Question

Finding the Derivative of a Parametric Curve Calculate the derivative \(\frac{d y}{d x}\) for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs. $$ \begin{array}{ll} \text { a. } x(t)=t^{2}-3, & y(t)=2 t-1, \quad-3 \leq t \leq 4 \\ \text { b. } x(t)=2 t+1, & y(t)=t^{3}-3 t+4, \quad-2 \leq t \leq 5 \\ \text { c. } x(t)=5 \cos t, & y(t)=5 \sin t, \quad 0 \leq t \leq 2 \pi \end{array} $$

Short answer

(a) \( \frac{dy}{dx} = \frac{1}{t} \), critical point: \( t=0 \). (b) \( \frac{dy}{dx} = \frac{3t^2 - 3}{2} \), critical points: \( t=\pm 1 \). (c) \( \frac{dy}{dx} = -\cot(t) \), critical points: \( t=0, \pi, 2\pi \).

Step-by-step solution

Differentiate Parametric Equations for (a)

For part (a), we have \( x(t) = t^2 - 3 \) and \( y(t) = 2t - 1 \). First, differentiate \( x(t) \) with respect to \( t \):\[ \frac{dx}{dt} = \frac{d}{dt}(t^2 - 3) = 2t. \] Then, differentiate \( y(t) \) with respect to \( t \):\[ \frac{dy}{dt} = \frac{d}{dt}(2t - 1) = 2. \]

Calculate Derivative \( \frac{dy}{dx} \) for (a)

Using the chain rule, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Substitute the derivatives from Step 1:\[ \frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}. \]

Find Critical Points for (a)

A critical point occurs when \( \frac{dy}{dx} = 0 \) or is undefined. Since \( \frac{1}{t} \) is undefined at \( t = 0 \), check if \( t = 0 \) is within the given interval \(-3 \leq t \leq 4\). Thus, \( t = 0 \) is a critical point.

Differentiate Parametric Equations for (b)

For part (b), \( x(t) = 2t + 1 \) and \( y(t) = t^3 - 3t + 4 \). Differentiate \( x(t) \) with respect to \( t \):\[ \frac{dx}{dt} = \frac{d}{dt}(2t + 1) = 2. \] Differentiate \( y(t) \) with respect to \( t \):\[ \frac{dy}{dt} = \frac{d}{dt}(t^3 - 3t + 4) = 3t^2 - 3. \]

Calculate Derivative \( \frac{dy}{dx} \) for (b)

Substitute the derivatives into the formula for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{3t^2 - 3}{2}. \]

Find Critical Points for (b)

Set \( \frac{dy}{dx} = 0 \) to find critical points:\[ 3t^2 - 3 = 0 \Rightarrow t^2 = 1 \Rightarrow t = \pm 1. \] Check both are within \(-2 \leq t \leq 5\), so \( t = 1 \) and \( t = -1 \) are critical points.

Differentiate Parametric Equations for (c)

For part (c), \( x(t) = 5\cos(t) \) and \( y(t) = 5\sin(t) \). Differentiate \( x(t) \):\[ \frac{dx}{dt} = -5\sin(t). \] Differentiate \( y(t) \):\[ \frac{dy}{dt} = 5\cos(t). \]

Calculate Derivative \( \frac{dy}{dx} \) for (c)

Substitute the derivatives into \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{5\cos(t)}{-5\sin(t)} = -\cot(t). \]

Find Critical Points for (c)

Critical points occur when \( \frac{dy}{dx} = 0 \) or is undefined. Since \( \cot(t) = \frac{\cos(t)}{\sin(t)} \), it is undefined where \( \sin(t) = 0 \), which is at \( t = 0, \pi, 2\pi \) within \( 0 \leq t \leq 2\pi \). These are the critical points.

Key concepts

Critical Points

In the context of parametric differentiation, critical points are significant because they indicate where the graph of a curve may have a local maximum, minimum, or a point of inflection. For a parametrically defined curve, a critical point occurs where the derivative \( \frac{dy}{dx} \) is either zero or undefined.
To find these critical points, we first calculate the derivative \( \frac{dy}{dx} \) using the derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). We then solve for the parameter \( t \) at points where \( \frac{dy}{dx} = 0 \) or is undefined, and verify that these \( t \) values lie within the given interval.
In each of the problems from the exercise, we located critical points by first evaluating the derivative \( \frac{dy}{dx} \) and analyzing where significant changes such as zero-slopes or vertical tangents occur.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. In the context of parametric equations, the chain rule simplifies the process of finding \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \).
Since both \( x \) and \( y \) are given as functions of a third variable, typically \( t \), we use the chain rule by first finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). We then use the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). By treating both derivatives as functions of \( t \), the chain rule effortlessly links the rates of change between the dependent variables, \( x \) and \( y \).
This process demonstrates the elegant power of the chain rule in simplifying seemingly complex relations into manageable pieces.

Derivatives of Trigonometric Functions

In parametric problems involving trigonometric functions, the derivatives play a significant role in exploring the curve's behavior. For example, in part (c) of the exercise, we dealt with the parametric equations defined by trigonometric functions: \( x(t) = 5\cos(t) \) and \( y(t) = 5\sin(t) \).
When differentiating these equations with respect to \( t \), we apply the derivatives of trigonometric functions. The derivative of \( \cos(t) \) is \(-\sin(t)\), and the derivative of \( \sin(t) \) is \( \cos(t) \). These principles simplify the differentiation process and allow us to find \( \frac{dy}{dx} \) by substitution after obtaining \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).
Understanding these derivatives is crucial for examining the nature and properties of curves generated by trigonometric relations.

Parametric Equations

Parametric equations provide a versatile way to describe curves in the plane, especially when the relationship between \( x \) and \( y \) is not straightforward. Each curve is described using a third variable, \( t \), making it easy to analyze more complex scenarios like motion or curved surfaces.
A key benefit of parametric equations is the capability to model interesting geometrical phenomena, such as circles or ellipses. The parametric form allows for separate control over each coordinate, offering greater flexibility in exploring the curve's dynamics.
For instance, part (c) of the exercise utilizes parametric equations of a circle defined by \( x(t) = 5\cos(t) \) and \( y(t) = 5\sin(t) \), creating a path that traces out a circle as \( t \) varies from \( 0 \) to \( 2\pi \). Using parametric forms, students can build a solid understanding of how to harness these equations effectively.