Calculus: Early Transcendentals

James Stewart

ISBN #9781285741550

8th Edition

6,422 Questions

2,819,387 Students Helped

Homework Questions

Summary

Learning Objectives

Key Concepts

Example Problems

Explanations

Common Mistakes

Summary

This section explains how limits are fundamental in defining the slope of a tangent line and instantaneous velocity. By considering the behavior of secant line slopes as a point on the curve is approached, we can precisely calculate these rates of change. The methods illustrated using algebraic, numerical, and graphical techniques bridge the gap between intuitive concepts and rigorous mathematical definitions.

Learning Objectives

Describe the limiting process as it applies to finding the slope of a tangent line.

Explain how the slope of a tangent line is defined as the limit of secant line slopes.

Apply numerical and graphical methods to estimate limits in various contexts such as instantaneous velocity and charge decay.

Connect the concept of limits to the computation of instantaneous rates of change.

Key Concepts

CONCEPT

DEFINITION

Limit

The value that a function (or sequence) approaches as the input (or index) approaches some value.

Tangent Line

A line that touches a curve at a point and has the same direction as the curve at that point. Its slope is defined as the limit of the slopes of secant lines approaching that point.

Secant Line

A line that intersects a curve at two or more points. The slope of a secant line is used to approximate the tangent line's slope.

Instantaneous Velocity

The rate of change of an object’s position at a specific instant in time, defined as the limit of the average velocity over increasingly small time intervals.

Average Velocity

The change in position divided by the change in time over a given interval.

Example Problems

Example 1

A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume $V$ of water remaining in the tank (in gallons) after $t$ minutes. $$ \begin{array}{|c|c|c|c|c|c|c|} \hline t(\mathrm{~min}) & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline V(\mathrm{gal}) & 694 & 444 & 250 & 111 & 28 & 0 \\ \hline \end{array} $$ (a) If $P$ is the point (15,250) on the graph of $V$, find the slopes of the secant lines $P Q$ when $Q$ is the point on the graph with $t=5,10,20,25,$ and 30 (b) Estimate the slope of the tangent line at $P$ by averaging the slopes of two secant lines. (c) Use a graph of the function to estimate the slope of the tangent line at $P$. (This slope represents the rate at which the water is flowing from the tank after 15 minutes.)

Example 2

A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after $ t $ minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. $$ \begin{array}{|l|c|c|c|c|c|} \hline t \text { (min) } & 36 & 38 & 40 & 42 & 44 \\ \hline \text { Heartbeats } & 2530 & 2661 & 2806 & 2948 & 3080 \\ \hline \end{array} $$ The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient's heart rate after 42 minutes using the secant line between the points with the given values of $ t $. (a) $ t = 36 $ and $ t = 42 $ (b) $ t = 38 $ and $ t = 42 $ (c) $ t = 40 $ and $ t = 42 $ (d) $ t = 42 $ and $ t = 44 $ What are your conclusions?

Example 3

The point $ P(2, -1) $ lies on the curve $ y = 1/(1-x) $. (a) If $ Q $ is the point $ (x, 1/(1-x)) $, use your calculator to find the slope of the secant line $ PQ $ (correct to six decimal places) for the following values of $ x $: (i) $ 1.5 $ (ii) $ 1.9 $ (iii) $ 1.99 $ (iv) $ 1.999 $ (v) $ 2.5 $ (vi) $ 2.1 $ (vii) $ 2.01 $ (viii) $ 2.001 $ (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at $ P(2, -1) $. (c) Using the slope from part (b), find an equation of the tangent line to the curve at $ P(2, -1) $.

Example 4

The point $ P(0.5, 0) $ lies on the curve $ y = \cos \pi x $. (a) If $ Q $ is the point $ (x, \cos \pi x) $, use your calculator to find the slope of the secant line $ PQ $ (correct to six decimal places) for the following values of $ x $: (i) $ 0 $ (ii) $ 0.4 $ (iii) $ 0.49 $ (iv) $ 0.499 $ (v) $ 1 $ (vi) $ 0.6 $ (vii) $ 0. 51 $ (viii) $ 0.501 $ (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at $ P(0.5, 0) $. (c) Using the slope from part (b), find an equation of the tangent line to the curve at $ P(0.5, 0) $. (d) Sketch the curve, two of the secant lines, and the tangent line.

Example 5

If a ball is thrown into the air with a velocity of $ 40 ft/s $, its height in feet $ t $ seconds later is given by $ y = 40t - 16t^2 $. (a) Find the average velocity for the time period beginning when $ t = 2 $ and lasting (i) 0.5 seconds (ii) 0.1 seconds (iii) 0.05 seconds (iv) 0.01 seconds (b) Estimate the instantaneous velocity when $ t = 2 $.

Step-by-Step Explanations

QUESTION

How do we determine the slope of the tangent line to the parabola y = x² at point P(1, 1)?

STEP-BY-STEP ANSWER:

Step 1: Choose a nearby point Q on the parabola, for example Q(x, x²) with x close to 1.
Step 2: Compute the slope of the secant line, mPQ = (x² - 1) / (x - 1).
Step 3: Simplify the expression to get mPQ = x + 1.
Step 4: As x approaches 1, the slope approaches 2.
Final Answer: The slope of the tangent line at P is 2.

Finding the Tangent Slope for y = x²

QUESTION

How do we estimate the instantaneous velocity of a ball after 5 seconds of free fall?

STEP-BY-STEP ANSWER:

Step 1: Use Galileo's law for free fall: s(t) = 4.9t².
Step 2: Calculate the average velocity over a small interval around t=5, such as from t=5 to t=5.1 seconds.
Step 3: Compute average velocity = [s(5.1) - s(5)] / 0.1.
Step 4: Observe that as the interval decreases, the computed average velocity approaches 49 m/s.
Final Answer: The instantaneous velocity at t = 5 seconds is approximately 49 m/s.

Estimating Instantaneous Velocity for a Falling Ball

Common Mistakes

Confusing the single-point tangent with a secant line that cuts the curve multiple times.
Assuming the limit must be computed exactly at the point rather than approaching it from both sides.
Overlooking the importance of selecting points sufficiently close to the point of tangency to approximate the limit.
Mixing up average velocity with instantaneous velocity without applying the limiting process.