Daniel J.

Evaluate the difference quotient for the given function. Simplify your answer.

$ f(x) = x^3 $ , $ \dfrac{f(a + h) - f(a)}{h} $

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.)

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?

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) $.

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.

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 $.