How tangent lines are a limit of secant lines, and where the derivative and rate of change fit into all this.

Introduction

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.

Secant and tangent lines

If two distinct points P, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis and Q, left parenthesis, x, start subscript, 1, end subscript, comma, y, start subscript, 1, end subscript, right parenthesis lie on the curve y, equals, f, left parenthesis, x, right parenthesis, the slope of the secant line connecting the two points is
msec=y1y0x1x0=f(x1)f(x0)x1x0\displaystyle m_{\sec} = \frac{y_1 - y_0} {x_1 - x_0} = \frac{f(x_1) - f(x_0)} {x_1 - x_0}\,.
If we let the point x, start subscript, 1, end subscript approach x, start subscript, 0, end subscript, then Q will approach P along the graph f. The slope of the secant line through points P and Q will gradually approach the slope of the tangent line through P as x, start subscript, 1, end subscript approaches x, start subscript, 0, end subscript. In the limit, the previous equation becomes
mtan=limx1x0f(x1)f(x0)x1x0\displaystyle m_{\tan} =\lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)} {x_1 - x_0}\,.
If we let h, equals, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, then x, start subscript, 1, end subscript, equals, x, start subscript, 0, end subscript, plus, h and h, right arrow, 0 as x, start subscript, 1, end subscript, right arrow, x, start subscript, 0, end subscript. We can rewrite the limit as
mtan=limh0f(x0+h)f(x0)h\displaystyle m_{\tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}.
When the limit exists, its value m, start subscript, tangent, end subscript is the slope of the tangent line to the graph of f at the point P, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.

Example 1

Find the slope of the tangent line to the graph of the function f, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript at the point left parenthesis, 2, comma, 8, right parenthesis.

Solution

Since 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
mtan=limh0f(x0+h)f(x0)h\displaystyle m_{\tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}
we get
mtan=limh0f(2+h)f(2)h=limh0(h3+6h2+12h+8)8h=limh0h3+6h2+12hh=limh0(h2+6h+12)=12.\begin{aligned} m_{\tan} &=\lim_{h \to 0} \frac{f(2 + h) - f(2)} {h}\\ \\ &=\lim_{h \to 0} \frac{\left(h^3 + 6h^2 + 12h + 8\right) - 8} {h}\\ \\ &=\lim_{h \to 0} \frac{h^3 + 6h^2 + 12h} {h}\\ \\ &=\lim_{h \to 0} \left(h^2 + 6h + 12\right)\\ \\ &= 12. \end{aligned}
Thus, the slope of the tangent line is 12. Recall from algebra that the point-slope form of the equation of the tangent line is
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
y, minus, 8, equals, 12, dot, left parenthesis, x, minus, 2, right parenthesis
which we can rewrite as
y, equals, 12, x, minus, 16.

Finding the slope at any point

Next we are interested in finding a formula for the slope of the tangent line at any point on the graph of f. Such a formula would be the same formula that we are using except we replace the constant x, start subscript, 0, end subscript by the variable x. This yields
mtan=limh0f(x+h)f(x)h\displaystyle m_{\tan} =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h}\,.
We denote this formula by
f(x)=limh0f(x+h)f(x)h\displaystyle f\,'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h}\,,
where f, space, prime, left parenthesis, x, right parenthesis is read "f prime of x." The next example illustrates its usefulness.

Example 2

If f, left parenthesis, x, right parenthesis, equals, x, start superscript, 2, end superscript, minus, 3, find f, space, prime, left parenthesis, x, right parenthesis and use the result to find the slopes of the tangent lines at x, equals, 2 and x, equals, minus, 1.
Tangent lines to f, left parenthesis, x, right parenthesis, equals, x, start superscript, 2, end superscript, minus, 3 at the points x, equals, minus, 1 and x, equals, 2.

Solution

Since
f(x)=limh0f(x+h)f(x)h\displaystyle f\,'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h},
then
f(x)=limh0[(x+h)23][x23]h=limh0x2+2xh+h23x2+3h=limh02xh+h2h=limh0(2x+h)=2x.\begin{aligned} f\,'(x) &=\lim_{h \to 0} \frac{\left[(x + h)^2 - 3\right] - \left[x^2 - 3\right]} {h}\\ \\ &=\lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 3 - x^2 + 3} {h}\\ \\ &=\lim_{h \to 0} \frac{2xh + h^2} {h}\\ \\ &=\lim_{h \to 0} (2x + h)\\ \\ &= 2x. \end{aligned}
To find the slope, we substitute x, equals, 2 and x, equals, minus, 1 into the result f, space, prime, left parenthesis, x, right parenthesis, We get
f, space, prime, left parenthesis, 2, right parenthesis, equals, 2, left parenthesis, 2, right parenthesis, equals, 4
and
f, space, 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, equals, 2 and x, equals, minus, 1 are 4 and minus, 2, respectively.

Example 3

Find the slope of the tangent line to the curve y, equals, 1, slash, x at the point left parenthesis, 1, comma, 1, right parenthesis.
Tangent line to the curve f, left parenthesis, x, right parenthesis, equals, 1, slash, x at the point x, equals, 1

Solution

Using the slope of the tangent line formula
f(x)=limh0f(x+h)f(x)h\displaystyle f\,'(x) =\lim_{h \to 0}\frac{f(x + h) - f(x)} {h}
and substituting y, equals, 1, slash, x gives us
y=limh0(1x+h)1xh=limh0x(x+h)x(x+h)h=limh0xxhhx(x+h)=limh0hhx(x+h)=limh01x(x+h)=1x2.\begin{aligned} y\,' &=\lim_{h \to 0} \dfrac{\left (\dfrac{1} {x+h} \right) - \dfrac{1} {x}} {h}\\ \\ &= \lim_{h \to 0} \frac{\dfrac{x - (x + h)} {x(x + h)}} {h}\\ \\ &= \lim_{h \to 0} \frac{x - x - h} {hx(x + h)}\\ \\ &= \lim_{h \to 0} \frac{-h} {hx(x + h)}\\ \\ &= \lim_{h \to 0} \frac{-1} {x(x + h)}\\ \\ &= \frac{-1} {x^2}. \end{aligned}
Substituting x, equals, 1 yields
y, space, prime, equals, start fraction, minus, 1, divided by, left parenthesis, 1, right parenthesis, start superscript, 2, end superscript, end fraction, equals, minus, 1.
Thus, the slope of the tangent line at x, equals, 1 for the curve y, equals, 1, slash, x is m, equals, minus, 1. To find the equation of the tangent line, we use the point-slope formula,
y, minus, y, start subscript, 0, end subscript, equals, m, dot, left parenthesis, x, minus, x, start subscript, 0, end subscript, right parenthesis,
where 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
y1=1(x1)y=x+1+1y=x+2.\begin{aligned} y - 1 &= -1 \cdot ( x - 1)\\ y&= -x + 1 + 1\\ y&= -x + 2. \end{aligned}

Average speed

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 miles in 4 hours, his speed is
start fraction, 120, space, m, i, l, e, s, divided by, 4, space, h, o, u, r, s, end fraction, equals, 30, space, m, i, slash, h, r.
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 miles at an average rate of 30 miles per hour for 4 hours does not necessarily do so at constant speed. It may have slowed down or sped up during the 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 divided by the time interval triangle, t during which the displacement occurs:
v, equals, start fraction, triangle, x, divided by, triangle, t, end fraction, equals, start fraction, x, start subscript, 1, end subscript, minus, x, start subscript, 0, end subscript, divided by, t, start subscript, 1, end subscript, minus, t, start subscript, 0, end subscript, end fraction.
The average speed is also the expression for the slope of a secant line connecting the two points. Figure 1 shows the start color purple, s, e, c, a, n, t, space, l, i, n, e, end color purple through the points start color pink, left parenthesis, t, start subscript, 0, end subscript, comma, x, start subscript, 0, end subscript, right parenthesis, end color pink and start color pink, left parenthesis, t, start subscript, 1, end subscript, comma, x, start subscript, 1, end subscript, right parenthesis, end color pink on the start color blue, 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 color blue.
Figure 1. The slope of the secant line through two points on the position curve gives the average speed.
Thus we conclude that the average speed of an object between time t, start subscript, 0, end subscript and t, start subscript, 1, end subscript is represented geometrically by the slope of the secant line connecting the two points left parenthesis, t, start subscript, 0, end subscript, comma, x, start subscript, 0, end subscript, right parenthesis and left parenthesis, t, start subscript, 1, end subscript, comma, x, start subscript, 1, end subscript, right parenthesis. If we choose t, start subscript, 1, end subscript close to t, start subscript, 0, end subscript, then the average speed will closely approximate the instantaneous speed at time t, start subscript, 0, end subscript.

Rates of change

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

Average rate of change

The average rate of change of the function f over the interval open bracket, x, start subscript, 0, end subscript, comma, x, start subscript, 1, end subscript, close bracket is
msec=f(x1)f(x0)x1x0\displaystyle m_{\sec} = \frac{f(x_1) - f(x_0)} {x_1 - x_0}.
Figure 2 shows the start color purple, s, e, c, a, n, t, space, l, i, n, e, end color purple through the points start color pink, left parenthesis, x, start subscript, 0, end subscript, comma, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, right parenthesis, end color pink and start color pink, left parenthesis, x, start subscript, 1, end subscript, comma, f, left parenthesis, x, start subscript, 1, end subscript, right parenthesis, right parenthesis, end color pink on the start color blue, g, r, a, p, h, space, o, f, space, f, end color blue. The slope of the secant line is the average rate of change msecm_{\sec}.
Figure 2. The slope of the secant line through two points on the graph of a function gives the function's average rate of change over the interval.

Instantaneous rate of change

The instantaneous rate of change of the function f at the point x, start subscript, 0, end subscript is
mtan=f(x0)=limx1x0f(x1)f(x0)x1x0\displaystyle m_{\tan} = f\,'(x_0) = \lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)} {x_1 - x_0}.
Figure 3 shows start color purple, t, a, n, g, e, n, t, space, l, i, n, e, end color purple through the point start color pink, left parenthesis, x, start subscript, 0, end subscript, comma, f, left parenthesis, x, start subscript, 0, end subscript, right parenthesis, right parenthesis, end color pink on the start color blue, g, r, a, p, h, space, o, f, space, f, end color blue. The slope of the tangent line is the instantaneous rate of change m, start subscript, tangent, end subscript.
Figure 3. The slope of the tangent line through a point on the graph of a function gives the function's instantaneous rate of change at that point.

Example 4

Suppose that y, equals, x, start superscript, 2, end superscript, minus, 3.
(a) Find the average rate of change of y with respect to x over the interval open bracket, 0, comma, 2, close bracket.
(b) Find the instantaneous rate of change of y with respect to x at the point x, equals, minus, 1.

Solution

(a) Applying the formula for average rate of change with f, left parenthesis, x, right parenthesis, equals, x, start superscript, 2, end superscript, minus, 3 and x, start subscript, 0, end subscript, equals, 0 and x, start subscript, 1, end subscript, equals, 2 yields
msec=f(x1)f(x0)x1x0=f(2)f(0)20=1(3)2=2\begin{aligned} m_{\sec} &= \frac{f(x_1) - f(x_0)} {x_1 - x_0}\\ \\ &= \frac{f(2) - f(0)} {2 - 0}\\ \\ &= \frac{1 - (-3)} {2}\\ \\ &= 2 \end{aligned}
This means the average rate of change over the interval open bracket, 0, comma, 2, close bracket is 2 units of increase in y for each unit of increase in x.
(b) From Example 2 above, we found that f, space, prime, left parenthesis, x, right parenthesis, equals, 2, x, so
mtan=f(x0)=f(1)=2(1)=2.\begin{aligned} m_{\tan} &= f\,'(x_0)\\ &= f\,'(-1)\\ &= 2(-1)\\ &= -2. \end{aligned}
This means that the instantaneous rate of change is negative. That is, y is decreasing at x, equals, minus, 1. It is decreasing at a rate of 2 units in y for each unit of increase in x.