5128_Ch03_pp098-184

Section 3.7 Implicit Differentiation 159
Normal line
A
Light ray
P
Tangent
Point of entry
B
Curve
of lens
surface
Implicit Differentiation Process
1. Differentiate both sides of the equation with respect to x.
2. Collect the terms with dydx on one side of the equation.
3. Factor out dydx.
4. Solve for dydx.
Figure 3.50 The profile of a lens, showing
the bending (refraction) of a ray of
light as it passes through the lens surface.
Normal
(–1, 2)
–1
2
y
Tangent
x 2 xy y 2 7
Figure 3.51 Tangent and normal lines to
the ellipse x 2 xy y 2 7 at the point
1, 2). (Example 4)
x
Lenses, Tangents, and Normal Lines
In the law that describes how light changes direction as it enters a lens, the important angles
are the angles the light makes with the line perpendicular to the surface of the lens at
the point of entry (angles A and B in Figure 3.50). This line is called the normal to the surface
at the point of entry. In a profile view of a lens like the one in Figure 3.50, the normal
is a line perpendicular to the tangent to the profile curve at the point of entry.
Profiles of lenses are often described by quadratic curves (see Figure 3.51). When
they are, we can use implicit differentiation to find the tangents and normals.
EXAMPLE 4
Tangent and normal to an ellipse
Find the tangent and normal to the ellipse x 2 xy y 2 7 at the point 1, 2.
(See Figure 3.51.)
SOLUTION
We first use implicit differentiation to find dydx:
x 2 xy y 2 7
d
x d x
2 d d
xy y d x d x
2 d
7
d x
2x ( x d y
y d x
dx
d
) x
2y d y
0
dx
2y x d y
y 2x
dx
Differentiate both sides
with respect to x . . .
. . . treating xy as a product
and y as a function of x.
Collect terms.
d y
y 2x
. Solve for dy/dx.
dx
2y x
We then evaluate the derivative at x 1, y 2 to obtain
d y
dx
|1,
y 2x
2
2y x
|1, 2
2 21
2 2 1
The tangent to the curve at 1, 2 is
4 5 .
y 2 4 x 1
5
y 4 5 x 1 4
.
5
continued