Defining and Differentiating Parametric Equations - AP Calculus BC Study Guide

The Core Idea: Defining and Differentiating Parametric Equations

Parametric equations describe the position of a point $(x,y)$ in a plane using a third variable, called a parameter, which is often denoted by $t$. The coordinates are expressed as functions of this parameter, $x=x(t)$ and $y=y(t)$. While $y$ is not given as an explicit function of $x$, we can still analyze the properties of the curve, such as its slope and concavity.

The fundamental challenge this topic addresses is how to find the rate of change of $y$ with respect to $x$ (the slope, $dy/dx$) and the rate of change of the slope (related to concavity, $d^{2}y/d{x}^2$) when both $x$ and $y$ depend on $t$. This is accomplished by relating the derivatives with respect to $x$ to the derivatives with respect to $t$ using the chain rule. The essential knowledge for this topic provides the specific formulas for calculating these first and second derivatives for parametrically defined curves.

Key Formulas

The derivatives of a curve defined parametrically by $x=x(t)$ and $y=y(t)$ are found using the following rules, which are applications of the chain rule.

1. The First Derivative ($dy/dx$)

The first derivative of $y$ with respect to $x$ gives the slope of the tangent line to the parametric curve. It is calculated by finding the ratio of the rate of change of $y$ with respect to $t$ to the rate of change of $x$ with respect to $t$.

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

This formula is valid provided that $\frac{dx}{dt}\ne 0$.

2. The Second Derivative ($d^{2}y/d{x}^2$)

The second derivative of $y$ with respect to $x$ is used to determine the concavity of the parametric curve. It is found by taking the derivative of the first derivative ($dy/dx$) with respect to $t$, and then dividing that result by $dx/dt$.

$$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$

Understanding the Second Derivative

A critical nuance in this topic is the correct calculation of the second derivative. It is a common mistake to assume that $\frac{d^{2}y}{d{x}^2}$ is simply the ratio of the second derivatives with respect to $t$. This is incorrect.

The formula $\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ arises from the chain rule. Recall that the second derivative is the derivative of the first derivative: $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$.

Since the expression for $\frac{dy}{dx}$ is a function of $t$, we cannot differentiate it directly with respect to $x$. We must apply the same logic used for the first derivative: to differentiate a function of $t$ with respect to $x$, we differentiate it with respect to $t$ and divide by $\frac{dx}{dt}$.

Applying this rule to the "function" $\frac{dy}{dx}$, we get:

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$

This confirms the formula for the second derivative. The process is sequential: first find $\frac{dy}{dx}$ as a function of $t$, then differentiate that function with respect to $t$, and finally, divide by the original $\frac{dx}{dt}$.

Core Concepts & Rules

• Parametric Derivatives: For a curve defined by $x=x(t)$ and $y=y(t)$, derivatives with respect to $x$ are found by using derivatives with respect to the parameter $t$.

• First Derivative (Slope): The slope of the tangent line, $\frac{dy}{dx}$, is the ratio $\frac{dy/dt}{dx/dt}$.

• Condition for Differentiability: The derivative $\frac{dy}{dx}$ exists as long as $\frac{dx}{dt}\ne 0$.

• Horizontal Tangents: A horizontal tangent may occur when $\frac{dy}{dt}=0$ (and $\frac{dx}{dt}\ne 0$).

• Vertical Tangents: A vertical tangent may occur when $\frac{dx}{dt}=0$ (and $\frac{dy}{dt}\ne 0$).

• Second Derivative (Concavity): The second derivative, $\frac{d^{2}y}{d{x}^2}$, determines the concavity of the curve. It is found by differentiating $\frac{dy}{dx}$ with respect to $t$ and then dividing the result by $\frac{dx}{dt}$.

Step-by-Step Example 1: Basic Differentiation

Consider the curve defined by the parametric equations $x(t)=t^3-3t$ and $y(t)=t^2-4$. Find $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$.

Step 1: Find the derivatives of $x$ and $y$ with respect to $t$.

$$\frac{dx}{dt}=\frac{d}{dt}(t^3-3t)=3t^2-3$$

$$\frac{dy}{dt}=\frac{d}{dt}(t^2-4)=2t$$

Step 2: Calculate the first derivative, $\frac{dy}{dx}$.

Using the formula $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$:

$$\frac{dy}{dx}=\frac{2t}{3t^2-3}$$

**Step 3: Calculate the derivative of $\frac{dy}{dx}$ with respect to t`.** This requires the quotient rule. Let $u = 2t and $v=3t^2-3$. Then $u^{\prime }=2$ and $v^{\prime }=6t$.

$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}=\frac{(2)(3t^2-3)-(2t)(6t)}{(3t^2-3{)}^2}$$

$$=\frac{6t^2-6-12t^2}{(3t^2-3{)}^2}=\frac{-6t^2-6}{(3t^2-3{)}^2}$$

Step 4: Calculate the second derivative, $\frac{d^{2}y}{d{x}^2}$.

Using the formula $\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}(dy/dx)}{dx/dt}$:

$$\frac{d^{2}y}{d{x}^2}=\frac{\frac{-6t^2-6}{(3t^2-3{)}^2}}{3t^2-3}=\frac{-6(t^2+1)}{(3(t^2-1){)}^3}=\frac{-6(t^2+1)}{27(t^2-1{)}^3}=\frac{-2(t^2+1)}{9(t^2-1{)}^3}$$

Step-by-Step Example 2: Exam-Style Application

A particle moves in the xy-plane with its position given by the parametric equations $x(t)=5\cos (t)$ and $y(t)=3\sin (t)$ for $0\le t\le 2\pi$.

(a) Find the equation of the tangent line to the curve at $t=\frac{\pi }{6}$.

(b) At what value(s) of t is the curve concave down?

Part (a): Equation of the Tangent Line

Step 1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

$$\frac{dx}{dt}=-5\sin (t)$$

$$\frac{dy}{dt}=3\cos (t)$$

Step 2: Find the slope $\frac{dy}{dx}$ as a function of t`.** Formula[11] **Step 3: Evaluate the slope at $t = \frac{\pi}{6}.

$$m=\frac{dy}{dx}{|}_{t=\pi /6}=-\frac{3}{5}\cot \left(\frac{\pi }{6}\right)=-\frac{3}{5}(\sqrt{3})=-\frac{3\sqrt{3}}{5}$$

Step 4: Find the coordinates of the point at $t=\frac{\pi }{6}$.

$$x\left(\frac{\pi }{6}\right)=5\cos \left(\frac{\pi }{6}\right)=5\left(\frac{\sqrt{3}}{2}\right)=\frac{5\sqrt{3}}{2}$$

$$y\left(\frac{\pi }{6}\right)=3\sin \left(\frac{\pi }{6}\right)=3\left(\frac{1}{2}\right)=\frac{3}{2}$$

The point is $(\frac{5\sqrt{3}}{2},\frac{3}{2})$.

Step 5: Write the equation of the tangent line using point-slope form.

$$y-y_1=m(x-x_1)$$

$$y-\frac{3}{2}=-\frac{3\sqrt{3}}{5}\left(x-\frac{5\sqrt{3}}{2}\right)$$

Part (b): Determine Concavity

Step 1: Find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$.

$$\frac{d}{dt}\left(-\frac{3}{5}\cot (t)\right)=-\frac{3}{5}(-{\csc }^2(t))=\frac{3}{5}{\csc }^2(t)$$

Step 2: Find $\frac{d^{2}y}{d{x}^2}$.

$$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}(dy/dx)}{dx/dt}=\frac{\frac{3}{5}{\csc }^2(t)}{-5\sin (t)}=\frac{3}{5{\sin }^2(t)}\cdot \frac{1}{-5\sin (t)}=-\frac{3}{25{\sin }^3(t)}$$

Step 3: Determine when the curve is concave down.

The curve is concave down when $\frac{d^{2}y}{d{x}^2}<0$.

$$-\frac{3}{25{\sin }^3(t)}<0$$

This inequality is true when the denominator is positive:

$$25{\sin }^3(t)>0⟹\sin (t)>0$$

For the interval $0\le t\le 2\pi$, $\sin (t)$ is positive when $0<t<\pi$.

Therefore, the curve is concave down for $t\in (0,\pi )$.

Using Your Calculator

The formulas for parametric derivatives are analytical and must be known. A calculator cannot derive these formulas for you. However, a graphing calculator is highly efficient for evaluating the first or second derivative at a specific numerical value of t, which is a common task on the exam.

Problem: Find the value of $\frac{d^{2}y}{d{x}^2}$ for the curve $x(t)=\ln (t)$ and $y(t)=t^3$ at $t=2$.

Analytical Steps (for context):

1. $\frac{dx}{dt}=\frac{1}{t}$ and $\frac{dy}{dt}=3t^2$.

2. $\frac{dy}{dx}=\frac{3t^2}{1/t}=3t^3$.

3. $\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(3t^3)=9t^2$.

4. $\frac{d^{2}y}{d{x}^2}=\frac{9t^2}{1/t}=9t^3$.

5. At $t=2$, $\frac{d^{2}y}{d{x}^2}=9(2{)}^3=72$.

Calculator Steps (TI-84 Style):

It is often easiest to compute the numerator and denominator of the derivative formulas separately.

1. Store the value of t:

   $2STO►X$ (It's common practice to use $X$ for the variable $t$ on the home screen).

2. Calculate $\frac{dy}{dx}$ at $t=2$:

   Use the numerical derivative command nDeriv( (found under MATH -> $8$).

   (nDeriv(X^3, X, 2)) / (nDeriv(ln(X), X, 2))

   This will return $24$, which is the value of $\frac{dy}{dx}$ at $t=2$.

3. Calculate $\frac{d^{2}y}{d{x}^2}$ at $t=2$:

   The formula is $\frac{\frac{d}{dt}(dy/dx)}{dx/dt}$. We need to numerically differentiate the expression for $\frac{dy}{dx}$, which is $3t^3$.

   (nDeriv(3X^3, X, 2)) / (nDeriv(ln(X), X, 2))

   This will return $72$.

AP Exam Quick Hit

Common Question Types

• Finding the slope at a point: Given $x(t)$ and $y(t)$, you will be asked to find the slope of the curve at a specific value of $t$ or at a point $(x,y)$ (which first requires you to solve for $t$).

  • Example: "Find the slope of the tangent line to the curve defined by $x=t^2,y=e^{3t}$ at $t=1$."

• Determining concavity: You will be asked to find the value of the second derivative $\frac{d^{2}y}{d{x}^2}$ at a given $t$ and use its sign to determine if the curve is concave up or concave down.

  • Example: "For the curve defined by $x=\sin (t),y=\cos (t)$, is the curve concave up or concave down at $t=\pi /4$?"

• Finding horizontal or vertical tangents: You will be asked to find the values of $t$ for which the tangent line to the curve is horizontal ($\frac{dy}{dt}=0,\frac{dx}{dt}\ne 0$) or vertical ($\frac{dx}{dt}=0,\frac{dy}{dt}\ne 0$).

  • Example: "Find all values of $t$ for which the curve given by $x=t^3-3t,y=t^2-2$ has a horizontal tangent."

Common Mistakes

• Incorrect Second Derivative Formula: The most frequent error is calculating $\frac{d^{2}y}{d{x}^2}$ as $\frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}$. This is fundamentally incorrect. Always use the formula $\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}(dy/dx)}{dx/dt}$.

• Forgetting to Divide by $\frac{dx}{dt}$ in the Second Derivative: A student might correctly find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ but then forget the final step of dividing this result by $\frac{dx}{dt}$.

• Algebraic Errors in Differentiation: The expression for $\frac{dy}{dx}$ is often a quotient, so finding its derivative with respect to $t$ requires careful application of the quotient rule, which is a common source of algebraic mistakes.

• Mixing up $t$ and $x$: When evaluating a derivative at a point $(x_0,y_0)$, students must first find the value of $t$ that generates this point before plugging it into the derivative formulas. Do not plug $x_0$ into the formulas for $\frac{dy}{dx}$ or $\frac{d^{2}y}{d{x}^2}$.