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

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Questions taken from the Advanced Placement Calculus test, as indicated.-
(2002BCB) The number of gallons, \(P(t)\), of a pollutant in a lake changes at the rate \(P'(t) = 1 - 3e^{-0.2\sqrt{t}}\) gallons per day, where \(t\) is measured in days. There are 50 gallons of the pollutant in the lake at time \(t = 0\). The lake is considered to be safe when it contains 40 gallons or less of pollutant.
- Is the amount of pollutant increasing at time \(t = 9\)? Why or why not?
- For what value of t will the number of gallons of pollutant be at its minimum? Justify your answer.
- Is the lake safe when the number of gallons of pollutant is at its minimum? Justify your answer.

## Solution

No, since \( P'(9) = 0.646\) is negative## Solution

\(P'(t) < 0\) on \([0, 30.1737) \). \( P'(t) > 0\) when \(t > 30.1737\). So \(P\) goes from falling to rising at \(t = 30.1737\), and \( P \) has a min there.## Solution

\(P(30.1737) = 50 + \int_{0}^{30.1737} P'(t) \, dt = 35.104\) gallons. Since this is less than 40, yes, it will be safe. -
(2003BCB) A blood vessel is 360 millimeters (mm) long with a circular cross section of varying diameter. The table below gives the measurements of the diameter of the blood vessel at selected points along the length of the blood vessel, where \(x\) represents the distance from one end of the blood vessel and \(B(x)\) is a twice-differentiable function that represents the diameter at that point.

Distance

\( x\)

(mm)060120180240300360Diameter

\(B(x)\)

(mm)24302830262426- Write an integral expression in terms of \(B(x)\) that represents the average radius, in mm, of the blood vessel between \(x = 0\) and \(x = 360\).
- Approximate the value of your answer from part (a) using the data from the table and a midpoint Riemann sum with three subintervals of equal length. Show the computations that lead to your answer.
- Explain why there must be at least one value, \( x\), for \( 0 < x < 360\), such that \(B''(x) = 0\).

## Solution

\( \frac{1}{360}\int_{0}^{360} \frac{B(x)}{2} \, dx \)## Solution

\( \frac{1}{2} (120 \cdot B(60) + 120 \cdot B(180) + 120 \cdot B(300)) = 60 (30 + 30 + 24)\). The \(\frac{1}{2}\) is because we're measuring radius, and given diameter. Now, since it's an average value we're trying to get, I have to divide by (360-0) to get \(\frac{30 + 30 + 24}{6}\).## Solution

Since \(B\) is twice-differentiable on \([0, 360], B\) and \(B'\) are continuous on that interval, so Rolle's theorem holds for both \(B\) and \(B'\). Since \(B(60) = B(180)\), by a corollary to Rolle's Thm, there must exist a \(c_1\) on \((60, 180) \) where \( B'(c_1) = 0.\) Similarly, since \(B(240) = B(360)\), by a corollary to Rolle's Thm, there must exist a \(c_2\) on \((240, 360) \) where \( B'(c_2) = 0.\) Then, applying Rolle's Thm to \(B',\) there must exist a \(c\) on \((c_1, c_2) \) where \( B''(c) = 0.\) - (2003BC) Let \(f\) be a function defined on the closed interval \(-3 \leq x \leq 4\) with \(f(0) = 3\). The graph of \(f'\), the derivative of \(f\), consists of one line segment with endpoints (-3, 1) and (0, -2), and a semicircle with radius 2, center (2, -2) which contains the points (0, -2), (2, 0), and (4, -2).
- On what intervals, if any, is \(f\) increasing? Justify your answer.
- Find the x-coordinate of each point of inflection of the graph of \(f\) on the open interval \(-3 < x < 4\). Justify your answer.
- Find an equation for the line tangent to the graph of \(f\) at the point (0, 3).
- Find \(f(-3)\) and \(f(4)\). Show the work that leads to your answers.

## Solution

The function \(f\) is increasing on [-3, -2] since \(f' > 0\) for \(-3 \leq x < -2\).## Solution

\(x = 0\) and \(x = 2\). \(f'\) changes from decreasing to increasing at \(x = 0\), and from increasing to decreasing at \(x = 2\).## Solution

\(f'(0) = -2\). The tangent line is \(y - 3 = -2(x - 0)\)## Solution

\(f(0) - f(-3) = \int_{-3}^0 f'(t) \, dt = \frac{1}{2}(1)(1) - \frac{1}{2}(2)(2) = -\frac{3}{2}\text{. }f(-3) = f(0) + \frac{3}{2} = \frac{9}{2}\). \(f(4) - f(0) = \int_0^4 f'(t) \, dt = - \left( 8 - \frac{1}{2}(2)^2\pi \right) = -8 + 2\pi\text{. }f(4) = f(0) - 9 + 2\pi = -5 + 2\pi\)