2.12
Implicit Differentiation
Until now we considered functions which were given explicitly. I.e., we were given an equation y=f(x), where f(x) is some instruction which assigns a value to x. The points on the graph of f are the points which satisfy the equation. Consider the equation
(x2+y2)2 =x2−y2.
(2.61)
The solutions of this equation form a curve in the plane called a lemniscate, see Figure 2.30. Parts of this curve look like the graph of a function, such as the points for which y ≥0. Without solving the equation fory, we still like to calculate the slope of curve at one of its points. This process is called implicit differentiation. -1 -0.5 0.5 1 -0.3 -0.2 -0.1 0.1 0.2 0.3 Figure 2.30: Lemniscate
Let us consider an easy situation which we have studied before.
Example 2.73. Find the slope of the tangent line to the unit circle (the
curve consisting of all points which satisfy the equation x2+y2= 1) at the point (1/2,√3/2).
Solution: We consider y as a function of x, and in this sense we write
y=y(x)22, and differentiate both sides of the equation. Apparently, dxdx2 = 2x. From the chain rule we deduce that dxdy2 = 2ydydx. That means that the derivative of the left hand side of the equation with respect toxis 2x+2ydydx. The derivative of the right hand side is zero. The derivative of the left and right hand side of the equation have to be the same, so that we get
2x+ 2ydy dx = 0.
Solving the equation for dydx, we find
dy dx =
−x y .
Plugging in the coordinates of the specified point, we find that
dy dx (1/2,√3/2) = √−1 3.
As we had to specify the x and the y coordinate of the point, we use a slightly different way to indicate at which point we evaluate the derivative.
♦
Example 2.74. Suppose you drop a circle of radius 1 into a parabola with
the equationy= 2x2. At which points will the circle touch the parabola?23
Solution: You see a picture of the problem in Figure 2.31. The crucial
observation in this example is, that the tangent line to the parabola and the circle will be the same at the point of contact.
Suppose the coordinates of the center of the circle are (0, a), then its equation is x2+ (y−a)2 = 1. Differentiating the equation of the parabola with respect tox, we find that dydx = 4x. Differentiating the equation of the circle with respect tox, we get
2x+ 2(y−a)dy
dx = 0.
Assuming that dydx is the same for both curves at the point of contact, we substitute dydx = 4xinto the second equation, cancel a factor 2, factor out an
x, and find:
x(1 + 4(y−a)) = 0.
22We can do this only for part of the curve asyis not really a function ofx. For most
xthere are two values ofywhich satisfy the equation.
23More sensibly, drop a ball of radius 1 into a cup whose vertical cross section is the
2.12. IMPLICIT DIFFERENTIATION 113 -1 -0.5 0.5 1 0.5 1 1.5 2 2.5 3
Figure 2.31: Ball in a Cup.
The ball it too large to fit into the parabola and touch at (0,0). So we may assume that x 6= 0. Solving the equation 1 + 4(y−a) = 0 for y, we find that the y coordinate of the point of contact is y = a−14. We substitute this expression into the equation of the circle and find that thexcoordinate of the point of contact is x=±√415. Substituting this into the equation of the parabola, we find that y= 158 at the point of contact. In summary, the circle touches the parabola in the points
(x, y) = ± √ 15 4 , 15 8 ! . ♦
Example 2.75. Find the slope of the tangent line to the lemniscate
(x2+y2)2 =x2−y2,
and find the coordinates of the points where the tangent line is horizontal.
Solution: You see a picture of the lemniscate in Figure 2.30. As in
Example 2.73, we equate the derivatives of the left and right hand side of the equation. We consider y as a function of x. Using standard rules of
differentiation, we find
2(x2+y2)(2x+ 2ydy
dx) = 2x−2y dy dx.
Cancelling a factor 2 and multiplying out part of the left hand side of the equation, we find
2x(x2+y2) + 2y(x2+y2)dy
dx =x−y dy dx.
Gathering all terms with a factor dydx on the left and those without on the right, we find the equation
(2y(x2+y2) +y)dy
dx =x(1−2(x
2+y2)).
Finally we get an explicit expression for dydx in terms ofx and y:
dy dx = x(1−2(x2+y2)) 2y(x2+y2) +y = x(1−2(x2+y2)) y(2(x2+y2) + 1).
Given any point (x, y) withy6= 0 on the lemniscate, we can plug it into the expression for dxdy and we get the slope of the curve at this point.
E.g, the point (x, y) = (12,12p−3 + 2√3) is a point on the lemniscate, and at this point the slope of the tangent line is
dy dx = 2−√2 √ 3p−3 + 2√3 .
This specific calculation takes a bit of arithmetic skill and effort to carry out.
The tangent line is horizontal whenever dydx = 0. A quick look at Fig- ure 2.30 tells us that we may ignore points where x = 0 or y = 0. That means that dydx = 0 whenever
1−2(x2+y2) = 0 or x2+y2= 1 2.
Substitute x2+y2 = 12, and y2 = 12 −x2 into the equation of the curve. Then we get an equation in one variable:
1 4 =x 2−1 2−x 2 or x2= 3 8 and y 2 = 1 8.
The points at which the tangent line to the lemniscate is horizontal are (x, y) = (± √ 6 4 ,± √ 2 4 )≈(±.6124,±.3536). ♦