# BC Calculus: Extra Essay Problems for Review

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

Questions were taken from Advanced Placement Calculus Tests as indicated.
1. (1994 AB5) (Calculator inactive) A circle is inscribed in a square. The circumference of the circle is increasing a at a constant rate of 6 inches per second. As the circle expands, the square expands to maintain the condition of tangency. (Note: A circle with radius r has circumference $$C = 2 \pi r$$ and area $$A = \pi r^2$$.)
1. Find the rate at which the perimeter of the square is increasing. Indicate units of measure.
2. Solution Find $$\frac{dP}{dt}$$ with units given $$\frac{dC}{dt} = 6$$. $$P = 8r \implies \frac{dP}{dt} = 8\frac{dr}{dt}$$. $$C = 2 \pi r \implies \frac{dC}{dt} = 6 = 2 \pi \frac{dr}{dt}$$. Therefore, $$\frac{dr}{dt} = \frac{6}{2 \pi } \implies \frac{dP}{dt} = \frac{8 \cdot 6}{2 \pi }$$ inches/second.
3. At the instant when the area of the circle is $$25 \pi$$ square inches, find the rate of increase in the area enclosed between the circle and the square. Indicate units of measure.
4. Solution Enclosed area $$E= s^2 - \pi r^2$$. Find $$\frac{dE}{dt}$$ with units when $$r = 5$$. $$\frac{dE}{dt} = 2s \frac{ds}{dt} - 2 \pi r\frac{dr}{dt}\text{. } \frac{dr}{dt} = \frac{6}{2 \pi }$$ from part (a) and since $$s = 2r\text{, } \frac{ds}{dt} = 2\frac{dr}{dt}$$. Finally, when $$r = 5, s = 10$$. So substituting, $$\frac{dE}{dt} = 2(10)\frac{2 \cdot 6}{2 \pi } - 2 \pi \frac{5 \cdot 6}{2 \pi }$$ square inches per second.
2. (1994 AB4 modified) (Calculator Active) A particle moves along the x-axis so that at any time $$0 \leq t < 7$$ its velocity is given by $$v(t) = t ^2 \sin {t}$$ .
1. Write an expression for the acceleration of the particle.
2. Solution $$a(t) = 2tsin(t) + t ^2cos (t)$$.
3. For what values of $$t$$ on $$0 \leq t < 7$$ is the particle moving to the right?
4. Solution The particle moves to the right when $$v(t) > 0$$. On the interval $$0 \leq t < 7\text{, }t ^2\sin{t} > 0$$ when $$t$$ is on $$(0, \pi )$$ and $$(2 \pi , 7)$$.
5. What is the minimum velocity of the particle on $$0 \leq t < 7$$? Show the analysis that leads to your conclusion.
6. Solution $$a(t)= 2t \sin{t} + t ^2 \cos {t} > 0$$ on $$(0, 2.2889)$$ and $$(5.0870, 7)$$, so $$v$$ is increasing on these intervals. $$a(t) < 0$$ on $$(2.2889, 5.0870)$$, so $$v$$ is decreasing on this interval. Therefore, $$v$$ can have a min at the left endpoint $$t = 0$$ (where $$v$$ becomes positive), or where $$a$$ goes from - to +, at 5.0870. $$V(0) = 0, v(5.0870) = -24.0829$$, the absolute min on this interval (but not at the right endpoint, where $$a > 0)$$.
3. (1994 AB1) (Calculator Inactive) Let $$f$$ be the function given by $$f(x) = 3x^4 + x^3 -21x^2$$.
1. Write an equation of the line tangent to the graph of $$f$$ at the point (2, -28)
2. Solution $$f'(x) = 12x^3 + 3x^2 -42x. f'(2) = 24$$. Tan line: $$y + 28 = 24(x - 2)$$
3. Find the absolute minimum value of $$f$$. Show the analysis that leads to your conclusion.
4. Solution $$f'(x) = 12x^3 + 3x^2 -42x = 3x(4x^2 + x -14) = 3x(4x - 7)(x + 2)$$. $$f'(x) = 0$$ when $$x = \{ -2, 0, \frac{7}{4} \}$$. $$f'(x) < 0$$ on $$(-\infty,-2)$$, and $$(0, \frac{7}{4})$$ so $$f$$ falls on those intervals. $$f'(x) > 0$$ on $$(-2, 0)$$ and ( $$\frac{7}{4}, \infty)$$, so $$f$$ rises there. Since $$\displaystyle \lim_{x \to \pm \infty} f(x) = \infty$$, there can be an abs. min, but no abs. max. The candidates for min are -2 and $$\frac{7}{4}$$. $$f(-2) = -44$$, $$f(\frac{7}{4}) = -\frac{7889}{256}$$ (which is a little under -30.) So the min is -44. (Sorry about the arithmetic that you have to do without a calculator, here.)
5. Find the x-coordinate of each point of inflection on the graph of $$f$$. Show the analysis that leads to your conclusion.
6. Solution $$f''(x) = 36x^2 + 6x -42 = 6(6x + x - 7) = 6(6x + 7)(x - 1)$$. $$f''(x) = 0\text{ at }x = -\frac{7}{6},$$ and 1. Since $$f''$$ changes sign at both those places, $$f$$ has points of inflection there.
4. (1994 AB3) (Calculator Inactive) Consider the curve defined by $$x^2 + xy + y^2 = 27$$.
1. Write an expression for the slope of the curve at any point $$(x, y)$$.
2. Solution $$2x + xy' + y + 2yy' = 0$$. $$y' = \frac{-2x - y}{x + 2y}$$
3. Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leads to your conclusion.
4. Solution X-intercepts at $$y = 0$$. $$x = \pm \sqrt{27}$$. At $$( \sqrt{27}, 0)\text{, }y' = -\frac{2\sqrt{27}}{\sqrt{27}} = -2$$. At $$( - \sqrt{27},0) \text{, }y' = (-2) \frac{-\sqrt{27}}{-\sqrt{27}} = -2$$. Yes, they're parallel as the slopes there are the same.
5. Find the points on the curve where the lines tangent to the curve are vertical.
6. Solution ertical when $$x + 2y = 0$$, or $$x = -2y$$. ( $$-2y) ^2 + (-2y)y + y^2 = 27 \implies 4y^2 -2y^2 + y^2 = 27 \implies 3y^2 = 27$$. so $$y = \pm 3$$. When $$y = 3, x + 2(3) = 0$$ so $$x = -6$$. When $$y = -3 \implies x + 2(-3) = 0$$, so $$x = 6$$. The two points are (6, -3) and (-6, 3).
5. (1995 AB3) (Calculator Active) Consider the curve defined by $$-8x^2 + 5xy + y^3 = -149$$.
1. Find $$\frac{dy}{dx}$$ .
2. Solution $$-16x + 5xy' + 5y + 3y^2y' = 0$$. $$\frac{dy}{dx} = \frac{16x - 5y}{5x + 3y^2}.$$
3. Write an equation for the line tangent to the curve at the point (4, -1).
4. Solution At (4, -1), $$y' = \frac{16 \cdot 4 + 5}{5 \cdot 4 + 3} = 3$$. Tan line: $$y + 1 = 3(x - 4)$$.
5. There is a number $$k$$ so that the point $$(4.2, k)$$ is on the curve. Using the tangent line found in part (b), approximate the value of $$k$$.
6. Solution $$y = 3(4.2 - 4) - 1 = -0.4$$
7. Write an equation that can be solved to find the actual value of $$k$$ so that the point $$(4.2, k)$$ is on the curve.
8. Solution $$-8(4.2)^2 + 5(4.2)k + k^3 = -149$$
9. Solve the equation in part (d) for the value of $$k$$.
10. Solution $$k = -.3727$$
6. (1995 AB5) (Calculator Inactive) A conical tank, vertex down, has a hole at its vertex. The conical tank is 12 feet in height, and has a diameter of 8 feet. Water is draining from that conical tank into a cylindrical tank that has a base with area $$400 \pi$$ square feet. The depth $$h$$, in feet, of the water in the conical tank is changing at the rate of $$h - 12$$ feet per minute. (The volume, $$V$$, of a cone with radius $$r$$ and height $$h$$ is $$\frac{1}{3} \pi r^2h$$ ).
1. Write an expression for the volume of water in the conical tank as a function of $$h$$.
2. Solution Using similar triangles, $$\frac{r}{4} =\frac{h}{12} \implies r = \frac{h}{3}\text{. }V = \frac{1}{3} \pi (\frac{h}{3})^2h = \pi \frac{h^3}{27}$$.
3. At what rate is the volume of water in the conical tank changing when $$h = 3$$? Indicate units of measure.
4. Solution $$\frac{dV}{dt} = \pi \frac{h^2}{9} \cdot \frac{dh}{dt}$$, and since $$\frac{dh}{dt} = h - 12$$, when $$h = 3, \frac{dh}{dt} = -9$$, and $$\frac{dV}{dt} = \frac{\pi (3)^2}{9} \cdot (-9) = -9 \pi$$ cubic feet/minute.
5. Let $$y$$ be the depth, in feet, of the water in the cylindrical tank. At what rate is $$y$$ changing when $$h = 3$$? Indicate units of measure.
6. Solution For the cylinder, $$V = 400 \pi h$$, so $$\frac{dV}{dt} = 400 \pi \frac{dh}{dt}$$. So, from part (b), $$\frac{dV}{dt} = 9 \pi$$ (positive because the water is coming in) and so $$9 \pi = 400 \pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{ 9 \pi}{400 \pi }$$ feet/minute.
7. The following is a graph of the derivative of $$f$$, not of the graph of $$f$$: The figure above shows the graph of $$f'$$, the derivative of a function $$f$$. The domain of $$f$$ is the set of all real numbers x such that $$-3 < x < 5$$.
1. For what values of $$x$$ does $$f$$ have a relative maximum? Why?
2. Solution $$f$$ has a relative max where $$f'$$ passes from + to -, which occurs at $$x = -2$$.
3. For what values of $$x$$ does $$f$$ have a relative minimum? Why?
4. Solution $$f$$ has a relative min where $$f'$$ passes from - to +, which occurs at $$x = 4$$.
5. On what intervals is the graph of $$f$$ concave upward? Use $$f'$$ to justify your answer.
6. Solution $$f$$ is concave up where $$f'$$ is rising. This occurs on [-1, 1] and [3, 5].
7. Suppose that $$f(1) = 0$$. On an xy-plane, draw a sketch that shows the general shape of the graph of the function $$f$$ on the open interval $$0 < x < 2.$$
8. Solution 