# BC Calculus: Assignment 32: FTC Applications

• Class: 5H: BC Calculus
• Author: Peter Atlas
• Text: Calculus Finney, Demana, Waits, Kennedy

Calculator Active

Questions taken from the Advanced Placement Calculus test, as indicated.
1. (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.
1. Is the amount of pollutant increasing at time $$t = 9$$? Why or why not?
2. Solution No, since $$P'(9) = 0.646$$ is negative
3. For what value of t will the number of gallons of pollutant be at its minimum? Justify your answer.
4. 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.
5. Is the lake safe when the number of gallons of pollutant is at its minimum? Justify your answer.
6. 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.
2. (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) 0 60 120 180 240 300 360 Diameter $$B(x)$$ (mm) 24 30 28 30 26 24 26
1. Write an integral expression in terms of $$B(x)$$ that represents the average radius, in mm, of the blood vessel between $$x = 0$$ and $$x = 360$$.
2. Solution $$\frac{1}{360}\int_{0}^{360} \frac{B(x)}{2} \, dx$$
3. 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.
4. 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 $$\frac{30 + 30 + 24}{6}$$.
5. Explain why there must be at least one value, $$x$$, for $$0 < x < 360$$, such that $$B''(x) = 0$$.
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 $$B''(c) = 0.$$

3. (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).
1. On what intervals, if any, is $$f$$ increasing? Justify your answer.
2. Solution The function $$f$$ is increasing on [-3, -2] since $$f' > 0$$ for $$-3 \leq x < -2$$.
3. Find the x-coordinate of each point of inflection of the graph of $$f$$ on the open interval $$-3 < x < 4$$. Justify your answer.
4. Solution $$x = 0$$ and $$x = 2$$. $$f'$$ changes from decreasing to increasing at $$x = 0$$, and from increasing to decreasing at $$x = 2$$.
5. Find an equation for the line tangent to the graph of $$f$$ at the point (0, 3).
6. Solution $$f'(0) = -2$$. The tangent line is $$y - 3 = -2(x - 0)$$
7. Find $$f(-3)$$ and $$f(4)$$. Show the work that leads to your answers.
8. Solution $$f(0) - f(-3) = \int_{-3}^0 f'(t) \, dt = \frac{1}{2}(1)(1) - \frac{1}{2}(2)(2) = -\frac{3}{2}\text{. }f(-3) = f(0) + \frac{3}{2} = \frac{9}{2}$$. $$f(4) - f(0) = \int_0^4 f'(t) \, dt = - \left( 8 - \frac{1}{2}(2)^2\pi \right) = -8 + 2\pi\text{. }f(4) = f(0) - 9 + 2\pi = -5 + 2\pi$$