1.4: The Calculus
Learning Objectives
A student will be able to:
 Use linear approximations to study the limit process.
 Compute approximations for the slope of tangent lines to a graph.
 Introduce applications of differential calculus.
Introduction
In this lesson we will begin our discussion of the key concepts of calculus. They involve a couple of basic situations that we will come back to time and again throughout the book. For each of these, we will make use of some basic ideas about how we can use straight lines to help approximate functions.
Let’s start with an example of a simple function to illustrate each of the situations.
Consider the quadratic function \begin{align*}f(x) = x^2.\end{align*}
Suppose we magnify our picture and zoom in on the point \begin{align*}(1,1).\end{align*}
We note that the curve now looks very much like a straight line. If we were to overlay this view with a straight line that intersects the curve at \begin{align*}(1,1),\end{align*}
We can make the following observations. First, this line would appear to provide a good estimate of the value of \begin{align*}f(x)\end{align*}
Tangent Line to a Graph
Continuing our discussion of the tangent line to \begin{align*}f(x)\end{align*}
Yes, we need to find the slope of the line. We would be able to find the slope if we knew a second point on the line. So let’s choose a point \begin{align*}P\end{align*}
\begin{align*}m = (x^2  1)/(x  1) = x + 1\end{align*}
In particular, for \begin{align*}x = 1.25\end{align*}
\begin{align*}& P(x, y) && m \\ & (1.2, 1.44) && 2.2 \\ & (1.15, 1.3225) && 2.15 \\ & (1.1, 1.21) && 2.1 \\ & (1.05, 1.1025) && 2.05 \\ & (1.005, 1.010025) && 2.005 \\ & (1.0001, 1.00020001) && 2.0001\end{align*}
As we get closer to \begin{align*}(1,1),\end{align*}
Let’s make a couple of observations about this process. First, we can interpret the process graphically as finding secant lines from \begin{align*}(1,1)\end{align*}
Second, in examining the sequence of slopes of these secants, we are systematically observing approximate slopes of the function as point \begin{align*}P\end{align*}
Applications of Differential Calculus
Maximizing and Minimizing Functions
Recall from Lesson 1.3 our example of modeling the number of Food Stamp recipients. The model was found to be \begin{align*}y = 0.5x^2 + 4x + 19\end{align*}
We note that the function appears to attain a maximum value about an \begin{align*}x\end{align*}value somewhere around \begin{align*}x=4.\end{align*} Using the process from the previous example, what can we say about the tangent line to the graph for that \begin{align*}x\end{align*} value that yields the maximum \begin{align*}y\end{align*} value (the point at the top of the parabola)? (Answer: the tangent line will be horizontal, thus having a slope of \begin{align*}0\end{align*}.)
Hence we can use calculus to model situations where we wish to maximize or minimize a particular function. This process will be particularly important for looking at situations from business and industry where polynomial functions provide accurate models.
Velocity of a Falling Object
We can use differential calculus to investigate the velocity of a falling object. Galileo found that the distance traveled by a falling object was proportional to the square of the time it has been falling:
\begin{align*}s(t) = 4.9t^2.\end{align*}
The average velocity of a falling object from \begin{align*}t = a\end{align*} to \begin{align*}t = b\end{align*} is given by \begin{align*}(s(b)s(a))/(b  a).\end{align*}
HW Problem #10 will give you an opportunity to explore this relationship. In our discussion, we saw how the study of tangent lines to functions yields rich information about functions. We now consider the second situation that arises in Calculus, the central problem of finding the area under the curve of a function \begin{align*}f(x)\end{align*}.
Area Under a Curve
First let’s describe what we mean when we refer to the area under a curve. Let’s reconsider our basic quadratic function \begin{align*}f(x) = x^2.\end{align*} Suppose we are interested in finding the area under the curve from \begin{align*}x = 0\end{align*} to \begin{align*}x = 1.\end{align*}
We see the crosshatched region that lies between the graph and the \begin{align*}x\end{align*}axis. That is the area we wish to compute. As with approximating the slope of the tangent line to a function, we will use familiar linear methods to approximate the area. Then we will repeat the iterative process of finding better and better approximations.
Can you think of any ways that you would be able to approximate the area? (Answer: One ideas is that we could compute the area of the square that has a corner at \begin{align*}(1,1)\end{align*} to be \begin{align*}A = 1\end{align*} and then take half to find an area \begin{align*}A = 1/2.\end{align*} This is one estimate of the area and it is actually a pretty good first approximation.)
We will use a variation of this covering of the region with quadrilaterals to get better approximations. We will do so by dividing the \begin{align*}x\end{align*}interval from \begin{align*}x= 0\end{align*} to \begin{align*}x=1\end{align*} into equal subintervals. Let’s start by using four such subintervals as indicated:
We now will construct four rectangles that will serve as the basis for our approximation of the area. The subintervals will serve as the width of the rectangles. We will take the length of each rectangle to be the maximum value of the function in the subinterval. Hence we get the following figure:
If we call the rectangles R1–R4, from left to right, then we have the areas
\begin{align*} R_1 & = \frac{1} {4} * f \left(\frac{1} {4}\right) = \frac{1} {64}, \\ R_2 & = \frac{1} {4} * f \left(\frac{1} {2}\right) = \frac{1} {16}, \\ R_3 & = \frac{1} {4} * f \left(\frac{3} {4}\right) = \frac{9} {64}, \\ R_4 & = \frac{1} {4} * f (1) = \frac{1} {4},\end{align*}
and \begin{align*}R_1 + R_2 + R_3 + R_4 = \frac{30} {64} = \frac{15} {32}.\end{align*}
Note that this approximation is very close to our initial approximation of \begin{align*}1/2.\end{align*} However, since we took the maximum value of the function for a side of each rectangle, this process tends to overestimate the true value. We could have used the minimum value of the function in each subinterval. Or we could have used the value of the function at the midpoint of each subinterval.
Can you see how we are going to improve our approximation using successive iterations like we did to approximate the slope of the tangent line? (Answer: we will subdivide the interval from \begin{align*}x=0\end{align*} to \begin{align*}x=1\end{align*} into more and more subintervals, thus creating successively smaller and smaller rectangles to refine our estimates.)
Example 1:
The following table shows the areas of the rectangles and their sum for rectangles having width \begin{align*}w= 1/8.\end{align*}
Rectangle \begin{align*}R_i\end{align*}  Area of \begin{align*}R_i\end{align*} 

\begin{align*}R_1\end{align*}  \begin{align*}\frac{1}{512}\end{align*} 
\begin{align*}R_2\end{align*}  \begin{align*}\frac{4}{512}\end{align*} 
\begin{align*}R_3\end{align*}  \begin{align*}\frac{9}{512}\end{align*} 
\begin{align*}R_4\end{align*}  \begin{align*}\frac{16}{512}\end{align*} 
\begin{align*}R_5\end{align*}  \begin{align*}\frac{25}{512}\end{align*} 
\begin{align*}R_6\end{align*}  \begin{align*}\frac{36}{512}\end{align*} 
\begin{align*}R_7\end{align*}  \begin{align*}\frac{49}{512}\end{align*} 
\begin{align*}R_8\end{align*}  \begin{align*}\frac{64}{512}\end{align*} 
\begin{align*}A = \sum \ R_i=\frac{195}{512}\end{align*}. This value is approximately equal to \begin{align*}.3803.\end{align*} Hence, the approximation is now quite a bit less than \begin{align*}.5.\end{align*} For sixteen rectangles, the value is \begin{align*}\frac{1432}{4096}\end{align*} which is approximately equal to \begin{align*}.34.\end{align*} Can you guess what the true area will approach? (Answer: using our successive approximations, the area will approach the value \begin{align*}1/3.\end{align*})
We call this process of finding the area under a curve integration of \begin{align*}f(x)\end{align*} over the interval \begin{align*}[0, 1].\end{align*}
Applications of Integral Calculus
We have not yet developed any computational machinery for computing derivatives and integrals so we will just state one popular application of integral calculus that relates the derivative and integrals of a function.
Example 2:
There are quite a few applications of calculus in business. One of these is the cost function \begin{align*}C(x)\end{align*} of producing \begin{align*}x\end{align*} items of a product. It can be shown that the derivative of the cost function that gives the slope of the tangent line is another function that that gives the cost to produce an additional unit of the product. This is called the marginal cost and is a very important piece of information for management to have. Conversely, if one knows the marginal cost as a function of \begin{align*}x,\end{align*} then finding the area under the curve of the function will give back the cost function \begin{align*}C(x).\end{align*}
Lesson Summary
 We used linear approximations to study the limit process.
 We computed approximations for the slope of tangent lines to a graph.
 We analyzed applications of differential calculus.
 We analyzed applications of integral calculus.
Review Questions
 For the function \begin{align*}f(x) = x^2\end{align*} approximate the slope of the tangent line to the graph at the point \begin{align*}(3,9).\end{align*} (a) Use the following set of \begin{align*}x\end{align*}values to generate the sequence of secant line slopes: \begin{align*}x=2.9,2.95,2.975,2.995,2.999.\end{align*}(b) What value does the sequence of slopes approach?
 Consider the function \begin{align*}f(x)=x^2.\end{align*}
 For what values of \begin{align*}x\end{align*} would you expect the slope of the tangent line to be negative?
 For what value of \begin{align*}x\end{align*} would you expect the tangent line to have slope \begin{align*}m = 0\end{align*}?
 Give an example of a function that has two different horizontal tangent lines?
 Consider the function \begin{align*}p(x)=x^3 x.\end{align*} Generate the graph of \begin{align*}p(x)\end{align*} using your calculator.
 Approximate the slope of the tangent line to the graph at the point \begin{align*}(2,6).\end{align*} Use the following set of \begin{align*}x\end{align*}values to generate the sequence of secant line slopes. \begin{align*}x = 2.1, 2.05, 2.005, 2.001, 2.0001.\end{align*}
 For what values of \begin{align*}x\end{align*} do the tangent lines appear to have slope of \begin{align*}0\end{align*}? (Hint: Use the calculate function in your calculator to approximate the \begin{align*}x\end{align*}values.)
 For what values of \begin{align*}x\end{align*} do the tangent lines appear to have positive slope?
 For what values of \begin{align*}x\end{align*} do the tangent lines appear to have negative slope?
 The cost of producing \begin{align*}x\end{align*} \begin{align*}HiFi\end{align*} stereo receivers by Yamaha each week is modeled by the following function: \begin{align*}C(x) = 850 + 200x .3x^2.\end{align*}
 Generate the graph of \begin{align*}C(x)\end{align*} using your calculator. (Hint: Change your viewing window to reflect the high \begin{align*}y\end{align*} values.)
 For what number of units will the function be maximized?
 Estimate the slope of the tangent line at \begin{align*}x = 200,300,400.\end{align*}
 Where is marginal cost positive?
 Find the area under the curve of \begin{align*}f(x) = x^2\end{align*} from \begin{align*}x = 1\end{align*} to \begin{align*}x = 3.\end{align*} Use a rectangle method that uses the minimum value of the function within subintervals. Produce the approximation for each case of the subinterval cases.
 four subintervals.
 eight subintervals.
 Repeat part a. using a MidPoint Value of the function within each subinterval.
 Which of the answers in a.  c. provide the best estimate of the actual area?
 Consider the function \begin{align*}p(x) = x^3 + 4x.\end{align*}
 Find the area under the curve from \begin{align*}x = 0\end{align*} to \begin{align*}x = 1.\end{align*}
 Can you find the area under the curve from \begin{align*}x=1\end{align*} to \begin{align*}x=0.\end{align*} Why or why not? What is problematic for this computation?
 Find the area under the curve of \begin{align*}f(x) = \sqrt{x}\end{align*} from \begin{align*}x = 1\end{align*} to \begin{align*}x = 4.\end{align*} Use the Max Value rectangle method with six subintervals to compute the area.
 The Eiffel Tower is \begin{align*}320\;\mathrm{meters}\end{align*} high. Suppose that you drop a ball off the top of the tower. The distance that it falls is a function of time and is given by \begin{align*}s(t) = 4.9t^2.\end{align*}
Find the velocity of the ball after \begin{align*}4\;\mathrm{seconds}\end{align*}. (Hint: the average velocity for a time interval is average velocity \begin{align*}=\end{align*} change in distance/change in time. Investigate the average velocity for \begin{align*}t\end{align*} intervals close to \begin{align*}t = 4\end{align*} such as \begin{align*}3.9 \le t \le 4\end{align*} and closer and see if a pattern is evident.)
Review Answers
a. The slope approximations are 5.9, 5.95, 5.975, 5.995, and 5.999.
\begin{align*}P(X,Y) = P(X,x^2)\end{align*}  \begin{align*}m \left(= \frac{9y}{3x}\right)\end{align*} 

(2.9, 8.41)  5.9 
(2.95, 8.7025)  5.95 
(2.975, 8.850625)  5.975 
(2.995, 8.970025)  5.995 
(2.999, 8.994001)  5.999 
b. \begin{align*}m = 6\end{align*}
a. The slope approximations are 11.61, 11.3025, 11.030025, 11.006001, and 11.0006. The slope tends toward \begin{align*}m = 11.\end{align*}
\begin{align*}P(X,Y) = P(X,x^3x)\end{align*}  \begin{align*}m \left(= \frac{6y}{2x}\right)\end{align*} 

(2.1, 7.161)  11.61 
(2.05, 6.565125)  11.3025 
(2.005, 6.055150125)  11.030025 
(2.001, 6.011006001)  11.006 
(2.0001, 6.00110006)  11.0006 
b. \begin{align*}x = \pm 0.57.\end{align*}
c. \begin{align*}x < 0.57\end{align*}, \begin{align*}x > 0.57.\end{align*}
d. \begin{align*}0.57 < x < 0.57.\end{align*}

 Not included.
 \begin{align*}x \approx 333\end{align*}
 At \begin{align*}x=200\end{align*}, \begin{align*}m=80\end{align*}. At \begin{align*}x=300\end{align*}, \begin{align*}m=20\end{align*}. At \begin{align*}x=400\end{align*}, \begin{align*}m=40\end{align*}.
 For all \begin{align*}x<335.\end{align*}
 The approximate area is 6.75.
 The approximate area is 7.69.
 The approximate area is 8.62.
 Part c provides the best approximation, since the actual area is 8.67.
 1.75
 The graph drops below the \begin{align*}x\end{align*}axis into the third quadrant. Hence we are not finding the area below the curve but actually the area between the curve and the \begin{align*}x\end{align*}axis. But note that the curve is symmetric about the origin. Hence the region from \begin{align*}x = 1\end{align*} to \begin{align*}x = 0\end{align*} will have the same area as the region from \begin{align*}x = 0\end{align*} to \begin{align*}x = 1.\end{align*}
 The approximate area is 4.91.
 \begin{align*}39.2 \frac{\mathrm{m}}{\mathrm{sec}}\end{align*}.