James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Problems Plus
Before you look at the examples, cover up the solutions and try them yourself first.
ExamplE 1 How many lines are tangent to both of the parabolas y − 21 2 x 2 and
y − 1 1 x 2 ? Find the coordinates of the points at which these tangents touch the
parabolas.
y
1
_1
Q
FIGURE 1
P
x
SOLUTION To gain insight into this problem, it is essential to draw a diagram. So we
sketch the parabolas y − 1 1 x 2 (which is the standard parabola y − x 2 shifted 1 unit
upward) and y − 21 2 x 2 (which is obtained by reflecting the first parabola about the
x-axis). If we try to draw a line tangent to both parabolas, we soon discover that there
are only two possibilities, as illustrated in Figure 1.
Let P be a point at which one of these tangents touches the upper parabola and let a
be its x-coordinate. (The choice of notation for the unknown is important. Of course we
could have used b or c or x 0 or x 1 instead of a. However, it’s not advisable to use x in
place of a because that x could be confused with the variable x in the equation of the
parabola.) Then, since P lies on the parabola y − 1 1 x 2 , its y-coordinate must be
1 1 a 2 . Because of the symmetry shown in Figure 1, the coordinates of the point Q
where the tangent touches the lower parabola must be s2a, 2s1 1 a 2 dd.
To use the given information that the line is a tangent, we equate the slope of the
line PQ to the slope of the tangent line at P. We have
m PQ − 1 1 a 2 2 s21 2 a 2 d
a 2 s2ad
− 1 1 a 2
a
y
3≈ ≈
1
≈
2
0.3≈
0.1≈
If f sxd − 1 1 x 2 , then the slope of the tangent line at P is f 9sad − 2a. Thus the condition
that we need to use is that
1 1 a 2
a
− 2a
Solving this equation, we get 1 1 a 2 − 2a 2 , so a 2 − 1 and a − 61. Therefore the
points are (1, 2) and s21, 22d. By symmetry, the two remaining points are s21, 2d and
s1, 22d. ■
270
0
FIGURE 2
y
FIGURE 3
y=ln x
0 a
y=ln x
y=c≈
c=?
x
x
ExamplE 2 For what values of c does the equation ln x − cx 2 have exactly one
solution?
SOLUTION One of the most important principles of problem solving is to draw a diagram,
even if the problem as stated doesn’t explicitly mention a geometric situation.
Our present problem can be reformulated geometrically as follows: For what values of
c does the curve y − ln x intersect the curve y − cx 2 in exactly one point?
Let’s start by graphing y − ln x and y − cx 2 for various values of c. We know that,
for c ± 0, y − cx 2 is a parabola that opens upward if c . 0 and downward if c , 0.
Figure 2 shows the parabolas y − cx 2 for several positive values of c. Most of them
don’t intersect y − ln x at all and one intersects twice. We have the feeling that there
must be a value of c (somewhere between 0.1 and 0.3) for which the curves intersect
exactly once, as in Figure 3.
To find that particular value of c, we let a be the x-coordinate of the single point of
intersection. In other words, ln a − ca 2 , so a is the unique solution of the given equation.
We see from Figure 3 that the curves just touch, so they have a common tangent
line when x − a. That means the curves y − ln x and y − cx 2 have the same slope
when x − a. Therefore
1
a − 2ca
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