, = , and so = approx .

You are standing at ( ) and receiving light from two sources: source A located at ( ) and source B located at ( ). You want to start walking in the direction which keeps the total index( light intensity) light intensity from the two sources as constant as possible. Assume that light intensity from a source is proportional to the strength of the source and inversely proportional to the square of the distance from the source. (a) If A and B are equally strong, show that you should walk horizontally, i.e., set out along the line . (b) If B is twice as strong as A, find the line along which you should move.

We are interested here in the curve of constant total light intensity passing through the point (0,1). That is, we want an equation satisfied by all points ( ) where the light intensity from A plus the light intensity from B remains constant. Since light intensity is inversely proportional to the square of the distance from the source, this means that

+ .

Here the proportionality constant K depends on the strength of the source; since A and B have the same strength in part (a), the constants are the same. In part (b) we change the second K to 2*K, because B is twice as strong as A.

Next, we use the distance formula in the denominators: the square of the
distance from ( ) to ( ) is , and the square of the distance from ( ) to ( ) is . For convenience, we also divide through by K, so that there are no unknown constants on the left. The result is

.

The question asks for the direction to go in from the point ( ) that best keeps to the curve of constant total light intensity. This amounts to asking for the tangent line to the curve at ( ). To find , we take of both sides, and then substitute the point ( ) = ( ) where we want to find . It is important to note that the fact that you happen to be walking from the point ( ) is not used until (it after) performing the implicit differentiation. When we take , we use the chain rule, writing the first term on the left as , and similarly for the second term on the left (it is more efficient to use the rule for than to use the quotient rule):

.

(Notice that the constants C and K have disappeared, because the derivative of any constant is 0.) Substituting x = 0, y = 1 gives:

.

Solving this, we find that , i.e., the curve of constant total light intensity has horizontal tangent line at the point (0,1).

(b) When the source B is twice as strong as the source A, the effect is to replace the second by , and this means having a rather than always in the numerator of the second term on the left. Our last equation is now

.

We can clear denominators by multiplying through by , obtaining:

, i.e., .

Thus, the tangent line has slope and y-intercept , so its equation is .