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Lesson 9 of 24
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# Preview of Calculus

As we embark on our study of calculus, we shall see how its development arose from common solutions to practical problems in areas such as engineering physics—like the space travel problem posed in the chapter opener. Two key problems led to the initial formulation of calculus: (1) the tangent problem, or how to determine the slope of a line tangent to a curve at a point; and (2) the area problem, or how to determine the area under a curve.

# The Tangent Problem and Differential Calculus

Rate of change is one of the most critical concepts in calculus. We begin our investigation of rates of change by looking at the graphs of the three lines , and , shown in (Figure).

As we move from left to right along the graph of , we see that the graph decreases at a constant rate. For every 1 unit we move to the right along the -axis, the -coordinate decreases by 2 units. This rate of change is determined by the slope (−2) of the line. Similarly, the slope of 1/2 in the function  tells us that for every change in  of 1 unit there is a corresponding change in  of 1/2 unit. The function  has a slope of zero, indicating that the values of the function remain constant. We see that the slope of each linear function indicates the rate of change of the function.

Compare the graphs of these three functions with the graph of  ((Figure)). The graph of  starts from the left by decreasing rapidly, then begins to decrease more slowly and level off, and then finally begins to increase—slowly at first, followed by an increasing rate of increase as it moves toward the right. Unlike a linear function, no single number represents the rate of change for this function. We quite naturally ask: How do we measure the rate of change of a nonlinear function?

We can approximate the rate of change of a function  at a point  on its graph by taking another point  on the graph of , drawing a line through the two points, and calculating the slope of the resulting line. Such a line is called a secant line. (Figure) shows a secant line to a function  at a point .

We formally define a secant line as follows:

### Definition

The secant to the function  through the points  and  is the line passing through these points. Its slope is given by.

The accuracy of approximating the rate of change of the function with a secant line depends on how close  is to . As we see in (Figure), if  is closer to , the slope of the secant line is a better measure of the rate of change of  at .

The secant lines themselves approach a line that is called the tangent to the function  at  ((Figure)). The slope of the tangent line to the graph at  measures the rate of change of the function at . This value also represents the derivative of the function  at , or the rate of change of the function at . This derivative is denoted by . Differential calculus is the field of calculus concerned with the study of derivatives and their applications.

For an interactive demonstration of the slope of a secant line that you can manipulate yourself, visit this applet

(Figure) illustrates how to find slopes of secant lines. These slopes estimate the slope of the tangent line or, equivalently, the rate of change of the function at the point at which the slopes are calculated.

### Finding Slopes of Secant Lines

Estimate the slope of the tangent line (rate of change) to  at  by finding slopes of secant lines through  and each of the following points on the graph of .

#### Solution

Use the formula for the slope of a secant line from the definition.

The point in part b. is closer to the point , so the slope of 2.5 is closer to the slope of the tangent line. A good estimate for the slope of the tangent would be in the range of 2 to 2.5 ((Figure)).

The secant lines to  at  through (a)  and (b)  provide successively closer approximations to the tangent line to  at .

Estimate the slope of the tangent line (rate of change) to  at  by finding slopes of secant lines through  and the point  on the graph of .

2.25

#### Hint

Use (Figure) and (Figure) as a solving guide.

We continue our investigation by exploring a related question. Keeping in mind that velocity may be thought of as the rate of change of position, suppose that we have a function, , that gives the position of an object along a coordinate axis at any given time . Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a given time ? We start by approximating the instantaneous velocity with an average velocity. First, recall that the speed of an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled. We define the average velocity of an object over a time period to be the change in its position divided by the length of the time period.

### Definition

Let  be the position of an object moving along a coordinate axis at time . The average velocity of the object over a time interval  where  (or  if ) is.

As  is chosen closer to , the average velocity becomes closer to the instantaneous velocity. Note that finding the average velocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function. Furthermore, to find the slope of a tangent line at a point , we let the -values approach  in the slope of the secant line. Similarly, to find the instantaneous velocity at time , we let the -values approach  in the average velocity. This process of letting  or  approach  in an expression is called taking a limit. Thus, we may define the instantaneous velocity as follows.

### Definition

For a position function , the instantaneous velocity at a time  is the value that the average velocities approach on intervals of the form  and  as the values of  become closer to , provided such a value exists.

(Figure) illustrates this concept of limits and average velocity.

### Finding Average Velocity

A rock is dropped from a height of 64 ft. It is determined that its height (in feet) above ground  seconds later (for ) is given by . Find the average velocity of the rock over each of the given time intervals. Use this information to guess the instantaneous velocity of the rock at time .

#### Solution

Substitute the data into the formula for the definition of average velocity.

The instantaneous velocity is somewhere between −15.84 and −16.16 ft/sec. A good guess might be −16 ft/sec.

An object moves along a coordinate axis so that its position at time  is given by . Estimate its instantaneous velocity at time  by computing its average velocity over the time interval .

12.006001

Use .

# The Area Problem and Integral Calculus

We now turn our attention to a classic question from calculus. Many quantities in physics—for example, quantities of work—may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area between the graph of a function and the -axis over an interval ((Figure))?

As in the answer to our previous questions on velocity, we first try to approximate the solution. We approximate the area by dividing up the interval  into smaller intervals in the shape of rectangles. The approximation of the area comes from adding up the areas of these rectangles ((Figure)).

As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the area between the graph of  and the -axis over the interval . Once again, we find ourselves taking a limit. Limits of this type serve as a basis for the definition of the definite integral. Integral calculus is the study of integrals and their applications.

### Estimation Using Rectangles

Estimate the area between the -axis and the graph of  over the interval  by using the three rectangles shown in (Figure).

#### Solution

The areas of the three rectangles are 1 unit2, 2 unit2, and 5 unit2. Using these rectangles, our area estimate is 8 unit2.

Estimate the area between the -axis and the graph of  over the interval  by using the three rectangles shown here:

16 unit2

#### Hint

Use (Figure) as a guide.

# Other Aspects of Calculus

So far, we have studied functions of one variable only. Such functions can be represented visually using graphs in two dimensions; however, there is no good reason to restrict our investigation to two dimensions. Suppose, for example, that instead of determining the velocity of an object moving along a coordinate axis, we want to determine the velocity of a rock fired from a catapult at a given time, or of an airplane moving in three dimensions. We might want to graph real-value functions of two variables or determine volumes of solids of the type shown in (Figure). These are only a few of the types of questions that can be asked and answered using multivariable calculus. Informally, multivariable calculus can be characterized as the study of the calculus of functions of two or more variables. However, before exploring these and other ideas, we must first lay a foundation for the study of calculus in one variable by exploring the concept of a limit.

### Key Concepts

• Differential calculus arose from trying to solve the problem of determining the slope of a line tangent to a curve at a point. The slope of the tangent line indicates the rate of change of the function, also called the derivative. Calculating a derivative requires finding a limit.
• Integral calculus arose from trying to solve the problem of finding the area of a region between the graph of a function and the -axis. We can approximate the area by dividing it into thin rectangles and summing the areas of these rectangles. This summation leads to the value of a function called the integral. The integral is also calculated by finding a limit and, in fact, is related to the derivative of a function.
• Multivariable calculus enables us to solve problems in three-dimensional space, including determining motion in space and finding volumes of solids.

# Key Equations

• Slope of a Secant Line
• Average Velocity over Interval