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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 1.4 The Tangent and Velocity Problems

Subsection 1.4.1 Before Class

https://mymedia.ou.edu/media/1.4-1/1_dfuhtyzs
Figure 5. Pre-Class Video 1

Subsubsection 1.4.1.1 Tangent Lines & Secant Lines

Recall that the slope of the line between two points on the curve \(f(x)\) is given in a few ways:
\begin{equation*} \dfrac{\text{rise}}{\text{run}} \qquad \dfrac{y_2-y_1}{x_2-x_1} \qquad \dfrac{f(x_2)-f(x_1)}{x_2-x_1} \end{equation*}
Definition 1.4.1. Tangent/Secant Lines.
The tangent line to a curve \(f(x)\) is a line which intersects the curve at only one point. A secant line intersects a curve in at least two points.
Figure 6. Illustration of the transformation from a secant line to a tangent line. Use the slider to look at different values of \(h\text{;}\) when \(h = 0\text{,}\) the secant line becomes a tangent line.
Example 1.4.2.
The points \((2,4)\) and \((4,16)\) lie on the graph of \(f(x)\text{.}\)
  1. Find the slope of the secant line between these two points.
  2. Find the formula for the secant line between these points.
  3. Assume \(f(x) = x^2\text{.}\) Give a quick sketch of \(f(x)\text{,}\) along with the points given (you don’t have to be super precise). How “close” are we to having a tangent line? Is there a way to get “closer” to having a tangent line?
Solution.
  1. The slope is 6
  2. \(\displaystyle y = 6x-8\)
  3. The graph of \(f(x) = x^2\) on the interval \([-1,5]\text{,}\) with the points \((2,4)\) and \((4,16)\) plotted, as well as the secant line between the points.
    We aren’t very close to having a tangent line; we could get closer to a tangent line by making the points closer.
Example 1.4.3.
The function \(f(x) = x^2\) is graphed below.
A graph of \(f(x)=x^2\) with an \(x-\)window of \([0,2]\text{,}\) and a \(y-\)window of \([0,4]\text{.}\) There are grid lines every 0.1 unit on the \(x-\)axis, and every 0.2 units on the \(y-\)axis.
Approximate the slope of the tangent line to \(f\) at \(x = 1\) by finding the slope of the secant line through \((1,1)\) and the following points. Sketch a few secant lines on the graph above as well.
  1. \(\displaystyle (0.7,0.49)\)
  2. \(\displaystyle (0.8,0.64)\)
  3. \(\displaystyle (0,9,0.81)\)
  4. \(\displaystyle (1.3,1.69)\)
  5. \(\displaystyle (1.2,1.44)\)
  6. \(\displaystyle (1.1, 1.21)\)
Solution.
  1. 1.7
  2. 1.8
  3. 1.9
  4. 2.3
  5. 2.2
  6. 2.1
Answers vary for secant lines

Subsection 1.4.2 Pre-Class Activities

Example 1.4.4.

  1. In Example 1.4.3, what happened to the slopes as you moved from (a) to (c), and from (d) to (f)? Be descriptive.
  2. Example 1.4.4 does not give us an exact value for the tangent line. How could we improve the approximation we got?
Solution.
  1. Moving from (a) to (c), the slopes increase toward 2; moving from (d) to (f), the slopes decrease toward 2.
  2. We can improve the approximation by taking the secant line through points closer to \((1,1)\text{.}\)

Subsection 1.4.3 In Class

Example 1.4.5.

Use a calculator to estimate the slope of the tangent line to the curve \(f(x) = \sqrt{x}\) at \(x = 3\text{.}\) Record your answers will full decimal accuracy, and round your final answer to the hundredths place, if necessary.
\(x\) \(\dfrac{f(x)-f(3)}{x-3}\)
\(2.9\)
\(2.99\)
\(2.999\)
\(2.9999\)
\(2.99999\)
Left-hand slope \(\approx\)
\(x\) \(\dfrac{f(x)-f(3)}{x-3}\)
\(3.1\)
\(3.01\)
\(3.001\)
\(3.0001\)
\(3.00001\)
Right-hand slope \(\approx\)
Slope at \(x=3\approx\)
Solution.
\(x\) \(\dfrac{f(x)-f(3)}{x-3}\)
\(2.9\) 0.291122
\(2.99\) 0.288916
\(2.999\) 0.288699
\(2.9999\) 0.288678
\(2.99999\) 0.288675
Left-hand slope \(\approx\) 0.29
\(x\) \(\dfrac{f(x)-f(3)}{x-3}\)
\(3.1\) 0.286309
\(3.01\) 0.288435
\(3.001\) 0.288651
\(3.0001\) 0.288673
\(3.00001\) 0.288675
Right-hand slope \(\approx\) 0.29
Slope at \(x=3\approx 0.29\)

Subsubsection 1.4.3.1 Velocity

Suppose we want to find the average velocity of an object. We can use the same approach as we did above to find this. The average velocity of an object is given by
\begin{equation*} v_{avg} = \dfrac{\text{change in position}}{\text{change in time}} \end{equation*}
Question 1.4.6.
What connection might exist between \(v_{avg}\) and the work we did in the previous exercises?
Solution.
Answers vary
Example 1.4.7.
Suppose a ball is dropped from the top of Dale Hall Tower, which is 250m tall. Find the average velocity of the ball over the given intervals. Use the fact that free-fall of an object is given by the equation \(s(t) = 4.9t^2\) m.
  1. \(\displaystyle [0,6]\)
  2. \(\displaystyle [2,5]\)
  3. \(\displaystyle [3,3.5]\)
  4. \(\displaystyle [3,3.01]\)
  5. \(\displaystyle [3,3.0001]\)
Solution.
  1. 29.4 m/s
  2. 34.3 m/s
  3. 31.85 m/s
  4. 29.449 m/s
  5. 24.40049 m/s
Question 1.4.8.
In the previous example, what do you think the purpose was of looking at so many average velocities?
Solution.
We want to get a better idea of how the velocity changes as we look at smaller and smaller intervals.

Subsection 1.4.4 After Class Activities

Example 1.4.9.

The point \(P = \lrpar{\dfrac{1}{2},0}\) lies on the curve \(y = \cos \pi x\text{.}\)
  1. If \(Q\) is the point \((x,\cos \pi x)\text{,}\) find the slope of the secant line \(PQ\) (to six decimal places) for the following values of \(x\text{.}\) Show your work, and make sure yoru calculator is in radian mode.
    1. 0.4
    2. 0.49
    3. 0.4999
    4. 0.6
    5. 0.51
    6. 0.5001
  2. Using part (a), estimate the value of the slope of the tangent line to \(y\) at \(x = 0.5\) to two decimal places.
  3. Using part (b), find an equation for the tangent line to \(y\) at \(x = 0.5\text{.}\)
Solution.
    1. \(\displaystyle -3.090170\)
    2. \(\displaystyle -3.141076\)
    3. \(\displaystyle -3.141593\)
    4. \(\displaystyle -3.090170\)
    5. \(\displaystyle -3.151076\)
    6. \(\displaystyle -3.141593\)
  2. The slope is about 3.14
  3. \(\displaystyle y = -3.14x + 1.57\)

Example 1.4.10.

If a rock is thrown upward on Io with a velocity of 10 m/s, its height (in meters) \(t\) seconds later is given by \(y = 10t - 1.86\text{.}\)
  1. Find the average velocity over the given time intervals:
    1. \(\displaystyle [1,2]\)
    2. \(\displaystyle [1,1.5]\)
    3. \(\displaystyle [1,1.1]\)
    4. \(\displaystyle [1,1.01]\)
    5. \(\displaystyle [1,1.001]\)
  2. Estimate the instantaneous velocity when \(t = 1\text{.}\)
Solution.
    1. 10 m/s
    2. 10 m/s
    3. 10 m/s
    4. 10 m/s
    5. 10 m/s
  2. 10 m/s

Example 1.4.11.

Imagine you need to teach this section to a student in another section. How would you explain the connection between tangent lines and secant lines to them? Be descriptive!
Solution.
Answers vary
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