Tangent lines and rates of change

The position of a car driving down the street, the value of currency adjusted for inflation, the number of bacteria in a culture, and the AC voltage of an electric signal are all examples of quantities that change with time. In this section, we will study the rate of change of a quantity and how is it related geometrically to secant and tangent lines.

If two distinct points $P(x_0, y_0)$P, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis and $Q(x_1, y_1)$Q, left parenthesis, x, start subscript, 1, end subscript, comma, y, start subscript, 1, end subscript, right parenthesis lie on the curve $y = f(x)$y, equals, f, left parenthesis, x, right parenthesis, the slope of the secant line connecting the two points is

$\displaystyle m_{\sec} = \dfrac{y_1 - y_0} {x_1 - x_0} = \dfrac{f(x_1) - f(x_0)} {x_1 - x_0}\,$m, start subscript, \sec, end subscript, equals, start fraction, y, start subscript, 1, end subscript, minus, y, start subscript, 0, end subscript, divided by, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, end fraction, equals, start fraction, f, left parenthesis, x, start subscript, 1, end subscript, right parenthesis, minus, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, divided by, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, end fraction.

If we let the point $x_1$x, start subscript, 1, end subscript approach $x_0$x, start subscript, 0, end subscript, then $Q$Q will approach $P$P along the graph $f$f. The slope of the secant line through points $P$P and $Q$Q will gradually approach the slope of the tangent line through $P$P as $x_1$x, start subscript, 1, end subscript approaches $x_0$x, start subscript, 0, end subscript. In the limit, the previous equation becomes

$\displaystyle m_{\tan} =\lim_{x_1 \to x_0} \dfrac{f(x_1) - f(x_0)} {x_1 - x_0}\,$m, start subscript, tangent, end subscript, equals, limit, start subscript, x, start subscript, 1, end subscript, \to, x, start subscript, 0, end subscript, end subscript, start fraction, f, left parenthesis, x, start subscript, 1, end subscript, right parenthesis, minus, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, divided by, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, end fraction.

If we let $h = x_1 - x_0$h, equals, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, then $x_1 = x_0 + h$x, start subscript, 1, end subscript, equals, x, start subscript, 0, end subscript, plus, h and $h \rightarrow 0$h, right arrow, 0 as $x_1 \rightarrow x_0$x, start subscript, 1, end subscript, right arrow, x, start subscript, 0, end subscript. We can rewrite the limit as

$\displaystyle m_{\tan} =\lim_{h \to 0} \dfrac{f(x_0 + h) - f(x_0)} {h}$m, start subscript, tangent, end subscript, equals, limit, start subscript, h, \to, 0, end subscript, start fraction, f, left parenthesis, x, start subscript, 0, end subscript, plus, h, right parenthesis, minus, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, divided by, h, end fraction.

When the limit exists, its value $m_{\tan}$m, start subscript, tangent, end subscript is the slope of the tangent line to the graph of $f$f at the point $P(x_0, y_0)$P, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.

Find the slope of the tangent line to the graph of the function $f(x) = x^3$f, left parenthesis, x, right parenthesis, equals, x, cubed at the point $(2, 8)$left parenthesis, 2, comma, 8, right parenthesis.

Since $(x_0, y_0) = (2, 8)$left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, left parenthesis, 2, comma, 8, right parenthesis, using the slope of the tangent line formula

we get

Thus, the slope of the tangent line is $12$12. Recall from algebra that the point-slope form of the equation of the tangent line is

$\displaystyle y - y_0 = m_{\tan}\cdot(x - x_0)$y, minus, y, start subscript, 0, end subscript, equals, m, start subscript, tangent, end subscript, dot, left parenthesis, x, minus, x, start subscript, 0, end subscript, right parenthesis.

The point-slope formula gives us the equation

which we can rewrite as

$\displaystyle y = 12x - 16$y, equals, 12, x, minus, 16.

Next we are interested in finding a formula for the slope of the tangent line at any point on the graph of $f$f. Such a formula would be the same formula that we are using except we replace the constant $x_0$x, start subscript, 0, end subscript by the variable $x$x. This yields

$\displaystyle m_{\tan} =\lim_{h \to 0} \dfrac{f(x + h) - f(x)} {h}\,$m, start subscript, tangent, end subscript, equals, limit, start subscript, h, \to, 0, end subscript, start fraction, f, left parenthesis, x, plus, h, right parenthesis, minus, f, left parenthesis, x, right parenthesis, divided by, h, end fraction.

We denote this formula by

$\displaystyle f\,'(x) =\lim_{h \to 0} \dfrac{f(x + h) - f(x)} {h}\,$f, prime, left parenthesis, x, right parenthesis, equals, limit, start subscript, h, \to, 0, end subscript, start fraction, f, left parenthesis, x, plus, h, right parenthesis, minus, f, left parenthesis, x, right parenthesis, divided by, h, end fraction,

where $f\,'(x)$f, prime, left parenthesis, x, right parenthesis is read "$f$f prime of $x$x." The next example illustrates its usefulness.

If $f(x) = x^2 - 3$f, left parenthesis, x, right parenthesis, equals, x, squared, minus, 3, find $f\,'(x)$f, prime, left parenthesis, x, right parenthesis and use the result to find the slopes of the tangent lines at $x = 2$x, equals, 2 and $x = -1$x, equals, minus, 1.

Since

$\displaystyle f\,'(x) =\lim_{h \to 0} \dfrac{f(x + h) - f(x)} {h}$f, prime, left parenthesis, x, right parenthesis, equals, limit, start subscript, h, \to, 0, end subscript, start fraction, f, left parenthesis, x, plus, h, right parenthesis, minus, f, left parenthesis, x, right parenthesis, divided by, h, end fraction,

then

To find the slope, we substitute $x = 2$x, equals, 2 and $x = -1$x, equals, minus, 1 into the result $f\,'(x)$f, prime, left parenthesis, x, right parenthesis. We get

and

$\displaystyle f\,'(-1) = 2(-1) = -2$f, prime, left parenthesis, minus, 1, right parenthesis, equals, 2, left parenthesis, minus, 1, right parenthesis, equals, minus, 2.

Thus, slopes of the tangent lines at $x = 2$x, equals, 2 and $x = -1$x, equals, minus, 1 are $4$4 and $-2$minus, 2, respectively.

Find the slope of the tangent line to the graph of $f(x) = 1/x$f, left parenthesis, x, right parenthesis, equals, 1, slash, x at the point $(1, 1)$left parenthesis, 1, comma, 1, right parenthesis.

Using the slope of the tangent line formula

and substituting $f(x) = 1/x$f, left parenthesis, x, right parenthesis, equals, 1, slash, x gives us

Substituting $x = 1$x, equals, 1 yields

$f'(1)= \dfrac{-1} {(1)^2}= -1$f, prime, left parenthesis, 1, right parenthesis, equals, start fraction, minus, 1, divided by, left parenthesis, 1, right parenthesis, squared, end fraction, equals, minus, 1.

Thus, the slope of the tangent line at $x = 1$x, equals, 1 for the graph of $f(x) = 1/x$f, left parenthesis, x, right parenthesis, equals, 1, slash, x is $m = -1$m, equals, minus, 1. To find the equation of the tangent line, we use the point-slope formula,

$\displaystyle y - y_0 = m\cdot(x - x_0)$y, minus, y, start subscript, 0, end subscript, equals, m, dot, left parenthesis, x, minus, x, start subscript, 0, end subscript, right parenthesis,

where $(x_{0}, y_{0}) = (1, 1)$left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, left parenthesis, 1, comma, 1, right parenthesis. The equation of the tangent line is

The primary concept of differential is calculating the rate of change of one quantity with respect to another. For example, speed is defined as the rate of change of the distance traveled with respect to time. If a car travels $120$120 miles in $4$4 hours, his speed is

$\dfrac{120\text{ miles}}{4\text{ hours}} = 30 \text{ mi/hr}$start fraction, 120, start text, space, m, i, l, e, s, end text, divided by, 4, start text, space, h, o, u, r, s, end text, end fraction, equals, 30, start text, space, m, i, slash, h, r, end text.

This speed is called the average speed or the average rate of change of distance with respect to time. Of course, a car that travels $120$120 miles at an average rate of $30$30 miles per hour for $4$4 hours does not necessarily do so at constant speed. It may have slowed down or sped up during the $4$4 hour period.

However, if the car hits a tree, it would not be its average speed that determines the resulting damage but its speed at the instant of the collision. So here we have two distinct kinds of speeds, average speed and instantaneous speed.

The average speed of an object is defined as the object’s displacement $\triangle x$triangle, x divided by the time interval $\triangle t$triangle, t during which the displacement occurs:

The average speed is also the expression for the slope of a secant line connecting the two points. Figure 1 shows the $\purple{\text{secant line}}$start color #9d38bd, start text, s, e, c, a, n, t, space, l, i, n, e, end text, end color #9d38bd through the points $\pink{(t_0, x_0)}$start color #ff00af, left parenthesis, t, start subscript, 0, end subscript, comma, x, start subscript, 0, end subscript, right parenthesis, end color #ff00af and $\pink{(t_1, x_1)}$start color #ff00af, left parenthesis, t, start subscript, 1, end subscript, comma, x, start subscript, 1, end subscript, right parenthesis, end color #ff00af on the $\blue{\text{position-versus-time curve}}$start color #6495ed, start text, p, o, s, i, t, i, o, n, negative, v, e, r, s, u, s, negative, t, i, m, e, space, c, u, r, v, e, end text, end color #6495ed.

Thus we conclude that the average speed of an object between time $t_0$t, start subscript, 0, end subscript and $t_1$t, start subscript, 1, end subscript is represented geometrically by the slope of the secant line connecting the two points $(t_0, x_0)$left parenthesis, t, start subscript, 0, end subscript, comma, x, start subscript, 0, end subscript, right parenthesis and $(t_1, x_1)$left parenthesis, t, start subscript, 1, end subscript, comma, x, start subscript, 1, end subscript, right parenthesis. If we choose $t_1$t, start subscript, 1, end subscript close to $t_0$t, start subscript, 0, end subscript, then the average speed will closely approximate the instantaneous speed at time $t_0$t, start subscript, 0, end subscript.

The average rate of change of an arbitrary function $f$f on an interval is represented geometrically by the slope of the secant line to the graph of $f$f. The instantaneous rate of change of $f$f at a particular point is represented by the slope of the tangent line to the graph of $f$f at that point. Let's consider each case in more detail.

The average rate of change of the function $f$f over the interval $[x_0, x_1]$open bracket, x, start subscript, 0, end subscript, comma, x, start subscript, 1, end subscript, close bracket is

Figure 2 shows the $\purple{\text{secant line}}$start color #9d38bd, start text, s, e, c, a, n, t, space, l, i, n, e, end text, end color #9d38bd through the points $\pink{(x_0, f(x_0))}$start color #ff00af, left parenthesis, x, start subscript, 0, end subscript, comma, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, right parenthesis, end color #ff00af and $\pink{(x_1, f(x_1))}$start color #ff00af, left parenthesis, x, start subscript, 1, end subscript, comma, f, left parenthesis, x, start subscript, 1, end subscript, right parenthesis, right parenthesis, end color #ff00af on the $\blue{\text{graph of} \,f}$start color #6495ed, start text, g, r, a, p, h, space, o, f, end text, f, end color #6495ed. The slope of the secant line is the average rate of change $m_{\sec}$m, start subscript, \sec, end subscript.

The instantaneous rate of change of the function $f$f at the point $x_0$x, start subscript, 0, end subscript is

Figure 3 shows $\purple{\text{tangent line}}$start color #9d38bd, start text, t, a, n, g, e, n, t, space, l, i, n, e, end text, end color #9d38bd through the point $\pink{(x_0, f(x_0))}$start color #ff00af, left parenthesis, x, start subscript, 0, end subscript, comma, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, right parenthesis, end color #ff00af on the $\blue{\text{graph of} \,f}$start color #6495ed, start text, g, r, a, p, h, space, o, f, end text, f, end color #6495ed. The slope of the tangent line is the instantaneous rate of change $m_{\tan}$m, start subscript, tangent, end subscript.

Suppose that $y = x^2 - 3$y, equals, x, squared, minus, 3.

(a) Find the average rate of change of $y$y with respect to $x$x over the interval $[0, 2]$open bracket, 0, comma, 2, close bracket.

(b) Find the instantaneous rate of change of $y$y with respect to $x$x at the point $x = -1$x, equals, minus, 1.

(a) Applying the formula for average rate of change with $f(x) = x^2 - 3$f, left parenthesis, x, right parenthesis, equals, x, squared, minus, 3 and $x_0 = 0$x, start subscript, 0, end subscript, equals, 0 and $x_1 = 2$x, start subscript, 1, end subscript, equals, 2 yields

This means the average rate of change over the interval $[0, 2]$open bracket, 0, comma, 2, close bracket is 2 units of increase in $y$y for each unit of increase in $x$x.

(b) From Example 2 above, we found that $f\,'(x) = 2x$f, prime, left parenthesis, x, right parenthesis, equals, 2, x, so

This means that the instantaneous rate of change is negative. That is, $y$y is decreasing at $x = -1$x, equals, minus, 1. It is decreasing at a rate of $2$2 units in $y$y for each unit of increase in $x$x.