- Class: 5H: BC Calculus
- Author: Peter Atlas
- Text:
__Calculus__Finney, Demana, Waits, Kennedy

Some of the following questions were taken from Advanced Placement Calculus Tests. They are indicated where they have been.

- (1998 BC5 Calculator active) The temperature outside of a house during a 24-hour period is given by \(F(t) =
80 - 10 \cos {\left(\frac{ \pi t}{12} \right) }, 0 \leq t \leq 24\), where \(F(t)\) is measured in degrees Fahrenheit and \(t\) is measured in hours.
- Sketch the graph of \(F\) on the given domain.
- Find the average temperature, to the nearest degree Fahrenheit, between \(t = 6\) and \(t = 14\).
- An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of \(t\) was the air conditioner cooling the house?
- The cost of cooling the house accumulates at the rate of $0.05 per hour for each degree the outside temperature exceeds 78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 24-hour period?

## Solution

## Solution

\(\frac{1}{14 - 6} \int_{6}^{14} (80 - 10\cos{ \left( \frac{ \pi t}{12} \right)) } \, dt = 87.162\). Answer: 87 degrees F## Solution

\(80 - 10\cos{ \left( \frac{\pi t}{12} \right) } - 78 \geq 0 \implies 5.231 \leq t \leq 18.769\).## Solution

Let \(p = 5.231\) and \(q = 18.769\text{. Then } 0.05 \int_{p}^{q}(80 - 10\cos{ \left( \frac{ \pi t}{12} \right)} - 78) \, dt = 5.096\); Answer: $5.10 - (Calculator inactive) The graph of the velocity \(v(t)\), in feet per second, of a bicycle racing on a straight road, for \( 0 \leq t \leq 60\), is shown. Also given is a table of values for \(v(t)\), at 5 second intervals of time t.
\(t\) (seconds) \(v(t)\) (feet per second) 0051310161521202525353040354240444548504555446047- During what intervals of time is the acceleration of the bike positive? Give a reason for your answer.
- Find the average acceleration of the bike, in feet per second squared, over the interval \(0 \leq t \leq 60\).
- Approximate the acceleration of the bike, in feet per second squared, at \(t = 30\). Show the computations you used to arrive at your answer.
- Approximate \(\int_0^{60} v(t) \, dt \) with a Riemann sum, using the midpoints of six subintervals of equal length. Using correct units, explain the meaning of this integral.

## Solution

Acceleration is positive on (0, 45) and (55, 60) because the velocity is increasing on these intervals.## Solution

Av. accel \( = \frac{47}{60} \frac{\text{ft}}{\text{sec}^2}\)## Solution

\( \frac{v(35) - v(25)}{10} = \frac{7}{10} \frac{\text{ft}}{\text{sec}^2} \text{ or } \frac{v(35) - v(30)}{5} = \frac{2}{5} \frac{ \text{ft}}{\text{sec}^2} \text{ or } \frac{v(30) - v(25)}{5} = 1 \frac{\text{ft}}{\text{sec}^2} \)## Solution

10(13 + 21 + 35 + 42 + 48 + 44) = 2030 ft. The particle moved 2030 feet to in th epositive direction from t = 0 to 60 seconds. - (1994 AB 6, calculator active) Let \( F(x) = \int_0^x \sin{ \left( t^2 \right) } \, dt\) for \(0 \leq x \leq 3\).
- Use the trapezoidal rule with four equal subdivisions of the closed interval [0, 1] to approximate \(F(1)\).
- On what intervals is \(F\) increasing?
- If the average rate of change of \(F\) on the closed interval [1, 3] is \(k\), find \(\int_1^3 \sin{ \left( t^2 \right) } \, dt\) in terms of \(k\).

## Solution

\( \frac{1}{2} \left( \sin{(0^2)} + 2\sin{\left( \frac{1}{4} \right) ^2} + 2\sin{\left( \frac{1}{2} \right) ^2} + 2\sin{\left( \frac{3}{4} \right) ^2} + \sin(1^2) \right) \frac{1}{4} \)## Solution

\(F\) increases on [0, 1.772] and [2.506, 3] since \(\sin{\left( t^2 \right) } > 0\). (0, 1.772) and (2.506, 3]## Solution

\(\frac{1}{3 - 1} \int_{1}^{3} \sin{ \left( t^2 \right) } \, dt = k\), so \( \int_{1}^{3} \sin{ \left( t^2 \right) } \, dt = 2k\) - (2003 BC 5 Form B, calculator inactive) Let \(f\) be a function defined on the closed interval [0, 7] so that the graph of \(f\) consists of four line segments, the first connecting the origin with the point (2, 4), the second connecting the points (2, 4) and (4, 0), the third connecting the points (4, 0), and (5, -1), and the fourth connecting the points (5, -1) and (7, 1). Let \(g\) be the function given by \(g(x) = \int_2^x f(t) \, dt\).
- Find \(g(3), g'(3)\) and \(g''(3)\).
- Find the average rate of change of \(g\) on the interval \(0 \leq x \leq 3\).
- For how many values \(c\), where \(0 < c < 3\), is \(g'(c)\) equal to the average rate found in part (b)? Explain your reasoning.
- Find the x-coordinate of each point of inflection on the graph of \(g\) on the interval \(0 < x < 7\). Justify your answer.

## Solution

\(g(3) = 3; g'(3) = 2; g''(3) = -2\)## Solution

\(\frac{1}{3} \int_{0}^{3}f(t) \, dt = \frac{1}{3} \cdot \left( g(3) - g(0) \right) = \frac{7}{3}\)## Solution

There are two values of \(c\). We need \(\frac{7}{3} = g'(c) = f(c)\). The graph of \(f\) intersects the line \(y = \frac{7}{3}\) at two places between 0 and 3.## Solution

\(x = 2\) and \(x = 5\) because \(g' = f\) changes from increasing to decreasing at \(x = 2\) and from decreasing to increasing at \(x = 5\).