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

Questions were taken from Advanced Placement Calculus Tests as indicated.

- (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 \(A = \pi r^2\).)
- Find the rate at which the perimeter of the square is increasing. Indicate units of measure.
- 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.

## 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.## 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. - (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 \(v(t) = t ^2 \sin {t}\) .
- Write an expression for the acceleration of the particle.
- For what values of \(t\) on \(0 \leq t < 7\) is the particle moving to the right?
- What is the minimum velocity of the particle on \(0 \leq t < 7\)? Show the analysis that leads to your conclusion.

## Solution

\(a(t) = 2tsin(t) + t ^2cos (t)\).## 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 \( (2 \pi , 7)\).## 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 \(a > 0) \). - (1994 AB1) (Calculator Inactive) Let \(f\) be the function given by \(f(x) = 3x^4 + x^3 -21x^2\).
- Write an equation of the line tangent to the graph of \(f\) at the point (2, -28)
- Find the absolute minimum value of \(f\). Show the analysis that leads to your conclusion.
- Find the x-coordinate of each point of inflection on the graph of \(f\). Show the analysis that leads to your conclusion.

## Solution

\( f'(x) = 12x^3 + 3x^2 -42x. f'(2) = 24\). Tan line: \(y + 28 = 24(x - 2)\)## 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.)## 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. - (1994 AB3) (Calculator Inactive) Consider the curve defined by \(x^2 + xy + y^2 = 27\).
- Write an expression for the slope of the curve at any point \((x, y)\).
- 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.
- Find the points on the curve where the lines tangent to the curve are vertical.

## Solution

\(2x + xy' + y + 2yy' = 0\). \(y' = \frac{-2x - y}{x + 2y}\)## 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.## 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). - (1995 AB3) (Calculator Active) Consider the curve defined by \(-8x^2 + 5xy + y^3 = -149\).
- Find \( \frac{dy}{dx}\) .
- Write an equation for the line tangent to the curve at the point (4, -1).
- 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 \(k\).
- 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.
- Solve the equation in part (d) for the value of \(k\).

## Solution

\(-16x + 5xy' + 5y + 3y^2y' = 0\). \( \frac{dy}{dx} = \frac{16x - 5y}{5x + 3y^2}.\)## Solution

At (4, -1), \(y' = \frac{16 \cdot 4 + 5}{5 \cdot 4 + 3} = 3\). Tan line: \(y + 1 = 3(x - 4)\).## Solution

\(y = 3(4.2 - 4) - 1 = -0.4 \)## Solution

\(-8(4.2)^2 + 5(4.2)k + k^3 = -149\)## Solution

\(k = -.3727\) - (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 \(\frac{1}{3} \pi r^2h\) ).
- Write an expression for the volume of water in the conical tank as a function of \(h\).
- At what rate is the volume of water in the conical tank changing when \(h = 3\)? Indicate units of measure.
- 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.

## Solution

Using similar triangles, \( \frac{r}{4} =\frac{h}{12} \implies r = \frac{h}{3}\text{. }V = \frac{1}{3} \pi (\frac{h}{3})^2h = \pi \frac{h^3}{27}\).## 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.## 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. - 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 \(-3 < x < 5\).- For what values of \(x\) does \(f\) have a relative maximum? Why?
- For what values of \(x\) does \(f\) have a relative minimum? Why?
- On what intervals is the graph of \(f\) concave upward? Use \(f'\) to justify your answer.
- 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 \(0 < x < 2.\)

## Solution

\(f\) has a relative max where \(f'\) passes from + to -, which occurs at \(x = -2\).## Solution

\(f\) has a relative min where \(f'\) passes from - to +, which occurs at \(x = 4\).## Solution

\(f\) is concave up where \(f'\) is rising. This occurs on [-1, 1] and [3, 5].## Solution