Calculus Of One Real Variable – By Pheng Kim Ving Chapter 2: The Derivative – Section 2.1: Tangent Lines And Their Slopes 2.1 Tangent Lines And Their Slopes

 1. Tangent Lines And Their Slopes

Non-Vertical Tangent Lines

Thus, the equation of secant PQ is y = (6 + h)(x – 3) + 9. Similarly, T passes thru the point (3, 9) and has slope 6. Let
(
x, y) be an arbitrary point of T. The equation of T is ( y – 9)/(x – 3) = 6, or y = 6(x – 3) + 9. In general, the equation
of the line passing thru the point (
x0, y0) and having slope m is:

y = m(xx0) + y0.

 Fig. 1.1   Tangent T is limit of secant PQ as Q approaches P.   Note: Scales on the axes are different.

This equation is called the point-slope equation of the line ( because it involves a point and the slope). Recall that the
slope-intercept equation of a line is y = mx + b ( because it involves the slope and the y-intercept), where m is the slope
and b the y-intercept. We see that the equation y = mx + b is a special case of the equation y = m(xx0) + y0 where
x0 = 0 and thus y0 is the y-intercept and in the notation is replaced by the letter b.

Vertical Tangent Lines

The graph of y = x1/3 is illustrated in Fig. 1.2. As Q approaches O, the secant OQ approaches the y-axis. The same thing
occurs if Q approaches O from the part of the graph to the left of O. Thus, the graph has a vertical line – the y-axis in

 Fig. 1.2   Tangent line to graph of y = x1/3 at x = 0 is y-axis, a vertical line.

this case – as the tangent line at x = 0. The slope of that tangent line is:

then the graph of f has a vertical tangent line at x = x0. The equation of such a vertical tangent line is x = x0.

Note that horizontal tangent lines are classified as non-vertical tangent lines. Their common slope is 0. The equation of a
horizontal tangent line to the graph of y = f(x) at (x0, y0) is therefore y = y0.

Where There Are No Tangent Lines

The graph of y = |x| is sketched in Fig. 1.3. As Q1 approaches O, the line OQ1 stays the same, as part of the graph to
the right of O. As Q2 approaches O, the line OQ2 stays the same, as part of the graph to the left of O. So, if the graph

 Fig. 1.3   Graph of y = |x| has no tangent line at x = 0.

has a tangent line at x = 0, then there would be two distinct tangent lines there, the right and left parts of the graph,
which contradicts the uniqueness of a tangent line. It follows that the graph has no tangent line at x = 0. Now:

Definition 1.1

 Suppose the function f(x) is continuous at x = x0. Then the quotient:

If the tangent line to the graph of f(x) at x = a exists, then it's clear that it must be unique. So we define its slope by
using the (two-sided) limit, not one-sided limits, which may be different when they exist.

Definition 1.2

 The slope of a curve C at a point P is the slope of the tangent line to C at P if such a tangent line exists.

Example 1.1

a.  Find the tangent line to the graph of y = f(x) = x2 at x = 3.
b.  Find the tangent line to the graph of f at x = 0.
c.  Use a graphing calculator or software to sketch a graph of y = g(x) = x2/3.
d.  Does the graph of g have a tangent line at x = 0?

Solution
a.
When x = 3 we have y = 32 = 9. The slope of the tangent line at x = 3 is:

Thus the equation of the tangent line at x = 0 is y = 0(x – 0) + 0, or y = 0, which is the x-axis.

c.  A graph of g is sketched in Fig. 1.4.

 Fig. 1.4   y = x2/3.

d.  The difference quotient of g at x = 0 is:

Hence the graph of g has no tangent line at x = 0.
EOS

 2. Normal Lines

Slopes Of Perpendicular Lines

Suppose two lines T and N are perpendicular. If none of them is vertical (thus none of them is horizontal), as shown in
Fig. 2.1, then their slopes are negative reciprocals of each other. In order to see this, let the slope of T be m and that of

 Fig. 2.1   Line N is normal to curve C at point P.

N be n. Draw a vertical line cutting T at A and N at B. Angles PAH and BPH are equal because their sides are
perpendicular. So right triangles PAH and BPH are similar. It follows that HA/PH = PH/HB. Now, m = HA/PH and n
= HB/PH. Thus, m = –1/n, which is the same as n = –1/m.

Lines Normal To A Curve

A line is said to be normal to a curve at a point if it's perpendicular to the tangent line of the curve at that point. In Fig.
2.1, T is tangent to curve C at point P, and N is normal to curve C at point P. We have (slope of normal N ) =
–1/(slope of tangent T ).

Example 2.1

Find the equation of the normal line to the curve y = x2 at the point (1, 1).

Solution
Let f(x) = x2. The slope of the tangent line at the point (1, 1) is:

EOS

 Problems & Solutions

1.  Find the equation of the tangent line to each of the following curves at the indicated point.

Solution

b.  Let f(x) = ax2 + bx + c. The slope of the tangent line is:

When x = u we have y = au2 + bu + c. Thus the equation of the tangent line is y = (2au + b)(xu) + (au2 + bu +
c), or y = (2au + b)xau2 + c.

2.  Find the equation of the normal line to each of the following curves at the indicated point.

Solution

b.  Let f(x) = ax2 + bx + c. The slope of the tangent line is:

Solution

a.  dom( g ) is the set of all non-negative real numbers.

b.  The points are (0, 0), (1, 1), (4, 2), and (9, 3).

c.  The slope of the tangent line is:

The slope of the normal line is – 4, and thus its equation is y = – 4(x – 4) + 2, or:

y = – 4x + 18.

d.  The lines are drawn in the graph in part b.

4.  Determine the slope of the curve y = x2 – 1 at the point x = a. Find the equation of the tangent line with a slope of – 3

to that curve.

Solution

We'll show that the line y = – 3x – 13/4, which has a slope of –3, is tangent to the curve y = x2 – 1.

Let f(x) = x2 – 1. The slope of f at x = a is the same as the slope of the tangent line to f at x = a, so it is:

5.  Find all points on the curve y = x3 – 3x where the tangent line is parallel to the x-axis.

Solution

We'll show that the tangent lines to the curve y = x3 – 3x that are parallel the x-axis are at the points (1, –2) and (–1, 2).
Let g(x) = x3 – 3x. The slope of a tangent line to g at an arbitrary point x is: