Quadratic Models & Projectile Motion
Quadratic models represent processes featuring an optimum turning point (maximum or minimum) or gravitational projectile trajectories:
In projectile motion under Earth gravity ($g \approx 9.8\text{ m/s}^2$):
Exponential Growth and Decay ($f(x) = ka^x + c$)
Exponential functions model quantities whose rate of increase or decrease is proportional to the current amount:
Horizontal Asymptote Interpretation: As $x \to \infty$ in decay models ($a < 1$), $ka^x \to 0$, so $f(x) \to c$. In cooling models, $c$ represents the ambient room temperature that an object approaches asymptotically but never falls below.
Half-Life and Doubling Time Workflows
Key time constants in IB AI applications:
- Doubling Time ($T_d$): The time needed for population or investment to double ($f(t) = 2f(0)$).
- Half-Life ($T_{1/2}$): The time needed for radioactive material to decay to half of its initial amount ($f(t) = 0.5f(0)$).
- On your GDC, graph $Y_1 = f(t)$ and $Y_2 = \text{target value}$ and find the point of intersection using
2nd CALC > Intersect.
Model Comparison & Shifted Growth
- Compare residuals on the same observations and validation data, not just visual appearance. A quadratic adds curvature, but extra parameters can overfit noise. Always state the fitted domain interval and real-world units.
- For $f(t) = ka^t + c$, the quantity following constant percentage change is the excess $f(t) - c$. Half-life and doubling time apply to this excess above horizontal asymptote $c$, not generally to $f$ itself. The asymptote $y = c$ is approached from above if $k > 0$ and from below if $k < 0$.
- Example: Cooling from $80^\circ\text{C}$ to $20^\circ\text{C}$ with excess temperature halving every $10\text{ minutes}$: $T(t) = 20 + 60\left(\frac{1}{2}\right)^{t/10}$. After $10\text{ minutes}$, $T = 20 + 30 = 50^\circ\text{C}$, not $40^\circ\text{C}$.
A ball is kicked into the air. Its height in meters after $t$ seconds is modeled by $h(t) = -4.9t^2 + 19.6t + 1.5$.
- Find the maximum height reached by the ball.
- Find the time at which the ball reaches this maximum height.
Solution:
(b) Time of maximum occurs at vertex: $t = -\frac{b}{2a} = -\frac{19.6}{2(-4.9)} = \frac{19.6}{9.8} = 2.0\text{ seconds}$. [M1][A1]
(a) Maximum height $= h(2) = -4.9(2)^2 + 19.6(2) + 1.5$ [M1]
$= -19.6 + 39.2 + 1.5 = 21.1\text{ meters}$. [A1]
Using the model from Exercise 1, find the total time the ball remains in the air before hitting the ground.
Solution:
Set $h(t) = 0 \implies -4.9t^2 + 19.6t + 1.5 = 0$ [M1]
Using quadratic formula or GDC polynomial solver: [M1]
$t = \frac{-19.6 \pm \sqrt{19.6^2 - 4(-4.9)(1.5)}}{-9.8} \implies t \approx 4.075\text{ s}$ (reject negative root $t \approx -0.075$). [A1]
Total time in air is $4.08\text{ seconds}$ (to 3 s.f.).
A population of bacteria is modeled by $P(t) = 450(1.08)^t$, where $t$ is measured in hours.
- State the initial bacteria population.
- Find the hourly percentage growth rate.
Solution:
(a) When $t = 0$: $P(0) = 450(1.08)^0 = 450$ bacteria. [A1]
(b) Base $a = 1.08 = 1 + 0.08$. [M1]
The hourly growth rate is $8\%$ per hour. [A1]
For the bacteria population in Exercise 3 ($P(t) = 450(1.08)^t$), calculate the doubling time in hours.
Solution:
Double initial population: $P(t) = 2 \times 450 = 900$. [M1]
$450(1.08)^t = 900 \implies 1.08^t = 2$ [M1]
$t = \frac{\ln(2)}{\ln(1.08)}$ [M1]
$t = \frac{0.693147}{0.076961} \approx 9.006... \approx 9.01\text{ hours}$. [A1]
A cup of coffee cools according to the model $T(t) = 65(0.92)^t + 22$, where $T$ is temperature in $^\circ\text{C}$ and $t$ is time in minutes.
- Find the temperature of the coffee immediately after brewing ($t = 0$).
- Write down the equation of the horizontal asymptote of $T(t)$, and state its physical meaning.
Solution:
(a) $T(0) = 65(0.92)^0 + 22 = 65(1) + 22 = 87^\circ\text{C}$. [A1]
(b) Horizontal asymptote: $y = 22$ (or $T = 22$). [A1]
Physical meaning: Ambient room temperature ($22^\circ\text{C}$), which the coffee approaches as time progresses. [R2]
A radioactive isotope decays according to $M(t) = M_0 e^{-0.035t}$, where $t$ is in years. Find the half-life of the substance.
Solution:
Set $M(t) = 0.5M_0 \implies M_0 e^{-0.035t} = 0.5M_0$ [M1]
$e^{-0.035t} = 0.5 \implies -0.035t = \ln(0.5)$ [M1]
$t = \frac{\ln(0.5)}{-0.035} = \frac{-0.69315}{-0.035}$ [M1]
$t \approx 19.80... \approx 19.8\text{ years}$. [A1]
A farmer has $120\text{ meters}$ of fencing to enclose a rectangular paddock against a straight river bank (no fencing needed on the river side). Find the dimensions of the paddock that maximize the enclosed area.
Solution:
Let width perpendicular to river be $x$. Length along river is $L = 120 - 2x$. [M1]
Area $A(x) = x(120 - 2x) = 120x - 2x^2 = -2x^2 + 120x$. [A1]
Maximum area occurs at vertex: $x = -\frac{b}{2a} = -\frac{120}{2(-2)} = \frac{120}{4} = 30\text{ meters}$. [M1]
Dimensions: width $x = 30\text{ m}$, length $L = 120 - 2(30) = 60\text{ m}$. [A1]
Explain why an exponential growth model $P(t) = P_0 (1.05)^t$ cannot realistically model human city population growth indefinitely over centuries.
Solution:
Exponential growth assumes unlimited resources and constant growth rate without bound. [R1]
In reality, environmental carrying capacity, land availability, water, food supply, and social factors constrain growth, leading to a logistic S-curve plateau rather than unbounded exponential growth. [A2]
A quadratic function has vertex at $(4, 18)$ and passes through $(2, 10)$. Find the equation of the function in expanded standard form $f(x) = ax^2 + bx + c$.
Solution:
Vertex form: $f(x) = a(x - 4)^2 + 18$. [M1]
Substitute $(2, 10)$: $10 = a(2 - 4)^2 + 18 \implies 10 = 4a + 18 \implies 4a = -8 \implies a = -2$. [M1]
$f(x) = -2(x^2 - 8x + 16) + 18 = -2x^2 + 16x - 32 + 18$. [M1]
$f(x) = -2x^2 + 16x - 14$. [A1]
The value of a car depreciates according to $V(t) = 32000(0.85)^t$, where $t$ is time in years. Find the time taken for the car's value to drop below $10,000$.
Solution:
Set $V(t) = 10000 \implies 32000(0.85)^t = 10000$ [M1]
$0.85^t = \frac{10000}{32000} = 0.3125$ [M1]
$t = \frac{\ln(0.3125)}{\ln(0.85)} = \frac{-1.16315}{-0.16252} \approx 7.157...$ [M1]
$7.16\text{ years}$ (or during the 8th year). [A1]
Given data points $(0, 2)$, $(1, 5)$, $(2, 10)$, and $(3, 17)$, compare the linear model $L(x) = 5x + 1$ and the quadratic model $Q(x) = x^2 + 2$ using the sum of squared errors ($\text{SSE}$). State which model fits these observations better and note one limitation.
Solution:
For $L(x) = 5x + 1$, predicted values at $x = 0, 1, 2, 3$ are $1, 6, 11, 16$. Residuals ($y - \hat{y}$) are $1, -1, -1, 1$. [M1]
$\text{SSE}_L = 1^2 + (-1)^2 + (-1)^2 + 1^2 = 4$. [A1]
For $Q(x) = x^2 + 2$, predicted values are $2, 5, 10, 17$, matching every observed point perfectly: $\text{SSE}_Q = 0$. [M1][A1]
Evaluation: Model $Q$ fits the recorded sample points without error ($\text{SSE} = 0$). However, perfect fit on four points does not justify unrestrained extrapolation beyond $x \in [0, 3]$ without supplementary validation data. [R1]
Newton's law of cooling for a liquid is modeled by $T(t) = 20 + 60\left(\frac{1}{2}\right)^{t/10}$, where $T$ is temperature in $^\circ\text{C}$ after $t$ minutes. Find the time taken for the liquid to reach $35^\circ\text{C}$, and state the practical meaning of the horizontal asymptote.
Solution:
Set $T(t) = 35 \implies 20 + 60\left(\frac{1}{2}\right)^{t/10} = 35$ [M1]
Excess temperature over room environment: $60\left(\frac{1}{2}\right)^{t/10} = 15 \implies \left(\frac{1}{2}\right)^{t/10} = \frac{15}{60} = \frac{1}{4}$ [M1]
Since $\frac{1}{4} = \left(\frac{1}{2}\right)^2$, we have $\frac{t}{10} = 2 \implies t = 20\text{ minutes}$. [A1]
Asymptote: As $t \to \infty$, $\left(\frac{1}{2}\right)^{t/10} \to 0$, giving $T(t) \to 20^\circ\text{C}$. This horizontal asymptote represents the constant ambient room temperature. [R1]
Core Formulas (Topic 2.4 & 2.5)
| Quadratic Axis of Symmetry | $$x = -\frac{b}{2a}$$ |
| Quadratic Vertex Form | $$f(x) = a(x - h)^2 + k \quad \implies \text{Vertex at } (h, k)$$ |
| Exponential Model Form | $$f(x) = ka^x + c \quad \text{with asymptote } y = c$$ |
| Doubling Time / Half-Life | $$T_d = \frac{\ln(2)}{\ln(a)}, \qquad T_{1/2} = \frac{\ln(0.5)}{\ln(a)}$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Confusing vertex $x$ with vertex $y$: The question "find the maximum height" asks for $y$ ($k$), whereas "find when maximum height is reached" asks for $t$ ($h$).
- Trap 2: Forgetting the horizontal asymptote in $ka^x + c$: The asymptote is $y = c$, not $y = 0$. In cooling models, $T \to \text{room temp}$, not $0^\circ\text{C}$.
- Trap 3: Sign error in $a(x - h)^2$: If $f(x) = 3(x + 5)^2 + 4$, the vertex is $(-5, 4)$, not $(+5, 4)$.
- Trap 4: Discarding context-negative roots without explanation: When finding flight time $h(t) = 0$, explicitly state "reject negative root $t < 0$ as $t$ represents elapsed time".
- Trap 5: Using degrees instead of logarithms on GDC: Ensure log base or natural log calculations are evaluated correctly.
5-Question Concept Check
Score: - / 51. What is the $x$-coordinate of the vertex of $f(x) = -2x^2 + 12x - 5$?
2. What is the equation of the horizontal asymptote for $f(x) = 50(0.8)^x + 18$?
3. An investment grows by $6\%$ compound interest per year. What is the base $a$ in $V(t) = V_0 a^t$?
4. A quadratic curve has vertex at $(h, k)$ where $a < 0$. Does $(h, k)$ represent a local maximum or minimum?
5. A radioactive substance has a half-life of 10 days. If the initial mass is 80 grams, what mass remains after 30 days?