Second Derivatives of Parametric Equations - AP Calculus BC Study Guide

The Core Idea: Second Derivatives of Parametric Equations

In the study of parametrically defined curves, the first derivative, $\frac{dy}{dx}$, provides the slope of the tangent line at any point on the curve. Building upon this, the second derivative, $\frac{d^{2}y}{d{x}^2}$, describes the concavity of the curve. It measures how the slope of the tangent line is changing with respect to $x$. However, since the curve is defined by a parameter $t$, we cannot simply differentiate a second time with respect to $x$ directly.

The core idea of this topic is to establish a method for finding $\frac{d^{2}y}{d{x}^2}$ by leveraging the chain rule in the context of the parameter $t$. The process involves finding the rate of change of the first derivative ($\frac{dy}{dx}$) with respect to $t$, and then relating this back to the rate of change with respect to $x$. This is accomplished by dividing the derivative of the first derivative (with respect to $t$) by $\frac{dx}{dt}$.

Key Formulas

The calculation of the second derivative of a parametrically defined function relies on a two-step process, using the formula for the first derivative as a starting point.

1. First Derivative: The first derivative of $y$ with respect to $x$ is given by the ratio of the derivatives of $y$ and $x$ with respect to $t$.

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

2. Second Derivative: The second derivative of $y$ with respect to $x$ is found by taking the derivative of the first derivative ($\frac{dy}{dx}$) with respect to $t$, and then dividing that result by $\frac{dx}{dt}$.

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

Understanding the Process

The formula for the second derivative is a direct consequence of the chain rule. A common point of confusion is the structure of the numerator, $\frac{d}{dt}\left(\frac{dy}{dx}\right)$. This step requires careful execution.

First, you must compute the first derivative, $\frac{dy}{dx}$, which will result in an expression that is a function of $t$. Let's call this expression $f(t)$. So, $\frac{dy}{dx}=f(t)$.

The numerator of the second derivative formula is then the derivative of this new function $f(t)$ with respect to $t$, which is $f^{\prime }(t)$. This often requires using the quotient rule or other differentiation rules.

Finally, you must divide this result, $f^{\prime }(t)$, by the original $\frac{dx}{dt}$ that was calculated in the first step. It is critical to remember this final division. The second derivative is not simply the derivative of the first derivative with respect to $t$. It must be scaled by dividing by $\frac{dx}{dt}$ to properly account for the change with respect to $x$.

Core Concepts & Rules

• The second derivative of a parametrically defined curve, $\frac{d^{2}y}{d{x}^2}$, is used to determine the concavity of the curve in the $xy$-plane.

• The calculation of $\frac{d^{2}y}{d{x}^2}$ is a multi-step process that begins with finding the first derivative, $\frac{dy}{dx}$.

• The formula for the second derivative is $\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$.

• The numerator of the formula, $\frac{d}{dt}\left(\frac{dy}{dx}\right)$, requires you to differentiate the expression for the first derivative with respect to the parameter $t$.

• The denominator of the formula is $\frac{dx}{dt}$, the same expression used to calculate the first derivative.

• A positive value for $\frac{d^{2}y}{d{x}^2}$ at a point indicates the curve is concave up at that point. A negative value indicates the curve is concave down.

Step-by-Step Example 1: Basic Application

Problem: A curve is defined by the parametric equations $x(t)=t^2+1$ and $y(t)=t^3-4t$. Find $\frac{d^{2}y}{d{x}^2}$ and determine the concavity of the curve at $t=1$.

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

$$\frac{dx}{dt}=\frac{d}{dt}(t^2+1)=2t$$

$$\frac{dy}{dt}=\frac{d}{dt}(t^3-4t)=3t^2-4$$

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

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^2-4}{2t}$$

Step 3: Find the derivative of $\frac{dy}{dx}$ with respect to $t$.

We need to calculate $\frac{d}{dt}\left(\frac{3t^2-4}{2t}\right)$. We use the quotient rule:

$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{(6t)(2t)-(3t^2-4)(2)}{(2t{)}^2}$$

$$=\frac{12t^2-6t^2+8}{4t^2}$$

$$=\frac{6t^2+8}{4t^2}=\frac{3t^2+4}{2t^2}$$

Step 4: Apply the second derivative formula.

$$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{\frac{3t^2+4}{2t^2}}{2t}$$

$$\frac{d^{2}y}{d{x}^2}=\frac{3t^2+4}{4t^3}$$

Step 5: Evaluate at $t=1$ and determine concavity.

$${\frac{d^{2}y}{d{x}^2}|}_{t=1}=\frac{3(1{)}^2+4}{4(1{)}^3}=\frac{7}{4}$$

Since $\frac{d^{2}y}{d{x}^2}=\frac{7}{4}>0$, the curve is concave up at the point corresponding to $t=1$.

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

Problem: A particle moves in the $xy$-plane with position given by $x(t)=e^{2t}$ and $y(t)=\sin (t)$ for $t\ge 0$. Find an expression for $\frac{d^{2}y}{d{x}^2}$ in terms of $t$.

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

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

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

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

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\cos (t)}{2e^{2t}}$$

Step 3: Find the derivative of $\frac{dy}{dx}$ with respect to $t$.

We must differentiate $\frac{\cos (t)}{2e^{2t}}$ with respect to $t$. Using the quotient rule:

$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{(-\sin (t))(2e^{2t})-(\cos (t))(4e^{2t})}{(2e^{2t}{)}^2}$$

Factor out $-2e^{2t}$ from the numerator:

$$=\frac{-2e^{2t}(\sin (t)+2\cos (t))}{4e^{4t}}$$

Simplify the expression:

$$=\frac{-(\sin (t)+2\cos (t))}{2e^{2t}}$$

Step 4: Apply the second derivative formula.

Now, we divide the result from Step 3 by the original $\frac{dx}{dt}$ from Step 1.

$$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{\frac{-(\sin (t)+2\cos (t))}{2e^{2t}}}{2e^{2t}}$$

$$\frac{d^{2}y}{d{x}^2}=-\frac{\sin (t)+2\cos (t)}{4e^{4t}}$$

This is the final expression for the second derivative in terms of $t$.

Using Your Calculator

This topic is primarily tested analytically, and you are expected to compute the second derivative by hand. A calculator cannot compute $\frac{d^{2}y}{d{x}^2}$ for a parametric curve with a single built-in function.

However, a calculator can be useful for finding the numerical value of the second derivative at a specific point, which can help check an analytical answer. This is an advanced, multi-step process.

To find the value of $\frac{d^{2}y}{d{x}^2}$ at $t=C$:

1. Enter x(t) into Y1 and $y(t)$ into Y2.

2. The first derivative is $\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$. You can define this in Y3:

   Y3 = nDeriv(Y2, T, T) / nDeriv(Y1, T, T)

   (Note: nDeriv is often found in the MATH menu. The syntax is typically nDeriv(expression, variable, value)).

3. The second derivative is $\frac{d/dt(dy/dx)}{dx/dt}$. To calculate this at $t=C$, you would compute:

   nDeriv(Y3, T, C) / nDeriv(Y1, T, C)

   This command calculates the numerical derivative of your first derivative expression (Y3) at $t=C$ and divides it by the numerical derivative of your $x(t)$ expression (Y1) at $t=C$.

This process is complex and prone to entry errors, reinforcing that the primary skill being tested is the analytical application of the formula.

AP Exam Quick Hit

Common Question Types

• Find the Expression: Given $x(t)$ and $y(t)$, find a simplified expression for $\frac{d^{2}y}{d{x}^2}$ in terms of $t$. This is a direct test of the formula and your differentiation skills (quotient rule, chain rule, etc.).

  • Example: For the curve given by $x(t)=\ln (t)$ and $y(t)=4t^2$, find $\frac{d^{2}y}{d{x}^2}$.

• Determine Concavity at a Point: Given $x(t)$ and $y(t)$, find the value of $\frac{d^{2}y}{d{x}^2}$ at a specific value of $t$ and state whether the curve is concave up or concave down at that point.

  • Example: Is the curve given by $x(t)=2\cos (t)$, $y(t)=\sin (t)$ concave up or concave down at $t=\pi /4$?

• Use Given Derivative Values: You are given the values of $\frac{dx}{dt}$, $\frac{dy}{dt}$, and $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ at a specific time $t$ and asked to find $\frac{d^{2}y}{d{x}^2}$. This is a pure test of knowing the formula's structure.

  • Example: At $t=3$, a particle's motion is described by $\frac{dx}{dt}=-2$ and $\frac{dy}{dx}=4t-1$. Find the value of $\frac{d^{2}y}{d{x}^2}$ at $t=3$.

Common Mistakes

• The Incorrect Formula: The most frequent mistake is to calculate $\frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}$. This is fundamentally incorrect and will always result in a wrong answer.

• Forgetting the Final Division: Students correctly calculate $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ but then forget to divide this result by $\frac{dx}{dt}$.

• Quotient Rule Errors: The expression for $\frac{dy}{dx}$ is often a quotient. Students frequently make algebraic or sign errors when applying the quotient rule to find its derivative with respect to $t$.

• Using the Wrong Denominator: Accidentally dividing by $t$, $dt$, or some other term instead of the correct $\frac{dx}{dt}$. Remember that the denominator in the second derivative formula is the same $\frac{dx}{dt}$ used for the first derivative.