# 2.1: Tangent Lines and Rates of Change

**At Grade**Created by: CK-12

## Learning Objectives

A student will be able to:

- Demonstrate an understanding of the slope of the tangent line to the graph.
- Demonstrate an understanding of the instantaneous rate of change.

A car speeding down the street, the inflation of currency, 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 to the tangent lines on a curve.

## The Tangent Line

If two points \begin{align*}P(x_0, y_0)\end{align*}

\begin{align*} m_{sec} = \frac{y_1 - y_0} {x_1 - x_0} = \frac{f(x_1) - f(x_0)} {x_1 - x_0}& &(1)\end{align*}

Now if we let the point \begin{align*}x_1\end{align*}

\begin{align*} m_{tan} =\lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)} {x_1 - x_0}.& &(2)\end{align*}

If we let \begin{align*}h = x_1 - x_0,\end{align*}

\begin{align*} m_{tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}.\end{align*}

If the point \begin{align*}P(x_0, y_0)\end{align*}

\begin{align*} m_{tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}\end{align*}

provided that the limit exists.

Recall from algebra that the *point-slope* form for the tangent line is given by

\begin{align*}y - y_0 = m_{tan}(x - x_0).\end{align*}

**Example 1:**

Find the slope of the tangent line to the curve \begin{align*}f(x) = x^3\end{align*}

*Solution:*

Since \begin{align*}P(x_0, y_0) = (2, 8)\end{align*}

\begin{align*} m_{tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}\end{align*}

we get

\begin{align*}m_{tan} &=\lim_{h \to 0} \frac{f(2 + h) - f(2)} {h}\\ &=\lim_{h \to 0} \frac{(h^3 + 6h^2 + 12h + 8) - 8} {h}\\ &=\lim_{h \to 0} \frac{h^3 + 6h^2 + 12h} {h}\\ &=\lim_{h \to 0} (h^2 + 6h + 12)\\ &= 12.\end{align*}

Thus the slope of the tangent line is \begin{align*}12\end{align*}

\begin{align*}y - 8 = 12(x - 2)\end{align*}

or

\begin{align*}y = 12x - 16\end{align*}

Next we are interested in finding a formula for the slope of the tangent line at *any point* on the curve \begin{align*}f\end{align*}

\begin{align*} m_{tan} =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h}.\end{align*}

We denote this formula by

\begin{align*} f'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h},\end{align*}

where \begin{align*}f(x)\end{align*}

**Example 2:**

If \begin{align*}f(x) = x^2 - 3,\end{align*} find \begin{align*}f(x)\end{align*} and use the result to find the slope of the tangent line at \begin{align*}x = 2\end{align*} and \begin{align*}x = -1.\end{align*}

*Solution:*

Since

\begin{align*} f'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h},\end{align*}

then

\begin{align*} f'(x) &=\lim_{h \to 0} \frac{[(x + h)^2 - 3] - [x^2 - 3]} {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{align*}

To find the slope, we simply substitute \begin{align*}x = 2\end{align*} into the result \begin{align*}f(x)\end{align*},

\begin{align*}f'(x) &= 2x\\ f'(2) &= 2(2)\\ &= 4\end{align*}

and

\begin{align*}f'(x) &= 2x\\ f'(-1) &= 2(-1)\\ &= -2\end{align*}

Thus slopes of the tangent lines at \begin{align*}x = 2\end{align*} and \begin{align*}x = -1\end{align*} are \begin{align*}4\end{align*} and \begin{align*}-2,\end{align*} respectively.

**Example 3:**

Find the slope of the tangent line to the curve \begin{align*}y = \frac{1}{x}\end{align*} that passes through the point \begin{align*}(1, 1)\end{align*}.

*Solution:*

Using the slope of the tangent formula

\begin{align*} f'(x) =\lim_{h \to 0}\frac{f(x + h) - f(x)} {h}\end{align*}

and substituting \begin{align*}y = \frac{1}{x}\end{align*},

\begin{align*}y' &=\lim_{h \to 0} \frac{\left (\frac{1} {x+h} \right) - \frac{1} {x}} {h}\\ &= \lim_{h \to 0} \frac{\frac{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)}\\ &= \lim_{h \to 0} \frac{-1} {x(x + h)}\\ &= \frac{-1} {x^2}.\end{align*}

Substituting \begin{align*}x = 1\end{align*},

\begin{align*}y' &= \frac{-1} {1}\\ &= -1.\end{align*}

Thus the slope of the tangent line at \begin{align*}x = 1\end{align*} for the curve \begin{align*}y = \frac{1}{x}\end{align*} is \begin{align*}m = -1.\end{align*} To find the equation of the tangent line, we simply use the point-slope formula,

\begin{align*}y - y_0 = m(x - x_0),\end{align*}

where \begin{align*}(x_{0}, y_{0}) = (1, 1).\end{align*}

\begin{align*}y - 1 &= -1 ( x - 1)\\ &= -x + 1 + 1\\ &= -x + 2,\end{align*}

which is the equation of the tangent line.

## Average Rates of Change

The primary concept of calculus involves calculating the rate of change of a quantity with respect to another. For example, speed is defined as the rate of change of the distance travelled with respect to time. If a person travels \begin{align*}120\;\mathrm{miles}\end{align*} in four hours, his speed is \begin{align*}120/4 = 30 \;\mathrm{mi/hr}\end{align*}. This speed is called the *average speed* or the *average rate of change* of distance with respect to time. Of course the person who travels \begin{align*}120\;\mathrm{miles}\end{align*} at a rate of \begin{align*}30 \;\mathrm{mi/hr}\end{align*} for four hours does not do so continuously. He must have slowed down or sped up during the four-hour period. But it does suffice to say that he traveled for four hours at an average rate of \begin{align*}30\;\mathrm{miles}\end{align*} per hour. However, if the driver strikes a tree, it would not be his average speed that determines his survival but his speed at the *instant of the collision*. Similarly, when a bullet strikes a target, it is not the average speed that is significant but its *instantaneous speed* at the moment it strikes. 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 \begin{align*}\triangle x\end{align*} divided by the time interval \begin{align*}\triangle t\end{align*} during which the displacement occurs:

\begin{align*} v = \frac{\triangle x} {\triangle t} = \frac{x_1 - x_0} {t_1 - t_0}.\end{align*}

Notice that the points \begin{align*}(t_0, x_0)\end{align*} and \begin{align*}(t_1, x_1)\end{align*} lie on the position-versus-time curve, as Figure 1 shows. This expression is also the expression for the slope of a secant line connecting the two points. Thus we conclude that the average velocity of an object between time \begin{align*}t_0\end{align*} and \begin{align*}t_1\end{align*} is represented geometrically by the slope of the secant line connecting the two points \begin{align*}(t_0, x_0)\end{align*} and \begin{align*}(t_1, x_1)\end{align*}. If we choose \begin{align*}t_1\end{align*} close to \begin{align*}t_0\end{align*}, then the average velocity will closely approximate the instantaneous velocity at time \begin{align*}t_0\end{align*}.

**Figure 1**

Geometrically, the average rate of change is represented by the slope of a secant line and the instantaneous rate of change is represented by the slope of the tangent line (Figures 2 and 3).

**Average Rate of Change** (such as the *average velocity*)

The average rate of change of \begin{align*}x = f(t)\end{align*} over the time interval \begin{align*}[t_0, t_1]\end{align*} is the slope \begin{align*}m_{sec}\end{align*} of the secant line to the points \begin{align*}(t_0, f(t_0))\end{align*} and \begin{align*}(t_1, f(t_1))\end{align*} on the graph (Figure 2).

**Figure 2**

\begin{align*} m_{sec} = \frac{f(t_1) - f(t_0)} {t_1 - t_0}\end{align*}

**Instantaneous Rate of Change**

The instantaneous rate of change of \begin{align*}x = f(t)\end{align*} at the time \begin{align*}t_0\end{align*} is the slope \begin{align*}m_{tan}\end{align*} of the tangent line at the time \begin{align*}t_0\end{align*} on the graph.

**Figure 3**

\begin{align*} m_{tan} = f'(t_0) = \lim_{t_1 \to t_0} \frac{f(t_1) - f(t_0)} {t_1 - t_0}\end{align*}

**Example 4:**

Suppose that \begin{align*}y = x^2 - 3.\end{align*}

- Find the average rate of change of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*} over the interval \begin{align*}[0, 2]\end{align*}.
- Find the instantaneous rate of change of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*} at the point \begin{align*}x = -1.\end{align*}

*Solution:*

1. Applying the formula for Average Rate of Change with \begin{align*}f(x) = x^2 - 3\end{align*} and \begin{align*}x_0 = 0\end{align*} and \begin{align*}x_1 = 2\;\mathrm{yields}\end{align*}

\begin{align*}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{align*}

This means that the average rate of change of \begin{align*}y\end{align*} is \begin{align*}2\;\mathrm{units \ per \ unit}\end{align*} increase in \begin{align*}x\end{align*} over the interval \begin{align*}[0, 2]\end{align*}.

2. From the example above, we found that \begin{align*}f'(x) = 2x\end{align*}, so

\begin{align*}m_{tan} &= f'(x_0)\\ &= f'(-1)\\ &= 2(-1)\\ &= -2\end{align*}

This means that the instantaneous rate of change is negative. That is, \begin{align*}y\end{align*} is decreasing at \begin{align*}x = -1.\end{align*} It is decreasing at a rate of \begin{align*}2\;\mathrm{units \ per \ unit}\end{align*} increase in \begin{align*}x\end{align*}.

## Multimedia Links

For a video explaining instantaneous rates of change **(4.2)**, see Slopes of Tangents and Instantaneous Rates of Change (9:26).

For a video with an application regarding velocity **(4.2)**, see Calculus Help: Instantaneous Rates of Change (9:03).

The following applet illustrates how the slope of a secant line can become the slope of the tangent to a curve at a point as \begin{align*}h \to 0\end{align*}. Follow the directions on the page to explore how changing the distance between the two points makes the slope of the secant approach the slope of the tangent Slope at a Point Applet. Note that the function and the point of tangency can also be edited in this simulator.

## Review Questions

- Given the function \begin{align*}y = 1/2 \ x^2\end{align*} and the values of \begin{align*}x_0 = 3\end{align*} and \begin{align*}x_1 = 4\end{align*}, find
- The average rate of change of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*} over the interval \begin{align*}[x_0, x_1]\end{align*}.
- The instantaneous rate of change of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*} at \begin{align*}x_0\end{align*}.
- The slope of the tangent line at \begin{align*}x_1\end{align*}.
- The slope of the secant line between points \begin{align*}x_0\end{align*} and \begin{align*}x_1\end{align*}.
- Make a sketch of \begin{align*}y = 1/2 \ x^2\end{align*} and show the secant and tangent lines at their respective points.

- Repeat problem #1 for \begin{align*}f(x) = 1/x\end{align*} and the values \begin{align*}x_0 = 2\end{align*} and \begin{align*}x_1 = 3\end{align*}.
- Find the slope of the graph \begin{align*}f(x) = x^2 + 1\end{align*} at a general point \begin{align*}x\end{align*}. What is the slope of the tangent line at \begin{align*}x_0 = 6\end{align*}?
- Suppose that \begin{align*} y = 1 / \sqrt{x}\end{align*}.
- Find the average rate of change of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*} over the interval \begin{align*}[1,3]\end{align*}.
- Find the instantaneous rate of change of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*} at point \begin{align*}x = 1\end{align*}.

- A rocket is propelled upward and reaches a height (in meters) of \begin{align*}h(t) = 4.9t^2\end{align*} in \begin{align*}t\;\mathrm{seconds}\end{align*}.
- How high does it reach in \begin{align*}35 \;\mathrm{seconds}\end{align*}?
- What is the average velocity of the rocket during the first \begin{align*}35 \;\mathrm{seconds}\end{align*}?
- What is the average velocity of the rocket during the first \begin{align*}200 \;\mathrm{meters}\end{align*}?
- What is the instantaneous velocity of the rocket at the end of the \begin{align*}35 \;\mathrm{seconds}\end{align*}?

- A particle moves in the positive direction along a straight line so that after \begin{align*}t\;\mathrm{nanoseconds}\end{align*}, its traversed distance is given by \begin{align*} \chi(t) = 9.9t^3\;\mathrm{nanometers}\end{align*}.
- What is the average velocity of the particle during the first \begin{align*}2\;\mathrm{nanoseconds}\end{align*}?
- What is the instantaneous velocity of the particle at \begin{align*}t = 2\;\mathrm{nanoseconds}\end{align*}?

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## Date Created:

Feb 23, 2012## Last Modified:

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