Topics of Exam 2:

• Lecture 11: The Chain Rule; Derivative of the Natural Logarithmic Function
• Lecture 12: Higher Order Derivatives
• Lecture 13: Implicit Differentiation
• Lecture 14-15: Related Rates
• Lecture 16: Relative Extrema and Critical Numbers
• Lecture 17: Increasing and Decreasing Functions and the First Derivative Test
• Lecture 18: Concavity; Inflection Points and the Second Derivative Test

1. Problem 1. Find \( f'(1) \) if:

   \[ f(x) = \ln \sqrt[4]{\frac{4+4x}{4-x}} \]\[ \begin{aligned} f(x) &= \ln \left[ \frac{4+4x}{4-x} \right]^{\frac{1}{4}} \\ &= \frac{1}{4} \ln \left[ \frac{4+4x}{4-x} \right] \\ &= \frac{1}{4} \left[ \ln(4+4x) - \ln(4-x) \right] \\ &= \frac{1}{4} \ln(4+4x) - \frac{1}{4} \ln(4-x) \end{aligned} \]\[ f'(x) = \frac{1}{4} \cdot \frac{4}{4+4x} - \frac{1}{4} \cdot \frac{(-1)}{4-x} \]\[ \begin{aligned} f'(1) &= \frac{1}{4} \cdot \frac{4}{8} + \frac{1}{4} \cdot \frac{1}{3} \\ &= \frac{1}{8} + \frac{1}{12} \\ &= \frac{3}{24} + \frac{2}{24} \\ &= \frac{5}{24} \end{aligned} \]

2. Problem 2. Find the initial acceleration of a particle if its position after \( t \) seconds is given by:

   \[ s(t) = \ln(t^2 + 1) \]

   Express your final answer in \( \frac{\text{m}}{\text{s}^2} \).

   Ie, we need to find \( s''(0) \).

   \[ \begin{gathered} v(t) = s'(t) = \frac{2t}{t^2+1} \\[0.5em] a(t) = s''(t) = \frac{2(t^2+1) - 2t(2t)}{(t^2+1)^2} \\[0.5em] a(0) = \frac{2-0}{(0+1)^2} = 2\,\frac{\text{m}}{\text{s}^2} \end{gathered} \]

3. Problem 3. Given the algebraic curve \( x^2 + 4xy + y^2 = 13 \). What is the equation of the line tangent to the curve at the point \( (x,y) = (2,1) \)?

   Find \( \left. \frac{dy}{dx} \right|_{(x,y)=(2,1)} \):

   \[ \begin{gathered} x^2 + 4xy + y^2 = 13 \\[0.5em] \frac{d}{dx}\left( x^2 + 4xy + y^2 \right) = \frac{d}{dx}(13) \\[0.5em] 2x + 4y + 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \\[0.5em] 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - 4y \\[0.5em] \left[ 4x + 2y \right] \frac{dy}{dx} = -2x - 4y \\[0.5em] \frac{dy}{dx} = \frac{-2x - 4y}{4x + 2y} \\[0.5em] \left. \frac{dy}{dx} \right|_{(2,1)} = \frac{-2(2) - 4(1)}{4(2) + 2(1)} \\[0.5em] = -\frac{8}{10} \\[0.5em] = -\frac{4}{5} \end{gathered} \]

   Equation for a line with slope \( -\frac{4}{5} \) containing \( (2,1) \):

   \[ \begin{gathered} y - 1 = -\frac{4}{5}(x - 2) \\[0.5em] y = -\frac{4}{5}x + \frac{8}{5} + 1 \\[0.5em] y = -\frac{4}{5}x + \frac{13}{5} \end{gathered} \]

4. Problem 4. A baseball diamond is a square with 90 ft on each side. A player runs from first base to second base at 13 ft/sec. At what rate is the player's distance from home base increasing when he is half way from first to second base?

   

   Let \( y \) be the distance from 1^st base to the player and \( z \) the distance from home plate to the player. We know \( \frac{dy}{dt} = 0 \) and we need to know \( \frac{dz}{dt} \) when \( y = 45 \). \( y \) and \( z \) are related by the Pythagorean Theorem:

   \[ \begin{gathered} 90^2 + y^2 = z^2 \\[0.5em] \frac{d}{dt}\left[ 90^2 + y^2 \right] = \frac{d}{dt}\left[ z^2 \right] \\[0.5em] 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\[0.5em] \frac{dz}{dt} = \frac{y}{z}\frac{dy}{dt} \end{gathered} \]

   We also need to determine \( z \) when \( y = 45 \). As \( z \ge 0 \),

   \[ \begin{aligned} z &= \sqrt{90^2 + y^2} \\ &= \sqrt{90^2 + (45)^2} \\ &= \sqrt{90^2 + \tfrac{1}{4}90^2} \\ &= \sqrt{\tfrac{5}{4} \cdot 90^2} \\ &= 45\sqrt{5} \end{aligned} \]

Hence, when \( y = 45 \):

The distance from home plate and the player is increasing at a rate of \( \frac{13}{\sqrt{5}} \frac{m}{s} \).

5. Problem 5. (OPTIONAL...very optional) Repeat Problem 4, but replace 13ft/s with an arbitrary speed \( v \) ft/s and 90ft with an arbitrary length \( \ell \) ft.

   Let \( y \) be the distance from 1^st base to the player and \( z \) the distance from home plate to the player. We know \( \frac{dy}{dt} = v \) and we need to know \( \frac{dz}{dt} \) when \( y = \frac{1}{2}\ell \). \( y \) and \( z \) are related by the Pythagorean Theorem:

   \[ \begin{gathered} \ell^2 + y^2 = z^2 \\[0.5em] \frac{d}{dt}\left[ \ell^2 + y^2 \right] = \frac{d}{dt}\left[ z^2 \right] \\[0.5em] 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\[0.5em] \frac{dz}{dt} = \frac{y}{z}\frac{dy}{dt} \end{gathered} \]

   We also need to determine \( z \) when \( y = \frac{1}{2}\ell \). As \( z \ge 0 \),

   \[ \begin{aligned} z &= \sqrt{\ell^2 + y^2} \\ &= \sqrt{\ell^2 + \left(\tfrac{1}{2}\ell\right)^2} \\ &= \sqrt{\ell^2 + \tfrac{1}{4}\ell^2} \\ &= \sqrt{\tfrac{5}{4}\ell^2} \\ &= \tfrac{1}{2}\ell\sqrt{5} \end{aligned} \]

Hence, when \( y = \frac{1}{2}\ell \):

The distance from home plate and the player is increasing at a rate of \( \frac{v}{\sqrt{5}} \frac{m}{s} \).

6. Problem 6. Sand pours onto a surface at 19 cm³/s, forming a conical pile with a base diameter that is always equal to the pile's altitude. How fast is the altitude of the pile increasing when the pile is 5 cm high?

   [NOTE: The volume of a cone with base radius \( r \) and height \( h \) is \( V = \frac{1}{3}\pi r^2 h \)]

   

   Let \( r, h, \) and \( V \) be the radius, height, and volume of the conical pile, respectively. We know \( \frac{dV}{dt} = 19 \) and we want to know \( \frac{dh}{dt} \) when \( h = 5 \). We can relate \( r, h, \) and \( V \) by

   \[ V = \frac{1}{3}\pi r^2 h \]

   However, taking the derivative as is, we would need to use the product rule and hence we will have \( \frac{dr}{dt} \) in our expression. So, we use the "base diameter ... is always equal to the pile's altitude" clause to rewrite \( V \) solely as a function of \( h \). In symbols, the clause translates to \( 2r = h \), or \( r = \frac{h}{2} \). Hence,

   \[ \begin{aligned} V &= \frac{1}{3}\pi r^2 h \\ V &= \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h \\ V &= \frac{1}{12}\pi h^3 \\ \frac{dV}{dt} &= \frac{d}{dt}\left( \frac{1}{12}\pi h^3 \right)\\ \frac{dV}{dt} &= \frac{1}{12}\pi\left(3h^2\frac{dh}{dt}\right) \\ \frac{dV}{dt} &= \frac{\pi}{4}h^2\frac{dh}{dt} \\ \frac{dh}{dt} &= \frac{4}{\pi h^2}\frac{dV}{dt} \end{aligned} \]

Thus, when \( h = 5 \),

The height of the pile is increasing at a rate of \( \frac{76}{25\pi} \frac{cm}{s} \).

7. Problem 7. Find the critical numbers of \( y = x^2 e^x \).

   We need to determine when \( y' \) is \(0\) or undefined:

   \[ \begin{aligned} y' &= 2x e^x + x^2 e^x \\ &= x e^x (2 + x) \end{aligned} \]

   \( y' \) is never undefined, so we solve

   \[ x e^x (2 + x) \stackrel{\text{set}}{=} 0 \]

   to get \( x \) is either 0 or \( -2 \).

8. Problem 8. For the function \( f(x) = x^3 - 6x^2 - 15x + 4 \):

   1. When is \( f \) increasing and concave upward?
   2. When is \( f \) increasing and concave downward?
   3. When is \( f \) decreasing and concave upward?
   4. When is \( f \) decreasing and concave downward?
   5. State the locations (\( x \)-coordinates) of any relative maximums, relative minimums, and inflection points.

   We need to make a sign chart for both \( f' \) and \( f'' \).

   \[ \begin{aligned} f'(x) &= 3x^2 - 12x - 15 \\ &= 3(x^2 - 4x - 5) \\ &= 3(x - 5)(x + 1) \stackrel{\text{set}}{=} 0 \end{aligned} \]

   So, \( x = -1 \) and \( x = 5 \) are our points of interest (i.e., our critical numbers). Also,

   \[ \begin{aligned} f''(x) &= 6x - 12 \\ &= 6(x - 2) \stackrel{\text{set}}{=} 0 \end{aligned} \]

   So, \( x = 2 \) is a candidate for an inflection point. Using the sign chart below

   Sign chart for \( f' \)

   	\( (-\infty, -1) \)	\( (-1, 2) \)	\( (2, 5) \)	\( (5, \infty) \)
   Test Point	\( -100 \)	\( 0 \)	\( 3 \)	\( 100 \)
   Sign of \( x + 1 \)	\( - \)	\( + \)	\( + \)	\( + \)
   Sign of \( x - 5 \)	\( - \)	\( - \)	\( - \)	\( + \)
   Sign of \( f' \)	\( + \)	\( - \)	\( - \)	\( + \)
   Sign of \( f'' \)	\( - \)	\( - \)	\( + \)	\( + \)

We get the answers to each question:

1. On \( (5, \infty) \)
2. On \( (-\infty, -1) \)
3. On \( (2, 5) \)
4. On \( (-1, 2) \)
5. By the First Derivative Test, \( f \) has a relative max at \( x = -1 \) and a relative min at \( x = 5 \). Also, \( x = 2 \) is the location of an inflection point by definition.

Now, since we used the First Derivative Test to determine the locations of the relative extrema, this is unnecessary. However, I do want to demonstrate the Second Derivative Test again in the review. So, we will confirm our answer using the Second Derivative Test. Recall that:

Plugging in each critical number we obtain:

Therefore, by the 2nd Derivative Test, \(f\) has a relative max at \(x = -1\) and a relative min at \(x = 5\).