5128_Ch03_pp098-184
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Section 3.7 Implicit Differentiation 159<br />
Normal line<br />
A<br />
Light ray<br />
P<br />
Tangent<br />
Point of entry<br />
B<br />
Curve<br />
of lens<br />
surface<br />
Implicit Differentiation Process<br />
1. Differentiate both sides of the equation with respect to x.<br />
2. Collect the terms with dydx on one side of the equation.<br />
3. Factor out dydx.<br />
4. Solve for dydx.<br />
Figure 3.50 The profile of a lens, showing<br />
the bending (refraction) of a ray of<br />
light as it passes through the lens surface.<br />
Normal<br />
(–1, 2)<br />
–1<br />
2<br />
y<br />
Tangent<br />
x 2 xy y 2 7<br />
Figure 3.51 Tangent and normal lines to<br />
the ellipse x 2 xy y 2 7 at the point<br />
1, 2). (Example 4)<br />
x<br />
Lenses, Tangents, and Normal Lines<br />
In the law that describes how light changes direction as it enters a lens, the important angles<br />
are the angles the light makes with the line perpendicular to the surface of the lens at<br />
the point of entry (angles A and B in Figure 3.50). This line is called the normal to the surface<br />
at the point of entry. In a profile view of a lens like the one in Figure 3.50, the normal<br />
is a line perpendicular to the tangent to the profile curve at the point of entry.<br />
Profiles of lenses are often described by quadratic curves (see Figure 3.51). When<br />
they are, we can use implicit differentiation to find the tangents and normals.<br />
EXAMPLE 4<br />
Tangent and normal to an ellipse<br />
Find the tangent and normal to the ellipse x 2 xy y 2 7 at the point 1, 2.<br />
(See Figure 3.51.)<br />
SOLUTION<br />
We first use implicit differentiation to find dydx:<br />
x 2 xy y 2 7<br />
d<br />
x d x<br />
2 d d<br />
xy y d x d x<br />
2 d<br />
7<br />
d x<br />
2x ( x d y<br />
y d x<br />
dx<br />
d<br />
) x<br />
2y d y<br />
0<br />
dx<br />
2y x d y<br />
y 2x<br />
dx<br />
Differentiate both sides<br />
with respect to x . . .<br />
. . . treating xy as a product<br />
and y as a function of x.<br />
Collect terms.<br />
d y<br />
y 2x<br />
. Solve for dy/dx.<br />
dx<br />
2y x<br />
We then evaluate the derivative at x 1, y 2 to obtain<br />
d y<br />
<br />
dx<br />
|1,<br />
y 2x<br />
2<br />
2y x<br />
|1, 2<br />
2 21<br />
<br />
2 2 1<br />
The tangent to the curve at 1, 2 is<br />
4 5 .<br />
y 2 4 x 1<br />
5<br />
y 4 5 x 1 4<br />
.<br />
5<br />
continued