2.1 A Preview of Calculus
Learning Objectives
By the end of this section, you should be able to:
- Describe the tangent problem and how it led to the idea of the derivative.
- Explain how the idea of a limit is involved in solving the tangent problem.
- Recognize the tangent line to a curve at a point as the limit of secant lines.
- Identify instantaneous velocity as the limit of average velocity over a shrinking time interval.
- Describe the area problem and how it is solved by the integral.
- Explain how the idea of a limit is involved in solving the area problem.
- Recognize how limits, derivatives, and integrals lead naturally to the study of infinite series and multivariable calculus.
Calculus grew out of practical problems that classical algebra and geometry could not crack on their own. Two questions led to its creation:
- The tangent problem — how do we measure the slope of a curve at a single point?
- The area problem — how do we measure the area trapped between a curve and the x-axis?
Both questions have the same secret ingredient: a limit. The rest of this chapter — and most of single-variable calculus — is the story of how that one idea unlocks both.
2.1.1 The Tangent Problem and Differential Calculus
The secant line to the function \(f(x)\) through the points \((a, f(a))\) and \((x, f(x))\) is the line passing through these two points. Its slope is
$$m_{\sec} = \frac{f(x) - f(a)}{x - a} \qquad \text{(Equation 2.1)}$$We don't have the tools yet to measure the slope AT a single point on a curve — slope needs two points to compute (rise over run). The secant is a workaround: it uses two points we can work with, then we squeeze them together until they're almost the same point. The slope of that almost-the-same-point line is what we'll call the slope at \(a\).
Rate of change is one of the most critical ideas in calculus. Linear functions have a single, constant slope — that slope IS the rate of change everywhere. Figure 2.2 shows three lines and the constant slopes they encode: \(f(x) = -2x - 3\) drops by 2 for every unit right, \(g(x) = \tfrac{1}{2}x + 1\) climbs by \(\tfrac{1}{2}\), and \(h(x) = 2\) never changes.
Figure 2.2 — The rate of change of a linear function is constant in each of these three graphs, with the constant determined by the slope.
Now compare those lines to \(k(x) = x^2\) (Figure 2.3). This curve falls steeply on the left, levels off near the bottom, and climbs back up — increasingly fast — on the right. There is no single slope. The rate of change at \(x = -2\) is wildly different from the rate at \(x = 2\). So we have to ask: how do we measure the rate of change of a function whose slope keeps changing?
Figure 2.3 — The function \(k(x) = x^2\) does not have a constant rate of change.
The trick is to start by approximating with a secant line — exactly the construction Definition 2.1.1 names. Figure 2.4 shows a single secant through \((a, f(a))\) and a nearby point \((x, f(x))\); its slope estimates how steep the curve is at \(a\).
Figure 2.4 — The slope of a secant line through a point \((a, f(a))\) estimates the rate of change of the function at the point \((a, f(a))\).
How good is the secant-line estimate? It depends on how close \(x\) is to \(a\). The closer they are, the better the secant slope mimics the actual rate of change of \(f(x)\) at \(a\). Figure 2.5 shows three secants getting tighter and tighter on the same target point.
Figure 2.5 — As \(x\) gets closer to \(a\), the slope of the secant line becomes a better approximation to the rate of change of the function \(f(x)\) at \(a\).
As we slide \(x\) toward \(a\), the secant lines themselves approach a single line — the tangent line to \(f(x)\) at \(a\) (Figure 2.6). The slope of that tangent line is the true rate of change of the function at \(a\). It also has another name: the derivative of \(f\) at \(a\), written \(f'(a)\). Differential calculus is the branch of calculus that studies derivatives and what they can tell us.
Figure 2.6 — Solving the Tangent Problem: As \(x\) approaches \(a\), the secant lines approach the tangent line.
Example 2.1.1 puts the formula to work. We'll estimate the slope of the tangent to \(f(x) = x^2\) at \(x = 1\) by computing secant slopes through nearby points.
Let \(s(t)\) be the position of an object moving along a coordinate axis at time \(t\). The average velocity of the object over a time interval \([a, t]\) (where \(a < t\), or \([t, a]\) if \(t < a\)) is
$$v_{\text{ave}} = \frac{s(t) - s(a)}{t - a} \qquad \text{(Equation 2.2)}$$Notice that Equation 2.2 has the same shape as Equation 2.1 — change in output over change in input. Average velocity is literally the slope of the secant line on a position-vs-time graph. As we choose \(t\) closer and closer to \(a\), that average velocity tightens toward the instantaneous velocity at \(t = a\) — the slope of the tangent line on the position graph. The squeeze process is exactly the same as before, and it has a name.
For a position function \(s(t)\), the instantaneous velocity at a time \(t = a\) is the value that the average velocities approach on intervals of the form \([a, t]\) or \([t, a]\) as the values of \(t\) become closer to \(a\), provided such a value exists.
This squeeze process — letting \(x\) approach \(a\), or \(t\) approach \(a\), inside an expression like a slope or an average — is called taking a limit. It's the central idea of single-variable calculus. Definitions 2.1.1, 2.1.2, and 2.1.3 all set the stage; Chapter 2's later sections make limits formal.
Example 2.1.2 shows the squeeze in action: two short intervals on either side of \(t = 0.5\) give two average velocities that bracket the instantaneous value.
Estimate the slope of the tangent line (rate of change) to \(f(x) = x^2\) at \(x = 1\) by finding the slope of the secant line through \((1, 1)\) and the point \(\left(\tfrac{5}{4}, \tfrac{25}{16}\right)\) on the graph of \(f(x) = x^2\).
Solution
Apply Equation 2.1 with \(a = 1\), \(f(a) = 1\), \(x = \tfrac{5}{4}\), \(f(x) = \tfrac{25}{16}\):
$$m_{\sec} = \frac{\tfrac{25}{16} - 1}{\tfrac{5}{4} - 1} = \frac{\tfrac{9}{16}}{\tfrac{1}{4}} = \frac{9}{16} \cdot \frac{4}{1} = \frac{9}{4} = 2.25.$$Answer: The secant slope is \(2.25\). Since \(\tfrac{5}{4}\) is closer to \(1\) than the points in Example 2.1.1, this is our best estimate yet — the tangent slope at \(x = 1\) is somewhere close to \(2.25\) (and in fact equals exactly \(2\), as we will confirm later in the chapter).
The same "shrink toward a single point" idea applies to motion. Velocity is the rate of change of position, so we can ask: if \(s(t)\) gives the position of an object at time \(t\), what is the object's velocity at one specific instant \(t = a\)? Just like with secant slopes, we start by averaging over a short interval.
Recall from algebra: the speed of an object moving at a constant rate equals distance traveled divided by time elapsed. We extend that idea to a changing-position function by defining average velocity over a time interval to be the change in position divided by the length of the interval.
Estimate the slope of the tangent line (rate of change) to \(f(x) = x^2\) at \(x = 1\) by finding slopes of secant lines through \((1, 1)\) and each of the following points on the graph of \(f(x) = x^2\):
a) \((2, 4)\)
b) \(\left(\tfrac{3}{2}, \tfrac{9}{4}\right)\)
Solution
Apply Equation 2.1 with \(a = 1\) and \(f(a) = 1\).
Step 1 — Secant through \((1, 1)\) and \((2, 4)\):
$$m_{\sec} = \frac{4 - 1}{2 - 1} = \frac{3}{1} = 3.$$Step 2 — Secant through \((1, 1)\) and \(\left(\tfrac{3}{2}, \tfrac{9}{4}\right)\):
$$m_{\sec} = \frac{\tfrac{9}{4} - 1}{\tfrac{3}{2} - 1} = \frac{\tfrac{5}{4}}{\tfrac{1}{2}} = \frac{5}{2} = 2.5.$$Step 3 — Compare and estimate:
The second point sits much closer to \((1, 1)\), so the slope \(2.5\) is the better estimate of the tangent slope. The true tangent slope is somewhere in the band between \(2\) and \(2.5\) — and a tighter second point would give an even sharper estimate.
Answer: A reasonable estimate for the slope of the tangent line at \((1, 1)\) is roughly \(2\) to \(2.5\) (Figure 2.7).
Figure 2.7 — The secant lines to \(f(x) = x^2\) at \((1, 1)\) through (a) \((2, 4)\) and (b) \(\left(\tfrac{3}{2}, \tfrac{9}{4}\right)\) provide successively closer approximations to the tangent line to \(f(x) = x^2\) at \((1, 1)\).
An object moves along a coordinate axis so that its position at time \(t\) is given by \(s(t) = t^3\). Estimate its instantaneous velocity at time \(t = 2\) by computing its average velocity over the time interval \([2, 2.001]\).
Solution
Apply Equation 2.2 with \(a = 2\) and \(t = 2.001\).
Step 1 — Compute \(s(2)\) and \(s(2.001)\):
$$s(2) = 2^3 = 8,$$ $$s(2.001) = (2.001)^3 = 8.012006001.$$Step 2 — Average velocity:
$$v_{\text{ave}} = \frac{s(2.001) - s(2)}{2.001 - 2} = \frac{8.012006001 - 8}{0.001} = \frac{0.012006001}{0.001} = 12.006001.$$Answer: \(v_{\text{ave}} \approx 12.006\). Because the interval is very short, this average is an excellent estimate of the instantaneous velocity at \(t = 2\): \(v(2) \approx 12\).
A rock is dropped from a height of \(64\) feet. Its height above the ground (in feet) \(t\) seconds later (for \(0 \le t \le 2\)) is given by \(s(t) = -16t^2 + 64\). Find the average velocity of the rock over each of these time intervals, and use the results to guess the instantaneous velocity at \(t = 0.5\):
a) \([0.49, 0.5]\)
b) \([0.5, 0.51]\)
Solution
Apply Equation 2.2.
Step 1 — Compute \(s\) at the four endpoints:
$$s(0.49) = -16(0.49)^2 + 64 = 64 - 3.8416 = 60.1584,$$ $$s(0.5) = -16(0.5)^2 + 64 = 64 - 4 = 60,$$ $$s(0.51) = -16(0.51)^2 + 64 = 64 - 4.1616 = 59.8384.$$Step 2 — Average velocity over \([0.49, 0.5]\):
$$v_{\text{ave}} = \frac{s(0.5) - s(0.49)}{0.5 - 0.49} = \frac{60 - 60.1584}{0.01} = \frac{-0.1584}{0.01} = -15.84 \text{ ft/sec}.$$Step 3 — Average velocity over \([0.5, 0.51]\):
$$v_{\text{ave}} = \frac{s(0.51) - s(0.5)}{0.51 - 0.5} = \frac{59.8384 - 60}{0.01} = \frac{-0.1616}{0.01} = -16.16 \text{ ft/sec}.$$Step 4 — Estimate the instantaneous velocity:
The two average velocities bracket the instantaneous value: \(-15.84\) and \(-16.16\). The instantaneous velocity at \(t = 0.5\) must sit between them.
Answer: Best guess: \(v(0.5) \approx -16\) ft/sec. (The negative sign means the rock is falling — height is decreasing.)
2.1.2 The Area Problem and Integral Calculus
The second great question that gave birth to calculus is about area. Many physical quantities — work done by a force, distance traveled when speed is changing, electrical charge, probability — can be read off as an area trapped under a curve. So we ask: how do we find the area between the graph of a function and the x-axis over an interval (Figure 2.8)?
Figure 2.8 — The Area Problem: How do we find the area of the shaded region?
The plan looks familiar: start by approximating. Slice the interval \([a, b]\) into thin pieces and stand a rectangle on each piece, with the height set by the function's value somewhere in that slice. Add up the rectangle areas. That sum is a rough estimate of the area under the curve (Figure 2.9).
Figure 2.9 — The area of the region under the curve is approximated by summing the areas of thin rectangles.
As we slice thinner and thinner — width approaching zero — the rectangle-sum approaches the true area between the curve and the x-axis over \([a, b]\). It is, once again, a limit. The number that the sums approach is called the definite integral of \(f\) on \([a, b]\), and integral calculus is the branch of calculus built around it.
Estimate the area between the x-axis and the graph of \(f(x) = x^2 + 1\) over the interval \([0, 3]\) by using the three rectangles shown in the figure below.
Solution
This alternate figure shades rectangles using the RIGHT endpoint of each unit-width slice as the height. The heights become \(f(1) = 2\), \(f(2) = 5\), and \(f(3) = 10\).
Step 1 — Add the right-endpoint rectangle areas:
$$\text{Estimated area} = 1 \cdot 2 + 1 \cdot 5 + 1 \cdot 10 = 17 \text{ unit}^2.$$Answer: \(17\) square units. This time the estimate OVERSHOOTS the true area of \(12\) square units — because each rectangle uses the height at the RIGHT edge, where the curve is higher. Notice the true answer (\(12\)) sits neatly between the left-endpoint estimate (\(8\), from Example 2.1.3) and this right-endpoint estimate (\(17\)). Taking thinner rectangles closes the gap from both sides.
Estimate the area between the x-axis and the graph of \(f(x) = x^2 + 1\) over the interval \([0, 3]\) by using the three rectangles shown in Figure 2.10.
![Figure 2.10 — The area under y = x^2 + 1 on [0, 3] estimated by three left-endpoint rectangles.](../images/figure_2.10_light.png?v=20260516a)
Figure 2.10 — The area of the region under the curve of \(f(x) = x^2 + 1\) can be estimated using rectangles.
Solution
Step 1 — Read each rectangle's area from the figure:
The three rectangles in Figure 2.10 have heights set by the left endpoint of each unit-width slice: \(f(0) = 1\), \(f(1) = 2\), and \(f(2) = 5\). Each rectangle has width \(1\), so the areas are \(1 \cdot 1 = 1\) square unit, \(1 \cdot 2 = 2\) square units, and \(1 \cdot 5 = 5\) square units.
Step 2 — Sum the areas:
$$\text{Estimated area} = 1 + 2 + 5 = 8 \text{ unit}^2.$$Answer: The estimate is \(8\) square units. The true area under \(f(x) = x^2 + 1\) on \([0, 3]\) is \(12\) square units — the rectangle estimate undershoots because each rectangle uses the height at the LEFT edge of its slice (where the curve is lower).
2.1.3 Other Aspects of Calculus
So far we have only studied functions of one variable, where the graph lives in two dimensions. There is no reason to stop there. What if instead of an object moving along a coordinate axis, we want the velocity of a rock fired from a catapult, or of an airplane moving in three dimensions? What if we want to compute volumes of solid shapes — like the surface shown in Figure 2.11? These questions live in multivariable calculus, the branch of the subject that studies functions of two or more variables.
Figure 2.11 — We can use multivariable calculus to find the volume between a surface defined by a function of two variables and a plane.
Before we can climb to multivariable calculus, though, we have to build the single-variable foundation. That starts with the concept this section has gestured at over and over: the limit. The rest of Chapter 2 makes limits precise.
Problem Set 2.1
For the following exercises, points \(P(1, 2)\) and \(Q(x, y)\) are on the graph of the function \(f(x) = x^2 + 1\).
Problem 1. [T] Complete the following table with the appropriate values: the \(y\)-coordinate of \(Q\), the point \(Q(x, y)\), and the slope of the secant line passing through points \(P\) and \(Q\). Round your answer to eight significant digits.
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(1.1\) | a. | e. | i. |
| \(1.01\) | b. | f. | j. |
| \(1.001\) | c. | g. | k. |
| \(1.0001\) | d. | h. | l. |
Problem 2. Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to \(f\) at \(x = 1\).
Problem 3. Use the value in the preceding exercise to find an equation of the tangent line at point \(P\). Graph \(f(x)\) and the tangent line.
Solutions 1–3
Problem 1
Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = f(x) = x^2 + 1\), form the point \(Q(x, y)\), and apply the secant slope formula \(m_{\sec} = \dfrac{y - 2}{x - 1}\). The expression simplifies algebraically to \(m_{\sec} = \dfrac{x^2 - 1}{x - 1} = x + 1\) for \(x \ne 1\), which gives a quick sanity check on each numerical row.
Step 2 — Compute each row:
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(1.1\) | \(2.21\) | \((1.1,\ 2.21)\) | \(2.1000000\) |
| \(1.01\) | \(2.0201\) | \((1.01,\ 2.0201)\) | \(2.0100000\) |
| \(1.001\) | \(2.002001\) | \((1.001,\ 2.002001)\) | \(2.0010000\) |
| \(1.0001\) | \(2.00020001\) | \((1.0001,\ 2.00020001)\) | \(2.0001000\) |
Answer: Slopes column reads \(2.1\), \(2.01\), \(2.001\), \(2.0001\) — clearly trending toward \(2\).
Problem 2
Step 1 — Read the trend: The four secant slopes from the previous table are \(2.1\), \(2.01\), \(2.001\), \(2.0001\). Each is closer to \(2\) than the one above it.
Answer: The slope of the tangent line at \(x = 1\) is approximately \(\boxed{2}\).
Problem 3
Step 1 — Apply point-slope form: With slope \(m = 2\) at the point \(P(1, 2)\),
$$y - 2 = 2(x - 1).$$
Step 2 — Solve for \(y\):
$$y = 2x.$$
Answer: The tangent line at \(P(1, 2)\) is \(y = 2x\). To graph, plot the parabola \(f(x) = x^2 + 1\) and the line \(y = 2x\); they touch at \((1, 2)\) and the line lies tangent there (under the parabola elsewhere).
For the following exercises, points \(P(1, 1)\) and \(Q(x, y)\) are on the graph of the function \(f(x) = x^3\).
Problem 4. [T] Complete the following table with the appropriate values: the \(y\)-coordinate of \(Q\), the point \(Q(x, y)\), and the slope of the secant line passing through points \(P\) and \(Q\). Round your answer to eight significant digits.
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(1.1\) | a. | e. | i. |
| \(1.01\) | b. | f. | j. |
| \(1.001\) | c. | g. | k. |
| \(1.0001\) | d. | h. | l. |
Problem 5. Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to \(f\) at \(x = 1\).
Problem 6. Use the value in the preceding exercise to find an equation of the tangent line at point \(P\). Graph \(f(x)\) and the tangent line.
Solutions 4–6
Problem 4
Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = x^3\), form \(Q(x, y)\), and apply \(m_{\sec} = \dfrac{y - 1}{x - 1}\). Algebraically the formula reduces to \(m_{\sec} = x^2 + x + 1\) for \(x \ne 1\).
Step 2 — Compute each row:
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(1.1\) | \(1.331\) | \((1.1,\ 1.331)\) | \(3.3100000\) |
| \(1.01\) | \(1.030301\) | \((1.01,\ 1.030301)\) | \(3.0301000\) |
| \(1.001\) | \(1.003003001\) | \((1.001,\ 1.003003001)\) | \(3.0030010\) |
| \(1.0001\) | \(1.00030003\) | \((1.0001,\ 1.00030003)\) | \(3.0003000\) |
Answer: Slopes trend \(3.31 \to 3.03 \to 3.003 \to 3.0003\), converging on \(3\).
Problem 5
Step 1 — Read the trend: The four secant slopes from the previous table are \(3.31\), \(3.0301\), \(3.003001\), \(3.00030001\). They march down toward \(3\).
Answer: The slope of the tangent line at \(x = 1\) is approximately \(\boxed{3}\).
Problem 6
Step 1 — Apply point-slope form: With slope \(m = 3\) at \(P(1, 1)\),
$$y - 1 = 3(x - 1).$$
Step 2 — Solve for \(y\):
$$y = 3x - 2.$$
Answer: The tangent line at \(P(1, 1)\) is \(y = 3x - 2\). The line touches the curve \(f(x) = x^3\) at \((1, 1)\); elsewhere it sits below the curve for \(x > 1\) and above for \(x < 1\).
For the following exercises, points \(P(4, 2)\) and \(Q(x, y)\) are on the graph of the function \(f(x) = \sqrt{x}\).
Problem 7. [T] Complete the following table with the appropriate values: the \(y\)-coordinate of \(Q\), the point \(Q(x, y)\), and the slope of the secant line passing through points \(P\) and \(Q\). Round your answer to eight significant digits.
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(4.1\) | a. | e. | i. |
| \(4.01\) | b. | f. | j. |
| \(4.001\) | c. | g. | k. |
| \(4.0001\) | d. | h. | l. |
Problem 8. Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to \(f\) at \(x = 4\).
Problem 9. Use the value in the preceding exercise to find an equation of the tangent line at point \(P\).
Solutions 7–9
Problem 7
Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = \sqrt{x}\) (to 8 sig digs), form \(Q(x, y)\), and apply \(m_{\sec} = \dfrac{y - 2}{x - 4}\). Rationalizing the numerator gives \(m_{\sec} = \dfrac{1}{\sqrt{x} + 2}\), a useful sanity check.
Step 2 — Compute each row:
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(4.1\) | \(2.0248457\) | \((4.1,\ 2.0248457)\) | \(0.24845673\) |
| \(4.01\) | \(2.0024984\) | \((4.01,\ 2.0024984)\) | \(0.24984394\) |
| \(4.001\) | \(2.0002500\) | \((4.001,\ 2.0002500)\) | \(0.24998438\) |
| \(4.0001\) | \(2.0000250\) | \((4.0001,\ 2.0000250)\) | \(0.24999844\) |
Answer: Slopes trend \(0.2485 \to 0.2498 \to 0.24998 \to 0.249998\), zooming in on \(0.25\).
Problem 8
Step 1 — Read the trend: The four secant slopes head toward \(0.25 = \dfrac{1}{4}\). (Cross-check via the rationalized form: \(\dfrac{1}{\sqrt{4} + 2} = \dfrac{1}{4}\).)
Answer: Tangent slope at \(x = 4\) is approximately \(\boxed{\tfrac{1}{4} = 0.25}\).
Problem 9
Step 1 — Apply point-slope form: With slope \(m = \tfrac{1}{4}\) at \(P(4, 2)\),
$$y - 2 = \tfrac{1}{4}(x - 4).$$
Step 2 — Solve for \(y\):
$$y = \tfrac{1}{4}x + 1.$$
Answer: The tangent line at \(P(4, 2)\) is \(y = \tfrac{1}{4}x + 1\).
For the following exercises, points \(P(1.5, 0)\) and \(Q(\phi, y)\) are on the graph of the function \(f(\phi) = \cos(\pi \phi)\).
Problem 10. [T] Complete the following table with the appropriate values: the \(y\)-coordinate of \(Q\), the point \(Q(\phi, y)\), and the slope of the secant line passing through points \(P\) and \(Q\). Round your answer to eight significant digits.
| \(\phi\) | \(y\) | \(Q(\phi, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(1.4\) | a. | e. | i. |
| \(1.49\) | b. | f. | j. |
| \(1.499\) | c. | g. | k. |
| \(1.4999\) | d. | h. | l. |
Problem 11. Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to \(f\) at \(\phi = 1.5\).
Problem 12. Use the value in the preceding exercise to find an equation of the tangent line at point \(P\).
Solutions 10–12
Problem 10
Step 1 — Set up the secant slope formula: For each \(\phi\), compute \(y = \cos(\pi \phi)\) (to 8 sig digs), form \(Q(\phi, y)\), and apply \(m_{\sec} = \dfrac{y - 0}{\phi - 1.5}\). A useful identity: \(\cos(\pi \phi) = -\sin(\pi(\phi - 1.5))\), so \(m_{\sec} = \dfrac{-\sin(\pi (\phi - 1.5))}{\phi - 1.5}\). For small \(h = \phi - 1.5\), this is approximately \(-\pi\) — but it's negative because \(\phi - 1.5\) is negative in each row.
Step 2 — Compute each row:
| \(\phi\) | \(y\) | \(Q(\phi, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(1.4\) | \(-0.30901699\) | \((1.4,\ -0.30901699)\) | \(3.0901699\) |
| \(1.49\) | \(-0.031410759\) | \((1.49,\ -0.031410759)\) | \(3.1410759\) |
| \(1.499\) | \(-0.0031415875\) | \((1.499,\ -0.0031415875)\) | \(3.1415875\) |
| \(1.4999\) | \(-0.00031415924\) | \((1.4999,\ -0.00031415924)\) | \(3.1415924\) |
Answer: Slopes trend \(3.0902 \to 3.1411 \to 3.1416 \to 3.1416\), converging on \(\pi\).
Problem 11
Step 1 — Read the trend: The four secant slopes from the previous table approach \(\pi \approx 3.14159265\).
Answer: The slope of the tangent line at \(\phi = 1.5\) is approximately \(\boxed{\pi}\).
Problem 12
Step 1 — Apply point-slope form: With slope \(m = \pi\) at \(P(1.5, 0)\),
$$y - 0 = \pi(\phi - 1.5).$$
Step 2 — Tidy:
$$y = \pi \phi - \tfrac{3\pi}{2}.$$
Answer: The tangent line at \(P(1.5, 0)\) is \(y = \pi(\phi - 1.5)\), or equivalently \(y = \pi \phi - \tfrac{3\pi}{2}\).
For the following exercises, points \(P(-1, -1)\) and \(Q(x, y)\) are on the graph of the function \(f(x) = \tfrac{1}{x}\).
Problem 13. [T] Complete the following table with the appropriate values: the \(y\)-coordinate of \(Q\), the point \(Q(x, y)\), and the slope of the secant line passing through points \(P\) and \(Q\). Round your answer to eight significant digits.
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(-1.05\) | a. | e. | i. |
| \(-1.01\) | b. | f. | j. |
| \(-1.005\) | c. | g. | k. |
| \(-1.001\) | d. | h. | l. |
Problem 14. Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to \(f\) at \(x = -1\).
Problem 15. Use the value in the preceding exercise to find an equation of the tangent line at point \(P\).
Solutions 13–15
Problem 13
Step 1 — Set up the secant slope formula: For each \(x\), compute \(y = \dfrac{1}{x}\) (to 8 sig digs), form \(Q(x, y)\), and apply \(m_{\sec} = \dfrac{y - (-1)}{x - (-1)} = \dfrac{y + 1}{x + 1}\). The algebra collapses nicely: \(m_{\sec} = \dfrac{(1/x) + 1}{x + 1} = \dfrac{(1 + x)/x}{x + 1} = \dfrac{1}{x}\) for \(x \ne -1\) — so \(m_{\sec}\) equals \(y\) for each row.
Step 2 — Compute each row:
| \(x\) | \(y\) | \(Q(x, y)\) | \(m_{\sec}\) |
|---|---|---|---|
| \(-1.05\) | \(-0.95238095\) | \((-1.05,\ -0.95238095)\) | \(-0.95238095\) |
| \(-1.01\) | \(-0.99009901\) | \((-1.01,\ -0.99009901)\) | \(-0.99009901\) |
| \(-1.005\) | \(-0.99502488\) | \((-1.005,\ -0.99502488)\) | \(-0.99502488\) |
| \(-1.001\) | \(-0.99900100\) | \((-1.001,\ -0.99900100)\) | \(-0.99900100\) |
Answer: Slopes trend \(-0.9524 \to -0.9901 \to -0.99502 \to -0.99900\), heading toward \(-1\).
Problem 14
Step 1 — Read the trend: The four secant slopes head toward \(-1\). (Cross-check from the simplification above: \(m_{\sec} = \dfrac{1}{x}\), so as \(x \to -1\), \(m_{\sec} \to -1\).)
Answer: Tangent slope at \(x = -1\) is approximately \(\boxed{-1}\).
Problem 15
Step 1 — Apply point-slope form: With slope \(m = -1\) at \(P(-1, -1)\),
$$y - (-1) = -1(x - (-1)).$$
Step 2 — Simplify:
$$y + 1 = -(x + 1) \implies y = -x - 2.$$
Answer: The tangent line at \(P(-1, -1)\) is \(y = -x - 2\).
For the following exercises, the position function of a ball dropped from the top of a \(200\)-meter tall building is given by \(s(t) = 200 - 4.9 t^2\), where position \(s\) is measured in meters and time \(t\) is measured in seconds. Round your answer to eight significant digits.
Problem 16. [T] Compute the average velocity of the ball over the given time intervals:
a) \([4.99, 5]\)
b) \([5, 5.01]\)
c) \([4.999, 5]\)
d) \([5, 5.001]\)
Problem 17. Use the preceding exercise to guess the instantaneous velocity of the ball at \(t = 5\) sec.
Solutions 16–17
Problem 16
Step 1 — Compute \(s(t) = 200 - 4.9 t^2\) at the endpoints:
$$s(4.99) = 200 - 4.9(4.99)^2 = 200 - 122.01049 = 77.98951,$$
$$s(5) = 200 - 4.9(25) = 77.5,$$
$$s(4.999) = 200 - 4.9(4.999)^2 = 200 - 122.4510049 = 77.5489951,$$
$$s(5.001) = 200 - 4.9(5.001)^2 = 200 - 122.5490049 = 77.4509951.$$
Step 2 — Apply \(v_{\text{ave}} = \dfrac{s(t) - s(a)}{t - a}\) on each interval:
a) \([4.99, 5]\): \(v_{\text{ave}} = \dfrac{77.5 - 77.98951}{0.01} = -48.951000\) m/sec
b) \([5, 5.01]\): \(s(5.01) = 200 - 4.9(25.1001) = 77.00951\), so \(v_{\text{ave}} = \dfrac{77.00951 - 77.5}{0.01} = -49.049000\) m/sec
c) \([4.999, 5]\): \(v_{\text{ave}} = \dfrac{77.5 - 77.5489951}{0.001} = -48.995100\) m/sec
d) \([5, 5.001]\): \(v_{\text{ave}} = \dfrac{77.4509951 - 77.5}{0.001} = -49.004900\) m/sec
Answer: \(-48.951000\), \(-49.049000\), \(-48.995100\), \(-49.004900\) m/sec respectively. The negative sign means the ball is falling.
Problem 17
Step 1 — Read the bracket: From the previous exercise, intervals on either side of \(t = 5\) give average velocities that tighten in on a single value. Looking at the closest pair, \([4.999, 5]\) gives \(-48.9951\) and \([5, 5.001]\) gives \(-49.0049\) — they straddle \(-49\).
Step 2 — Confirm with the derivative formula: \(s'(t) = -9.8 t\), so \(s'(5) = -49\) exactly.
Answer: The instantaneous velocity at \(t = 5\) sec is \(\boxed{-49}\) m/sec.
For the following exercises, consider a stone tossed into the air from ground level with an initial velocity of \(15\) m/sec. Its height in meters at time \(t\) seconds is \(h(t) = 15 t - 4.9 t^2\).
Problem 18. [T] Compute the average velocity of the stone over the given time intervals:
a) \([1, 1.05]\)
b) \([1, 1.01]\)
c) \([1, 1.005]\)
d) \([1, 1.001]\)
Problem 19. Use the preceding exercise to guess the instantaneous velocity of the stone at \(t = 1\) sec.
Solutions 18–19
Problem 18
Step 1 — Compute \(h(1) = 15(1) - 4.9(1)^2 = 10.1\) m.
Step 2 — Apply \(v_{\text{ave}} = \dfrac{h(t) - h(1)}{t - 1}\) on each interval:
a) \([1, 1.05]\): \(h(1.05) = 15(1.05) - 4.9(1.1025) = 15.75 - 5.40225 = 10.34775\). \(v_{\text{ave}} = \dfrac{10.34775 - 10.1}{0.05} = 4.9550000\) m/sec.
b) \([1, 1.01]\): \(h(1.01) = 15.15 - 4.9(1.0201) = 15.15 - 4.99849 = 10.15151\). \(v_{\text{ave}} = \dfrac{0.05151}{0.01} = 5.1510000\) m/sec.
c) \([1, 1.005]\): \(h(1.005) = 15.075 - 4.9(1.010025) = 15.075 - 4.9491225 = 10.1258775\). \(v_{\text{ave}} = \dfrac{0.0258775}{0.005} = 5.1755000\) m/sec.
d) \([1, 1.001]\): \(h(1.001) = 15.015 - 4.9(1.002001) = 15.015 - 4.9098049 = 10.1051951\). \(v_{\text{ave}} = \dfrac{0.0051951}{0.001} = 5.1951000\) m/sec.
Answer: \(4.9550000\), \(5.1510000\), \(5.1755000\), \(5.1951000\) m/sec.
Problem 19
Step 1 — Read the trend: The average velocities on shrinking right-hand intervals climb toward \(5.2\).
Step 2 — Confirm with derivative: \(h'(t) = 15 - 9.8 t\), so \(h'(1) = 5.2\) exactly.
Answer: The instantaneous velocity at \(t = 1\) sec is \(\boxed{5.2}\) m/sec.
For the following exercises, consider a rocket shot into the air that then returns to Earth. The height of the rocket in meters is given by \(h(t) = 600 + 78.4 t - 4.9 t^2\), where \(t\) is measured in seconds.
Problem 20. [T] Compute the average velocity of the rocket over the given time intervals:
a) \([9, 9.01]\)
b) \([8.99, 9]\)
c) \([9, 9.001]\)
d) \([8.999, 9]\)
Problem 21. Use the preceding exercise to guess the instantaneous velocity of the rocket at \(t = 9\) sec.
Solutions 20–21
Problem 20
Step 1 — Compute \(h(9) = 600 + 78.4(9) - 4.9(81) = 908.7\) m.
Step 2 — Apply \(v_{\text{ave}}\) on each interval:
a) \([9, 9.01]\): \(h(9.01) = 600 + 706.384 - 4.9(81.1801) = 600 + 706.384 - 397.78249 = 908.60151\). \(v_{\text{ave}} = \dfrac{-0.09849}{0.01} = -9.8490000\) m/sec.
b) \([8.99, 9]\): \(h(8.99) = 600 + 704.816 - 4.9(80.8201) = 600 + 704.816 - 396.01849 = 908.79751\). \(v_{\text{ave}} = \dfrac{908.7 - 908.79751}{0.01} = -9.7510000\) m/sec.
c) \([9, 9.001]\): \(h(9.001) = 600 + 705.6784 - 4.9(81.018001) = 908.6901951\). \(v_{\text{ave}} = \dfrac{-0.0098049}{0.001} = -9.8049000\) m/sec.
d) \([8.999, 9]\): \(h(8.999) = 600 + 705.5216 - 4.9(80.982001) = 908.7097951\). \(v_{\text{ave}} = \dfrac{-0.0097951}{0.001} = -9.7951000\) m/sec.
Answer: \(-9.8490000\), \(-9.7510000\), \(-9.8049000\), \(-9.7951000\) m/sec.
Problem 21
Step 1 — Read the bracket: Intervals on either side of \(t = 9\) close in on \(-9.8\).
Step 2 — Confirm with derivative: \(h'(t) = 78.4 - 9.8 t\), so \(h'(9) = 78.4 - 88.2 = -9.8\) m/sec exactly.
Answer: The instantaneous velocity at \(t = 9\) sec is \(\boxed{-9.8}\) m/sec. (The rocket is on its way back down at \(t = 9\).)
For the following exercises, consider an athlete running a \(40\)-m dash. The position of the athlete is given by \(d(t) = \tfrac{t^3}{6} + 4 t\), where \(d\) is the position in meters and \(t\) is the time elapsed, measured in seconds.
Problem 22. [T] Compute the average velocity of the runner over the given time intervals:
a) \([1.95, 2.05]\)
b) \([1.995, 2.005]\)
c) \([1.9995, 2.0005]\)
d) \([2, 2.00001]\)
Problem 23. Use the preceding exercise to guess the instantaneous velocity of the runner at \(t = 2\) sec.
Solutions 22–23
Problem 22
Step 1 — Compute \(d(2) = \dfrac{2^3}{6} + 4(2) = \dfrac{4}{3} + 8 = \dfrac{28}{3} \approx 9.3333333\) m.
Step 2 — Apply \(v_{\text{ave}}\) on each interval:
a) \([1.95, 2.05]\): \(d(1.95) = \dfrac{7.414875}{6} + 7.8 = 9.0358125\); \(d(2.05) = \dfrac{8.615125}{6} + 8.2 = 9.6358542\). \(v_{\text{ave}} = \dfrac{0.6000417}{0.1} = 6.0004167\) m/sec.
b) \([1.995, 2.005]\): similar arithmetic gives \(v_{\text{ave}} \approx 6.0000042\) m/sec.
c) \([1.9995, 2.0005]\): \(v_{\text{ave}} \approx 6.0000000\) m/sec (to 8 sig digs).
d) \([2, 2.00001]\): \(d(2.00001) \approx 9.3333933\); \(v_{\text{ave}} = \dfrac{0.0000600}{0.00001} \approx 6.0000000\) m/sec.
Answer: \(6.0004167\), \(6.0000042\), \(6.0000000\), \(6.0000000\) m/sec.
Problem 23
Step 1 — Read the trend: The averages collapse to \(6\) immediately on shrinking intervals.
Step 2 — Confirm with derivative: \(d'(t) = \dfrac{t^2}{2} + 4\), so \(d'(2) = 2 + 4 = 6\) exactly.
Answer: The instantaneous velocity at \(t = 2\) sec is \(\boxed{6}\) m/sec.
For the following exercises, consider the function \(f(x) = |x|\).
Problem 24. Sketch the graph of \(f\) over the interval \([-1, 2]\) and shade the region above the x-axis.
Problem 25. Use the preceding exercise to find the approximate value of the area between the x-axis and the graph of \(f\) over the interval \([-1, 2]\) using rectangles. For the rectangles, use unit squares, and approximate both above and below the lines. Use geometry to find the exact answer.
Solutions 24–25
Problem 24
Step 1 — Describe the graph: \(f(x) = |x|\) is a V-shape with vertex at the origin. On \([-1, 0]\), the graph descends from \((-1, 1)\) to \((0, 0)\); on \([0, 2]\), it climbs from \((0, 0)\) to \((2, 2)\).
Step 2 — Shade: The region above the x-axis between the graph and the x-axis consists of two triangles — the left one bounded by \((-1, 0)\), \((0, 0)\), \((-1, 1)\); the right one bounded by \((0, 0)\), \((2, 0)\), \((2, 2)\).
Answer: The shaded region is a pair of right triangles sharing a vertex at the origin: a smaller \(1 \times 1\) triangle on the left and a larger \(2 \times 2\) triangle on the right.
Problem 25
Step 1 — Set up unit squares: Three unit-wide rectangles span \([-1, 0]\), \([0, 1]\), \([1, 2]\). For each, pick a height equal to the function's minimum (below estimate) or maximum (above estimate) on the slice.
Step 2 — Below estimate (minimum height per slice):
- \([-1, 0]\): \(\min = f(0) = 0\), area \(= 0\). - \([0, 1]\): \(\min = f(0) = 0\), area \(= 0\). - \([1, 2]\): \(\min = f(1) = 1\), area \(= 1\).
Below total \(= 1\) square unit.
Step 3 — Above estimate (maximum height per slice):
- \([-1, 0]\): \(\max = f(-1) = 1\), area \(= 1\). - \([0, 1]\): \(\max = f(1) = 1\), area \(= 1\). - \([1, 2]\): \(\max = f(2) = 2\), area \(= 2\).
Above total \(= 4\) square units.
Step 4 — Geometry for the exact answer: The region splits into two right triangles. Left triangle has legs \(1\) and \(1\); right triangle has legs \(2\) and \(2\).
$$\text{Area} = \tfrac{1}{2}(1)(1) + \tfrac{1}{2}(2)(2) = 0.5 + 2 = 2.5 \text{ square units}.$$
Answer: Below estimate \(= 1\), above estimate \(= 4\), exact area \(= 2.5\) square units (which sits inside the bracket, as expected).
For the following exercises, consider the function \(f(x) = \sqrt{1 - x^2}\). (Hint: this is the upper half of a circle of radius \(1\) centered at \((0, 0)\).)
Problem 26. Sketch the graph of \(f\) over the interval \([-1, 1]\).
Problem 27. Use the preceding exercise to find the approximate area between the x-axis and the graph of \(f\) over the interval \([-1, 1]\) using rectangles. For the rectangles, use squares \(0.4\) by \(0.4\) units, and approximate both above and below the lines. Use geometry to find the exact answer.
Solutions 26–27
Problem 26
Step 1 — Describe the graph: \(f(x) = \sqrt{1 - x^2}\) is the upper half of the unit circle centered at the origin. The graph passes through \((-1, 0)\), peaks at \((0, 1)\), and returns to the x-axis at \((1, 0)\). It is symmetric about the y-axis.
Answer: A semicircle (radius \(1\)) sitting above the x-axis with endpoints \((-1, 0)\) and \((1, 0)\).
Problem 27
Step 1 — Place \(0.4\)-wide rectangles across \([-1, 1]\): Five rectangles span the intervals \([-1, -0.6]\), \([-0.6, -0.2]\), \([-0.2, 0.2]\), \([0.2, 0.6]\), \([0.6, 1]\). Compute heights from \(f(x) = \sqrt{1 - x^2}\):
$$f(\pm 1) = 0, \quad f(\pm 0.6) = \sqrt{0.64} = 0.8, \quad f(\pm 0.2) = \sqrt{0.96} \approx 0.9798, \quad f(0) = 1.$$
Step 2 — Below estimate (lowest \(f\)-value in each slice; symmetry around \(x = 0\) means each slice's lowest value sits at its outer edge):
- \([-1, -0.6]\) and \([0.6, 1]\): height \(0\); each area \(0\). - \([-0.6, -0.2]\) and \([0.2, 0.6]\): height \(0.8\); each area \(0.8 \cdot 0.4 = 0.32\). - \([-0.2, 0.2]\): height \(\sqrt{0.96} \approx 0.9798\); area \(\approx 0.39192\).
Below total \(\approx 0 + 0.32 + 0.39192 + 0.32 + 0 \approx 1.0319\) square units.
Step 3 — Above estimate (highest \(f\)-value in each slice):
- \([-1, -0.6]\) and \([0.6, 1]\): height \(0.8\); each area \(0.32\). - \([-0.6, -0.2]\) and \([0.2, 0.6]\): height \(\approx 0.9798\); each area \(\approx 0.39192\). - \([-0.2, 0.2]\): height \(1\); area \(0.4\).
Above total \(\approx 0.32 + 0.39192 + 0.4 + 0.39192 + 0.32 \approx 1.8238\) square units.
Step 4 — Geometry for the exact answer: The region is exactly half of a unit circle.
$$\text{Area} = \tfrac{1}{2} \pi r^2 = \tfrac{\pi}{2} \approx 1.5708 \text{ square units}.$$
Answer: Below \(\approx 1.032\), above \(\approx 1.824\), exact \(= \dfrac{\pi}{2} \approx 1.5708\) square units.
For the following exercises, consider the function \(f(x) = -x^2 + 1\).
Problem 28. Sketch the graph of \(f\) over the interval \([-1, 1]\).
Problem 29. Approximate the area of the region between the x-axis and the graph of \(f\) over the interval \([-1, 1]\).
Solutions 28–29
Problem 28
Step 1 — Describe the graph: \(f(x) = -x^2 + 1\) is a downward-opening parabola with vertex \((0, 1)\). It crosses the x-axis at \(x = \pm 1\), giving a "dome" shape over the interval \([-1, 1]\).
Answer: A parabolic dome sitting on the x-axis with endpoints \((-1, 0)\) and \((1, 0)\) and peak \((0, 1)\).
Problem 29
Step 1 — Slice into 4 rectangles of width \(0.5\): Sub-intervals are \([-1, -0.5]\), \([-0.5, 0]\), \([0, 0.5]\), \([0.5, 1]\). The function values at the relevant points are
$$f(\pm 1) = 0, \quad f(\pm 0.5) = 0.75, \quad f(0) = 1.$$
Step 2 — Midpoint estimate (heights at \(x = -0.75, -0.25, 0.25, 0.75\)):
$$f(\pm 0.75) = 1 - 0.5625 = 0.4375, \quad f(\pm 0.25) = 1 - 0.0625 = 0.9375.$$
$$\text{Area}_{\text{mid}} \approx (0.4375 + 0.9375 + 0.9375 + 0.4375)(0.5) = 2.75 \cdot 0.5 = 1.375 \text{ square units}.$$
Step 3 — Left-endpoint estimate:
$$\text{Area}_{\text{left}} \approx (f(-1) + f(-0.5) + f(0) + f(0.5))(0.5) = (0 + 0.75 + 1 + 0.75)(0.5) = 1.25.$$
(Right-endpoint estimate is identical by symmetry: \(1.25\).)
Step 4 — Exact answer (integration):
$$\int_{-1}^{1}(1 - x^2)\,dx = \left[ x - \tfrac{x^3}{3} \right]_{-1}^{1} = \tfrac{2}{3} - \left(-\tfrac{2}{3}\right) = \tfrac{4}{3} \approx 1.3333.$$
Answer: Midpoint estimate \(\approx 1.375\), endpoint estimates \(\approx 1.25\), exact area \(= \tfrac{4}{3} \approx 1.333\) square units.
Key Terms
rate of change — how fast one quantity changes when another changes; numerically, the change in output divided by the change in input.
secant line — the straight line through two points \((a, f(a))\) and \((x, f(x))\) on the graph of \(f\); its slope \(\dfrac{f(x) - f(a)}{x - a}\) estimates the rate of change of \(f\) near \(a\).
tangent line — the limiting line that a family of secants through \((a, f(a))\) approaches as the second point slides toward \(a\); its slope is the exact rate of change of \(f\) at \(a\).
derivative — the slope of the tangent line at \((a, f(a))\); written \(f'(a)\), it is the instantaneous rate of change of \(f\) at the input \(a\).
differential calculus — the branch of calculus concerned with derivatives and what they reveal about how functions change.
tangent problem — the historical question of how to compute the slope of a curve at a single point; its solution gave rise to differential calculus.
average velocity — for a position function \(s(t)\) on a time interval \([a, t]\), the quantity \(v_{\text{ave}} = \dfrac{s(t) - s(a)}{t - a}\) — the slope of the secant line on the position graph.
limit — the value a function or formula approaches as its input is moved closer and closer to a target; the central concept that makes derivatives and integrals work.
instantaneous velocity — the value the average velocities approach as the time interval shrinks toward a single instant; the slope of the tangent line on the position graph at \(t = a\).
area problem — the historical question of how to compute the area trapped between a curve and the x-axis over an interval; its solution gave rise to integral calculus.
integral calculus — the branch of calculus concerned with the definite integral and its applications, including area, volume, total change, and accumulated quantities.
multivariable calculus — the study of calculus for functions of two or more variables, where graphs live in three or more dimensions.