Modelling with Functions

• Parametric Equations define a set of related variables, usually $x$ and $y$, in terms of one or more independent variables called parameters. In physical modelling, the parameter is almost always $t$, representing time, which allows the model to track the position of an object at any specific moment.

• The Horizontal Component is defined as $x = f(t)$, representing the lateral position of an object relative to an origin. This function describes how far left or right an object has moved as time progresses.

• The Vertical Component is defined as $y = g(t)$, representing the height or altitude of the object. Together, these two functions define a coordinate pair $(f(t), g(t))$ that traces a specific path or trajectory in a 2D plane.

• A Model's Domain is the restricted range of the parameter $t$ (e.g., $0 \le t \le T$). This range often corresponds to the duration of a physical event, such as the time a projectile is in the air or the time a vehicle takes to complete one lap of a track.

• The Chain Rule is the logical foundation for differentiating parametric equations. Since $y$ is a function of $t$ and $t$ is implicitly a function of $x$, the gradient of the curve is found by the ratio of their individual rates of change with respect to the parameter.

• The Reciprocal Property of derivatives, $\frac{dt}{dx} = 1 / \frac{dx}{dt}$, allows us to convert a rate of change with respect to time into a rate of change with respect to position. This is critical when we need to find how $y$ changes as $x$ changes, rather than how both change over time.

• Parametric Integration relies on the substitution method. To find the area under a curve, we transform the standard Cartesian integral $\int y \, dx$ into a parametric integral $\int y \frac{dx}{dt} \, dt$, effectively weighting the height $y$ by the horizontal velocity $\frac{dx}{dt}$ over the time interval.

Finding Gradients and Tangents

• Step 1: Differentiate both parametric equations independently to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$. Ensure that any trigonometric or exponential functions are handled using their respective calculus rules.
• Step 2: Combine these results using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. This expression gives the gradient of the curve at any point defined by the parameter $t$.
• Step 3: To find the equation of a tangent at a specific point, substitute the given value of $t$ into the original equations to find $(x_1, y_1)$ and into the gradient formula to find $m$. Then use $y - y_1 = m(x - x_1)$.

Calculating Enclosed Areas

• Step 1: Identify the limits of integration. If the limits are given in terms of $x$, solve the equation $x = f(t)$ to find the corresponding values of $t$.
• Step 2: Set up the integral $\int_{t_1}^{t_2} y(t) \cdot x'(t) \, dt$. It is vital to multiply the function $y(t)$ by the derivative of $x(t)$ before integrating.
• Step 3: Evaluate the integral. If the resulting area is negative, it usually indicates that the direction of integration (from $t_1$ to $t_2$) is opposite to the direction of increasing $x$ on the graph.

• Horizontal vs. Vertical Tangents: A horizontal tangent occurs when the vertical rate of change is zero ($\frac{dy}{dt} = 0$), provided $\frac{dx}{dt} \neq 0$. Conversely, a vertical tangent occurs when the horizontal rate of change is zero ($\frac{dx}{dt} = 0$), provided $\frac{dy}{dt} \neq 0$.

• Parametric vs. Cartesian Integration: In Cartesian integration, you integrate $y$ with respect to $x$. In parametric integration, you must account for the 'stretch' or 'compression' of the $x$-axis by including the $\frac{dx}{dt}$ term inside the integral.

• Check the Parameter Limits: Always verify if the question provides a specific range for $t$. If you are finding an area or a stationary point, ensure your values of $t$ fall within the defined domain of the model.

• The $dx/dt$ Multiplier: The most common mistake in parametric integration is forgetting to multiply $y$ by $\frac{dx}{dt}$. Always write down the general formula $A = \int y \frac{dx}{dt} \, dt$ before substituting specific functions.

• Units and Context: In modelling questions, $x$ and $y$ often have units (like meters). Ensure your final answers for distances or areas include the correct units, and check if the numerical results make physical sense (e.g., a negative time $t$ is usually invalid).

• Normal Gradients: If asked for a normal line, remember that the gradient of the normal is the negative reciprocal of the tangent gradient ($m_{normal} = -1 / m_{tangent}$). This is a frequent requirement in coordinate geometry problems.