Final Exam Practice

Sky-Diver Velocity Function:
    * Variables:
        * Velocity is represented by $v$ in units of $\text{feet per second}$ .
        * Altitude is represented by $a$ in units of $\text{feet}$ .
    * Data Points: The function $v(a)$ describes velocity based on altitude using the following values:
        * When $a = 1000\,a$ , $v(1000) = 50\,\text{ft/s}$ .
        * When $a = 2000\,a$ , $v(2000) = 100\,\text{ft/s}$ .
        * When $a = 3000\,a$ , $v(3000) = 200\,\text{ft/s}$ .
        * When $a = 4000\,a$ , $v(4000) = 200\,\text{ft/s}$ .
        * When $a = 5000\,a$ , $v(5000) = 200\,\text{ft/s}$ .
Water Tank Drainage Function:
    * Context: A tank starts completely full and drains at a constant rate.
    * Variables:
        * Height of water is $h$ measured in $\text{feet}$ .
        * Time is $t$ measured in $\text{minutes}$ .
    * Model Formula: The height function is $h(t) = 10 - 0.25t$ .
Graphical Function Definitions:
* Two separate functions are presented graphically for evaluation: $y = p(x)$ (depicted in blue) and $y = q(x)$ (depicted in green).

Graphic Evaluation (Question 1):
    * The goal is to determine specific output values from the provided graphs of $p$ and $q$ :
        * Finding $p(3)$ .
        * Finding $q(-3)$ .
Average Rate of Change (Question 2):
    * Interval: The problem asks for the average rate of change on the interval $[2, 4]$ , denoted as $AV_{[2,4]}$ .
    * Formula: $AV_{[a,b]} = \frac{h(b) - h(a)}{b - a}$ .
    * Calculations for $h(t) = 10 - 0.25t$ :
        * Evaluate at $t = 2$ : $h(2) = 10 - 0.25(2) = 10 - 0.5 = 9.5\,\text{feet}$ .
        * Evaluate at $t = 4$ : $h(4) = 10 - 0.25(4) = 10 - 1 = 9\,\text{feet}$ .
        * Compute the average rate: $AV_{[2,4]} = \frac{9 - 9.5}{4 - 2} = \frac{-0.5}{2} = -0.25$ .
    * Contextual Meaning: The value $AV_{[2,4]} = -0.25\,\text{feet/minute}$ indicates that between minute $2$ and minute $4$ , the height of the water in the tank was decreasing at an average rate of $0.25\,\text{feet}$ per minute.
Evaluating Composite Functions (Question 4):
* The task is to evaluate the nested function $p(q(0))$ .
* This requires first finding the value of $q(0)$ from the green graph and then using that result as the input for function $p$ on the blue graph.

Linear Modeling of Velocity (Question 3):
    * To find the formula for $v(a)$ in point-slope form ( $y - y_1 = m(x - x_1)$ ), the user must identify a specific interval and its slope.
    * Slope Calculation: $m = \frac{v(a_2) - v(a_1)}{a_2 - a_1}$ .
    * Identification: The user must clearly identify the specific point $(a_1, v_1)$ used for the formula and the calculated slope $m$ .
Algebraic Function Transformations (Question 6):
    * Objective: Express function $j(x)$ (red graph) as a transformation of the parent function $k(x)$ (blue graph).
    * General Transformation Form: $j(x) = c \times k(x - b) + a$ .
    * Transformation Parameters:
        * $c$ : Represents the vertical stretch or compression factor.
        * $b$ : Represents the horizontal shift (left or right).
        * $a$ : Represents the vertical shift (up or down).

Logistic Function Model (Question 5):
    * The number of sick people, $S$ , on a cruise ship is modeled by the function $g(t)$ , where $t$ is the day:
        * $S = g(t) = \frac{2500}{1 + 499e^{-0.1t}}$ .
    * Existence of an Inverse: The function $g$ has an inverse function $g^{-1}$ because it is a strictly monotonic (increasing) function on its domain, meaning it passes the Horizontal Line Test.
    * Derivation of the Inverse Formula $t = g^{-1}(S)$ :
        1. Start with $S = \frac{2500}{1 + 499e^{-0.1t}}$ .
        2. Rearrange to isolate the denominator: $1 + 499e^{-0.1t} = \frac{2500}{S}$ .
        3. Subtract 1 from both sides: $499e^{-0.1t} = \frac{2500}{S} - 1$ .
        4. Isolate the exponential term: $e^{-0.1t} = \frac{\frac{2500}{S} - 1}{499}$ .
        5. Take the natural logarithm of both sides: $-0.1t = \ln\left(\frac{\frac{2500}{S} - 1}{499}\right)$ .
        6. Solve for $t$ : $t = -10\ln\left(\frac{2500 - S}{499S}\right)$ .

Circular Geometry Context (Question 7):
    * Circle Properties:
        * Center: $(3, 4)$ .
        * Circumference: $18$ .
    * Deriving the Radius ( $r$ ):
        * $C = 2\pi r \implies 18 = 2\pi r \implies r = \frac{9}{\pi}$ .
    * Height Function ( $H = f(d)$ ):
        * Tracks the y-coordinate ( $H$ ) as a point traverses the circle counterclockwise.
        * The input $d$ represents the distance (arc length) along the circle starting from point $P_0$ .
    * Symmetry and Coordinates:
        * $P_0$ (Starting point): Typically starting at the "3 o'clock" position relative to the center, coordinates are $(3 + \frac{9}{\pi}, 4)$ .
        * $P_1$ (Quarter circle, $d = 4.5$ ): Coordinates are $(3, 4 + \frac{9}{\pi})$ .
        * $P_2$ (Half circle, $d = 9$ ): Coordinates are $(3 - \frac{9}{\pi}, 4)$ .
    * Function Parameters:
        * Midline: The average height, which is the y-coordinate of the circle's center: $y = 4$ .
        * Amplitude: The radius of the circle: $A = \frac{9}{\pi}$ .
        * Period: The distance for one full cycle, which matches the circumference: $P = 18$ .
    * Graphical Representation: The graph of $H = f(d)$ is a trigonometric sine or cosine wave oscillating around the midline $y = 4$ with a peak at $4 + \frac{9}{\pi}$ and a trough at $4 - \frac{9}{\pi}$ .