Finding Arc Lengths of Curves Given by Parametric Equations - AP Calculus BC Study Guide

The Core Idea: Finding Arc Lengths of Curves Given by Parametric Equations

In calculus, we often need to determine the length of a curve in the plane. While finding the length of a simple line segment is straightforward, calculating the length of a curved path requires integration. This topic addresses how to find the arc length of a curve defined by a set of parametric equations, $x=x(t)$ and $y=y(t)$, over a given interval for the parameter $t$.

The fundamental concept is to approximate the curve as a series of infinitesimally small straight line segments. The length of each tiny segment, $ds$, can be found using the Pythagorean theorem with the differential changes in x and y, $dx$ and $dy$. By summing up the lengths of all these microscopic segments using a definite integral, we can find the exact length of the entire curve from a starting point to an ending point, as defined by the interval of the parameter $t$.

Key Formulas

The length of a smooth curve defined by the parametric functions $x=x(t)$ and $y=y(t)$ for $a\le t\le b$ is given by the definite integral:

$$L={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$

Alternatively, using prime notation for the derivatives with respect to $t$:

$$L={\int }_a^b\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}dt$$

• $L$ represents the arc length.

• $x^{\prime }(t)$ and $y^{\prime }(t)$ are the derivatives of the position functions with respect to the parameter $t$.

• $a$ and $b$ are the starting and ending values for the parameter $t$.

Understanding the Formula

The expression inside the integral, $\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$, has a significant physical interpretation. If $x(t)$ and $y(t)$ represent the position of a particle at time $t$, then $x^{\prime }(t)$ is its horizontal velocity and $y^{\prime }(t)$ is its vertical velocity. The expression $\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$ is the magnitude of the velocity vector, which is the speed of the particle at time $t$.

Therefore, the arc length formula is simply the integral of the speed over a time interval. Integrating speed with respect to time gives the total distance traveled. For a curve that is traversed only once without changing direction, the total distance traveled is equal to the arc length. The formula applies to "smooth" curves, which means that $x^{\prime }(t)$ and $y^{\prime }(t)$ are continuous and are not simultaneously zero on the interval of integration.

Core Concepts & Rules

• Arc Length as an Integral: The length of a parametrically defined curve is calculated using a specific definite integral.

• Required Components: To use the formula, you need the parametric equations $x(t)$ and $y(t)$, and the interval for the parameter $t$, from $t=a$ to $t=b$.

• Derivatives are Key: The first step is always to find the derivatives of both parametric functions with respect to the parameter $t$, i.e., $x^{\prime }(t)$ and $y^{\prime }(t)$.

• The Integrand: The function being integrated is the square root of the sum of the squares of the derivatives. This represents the rate of change of arc length with respect to $t$, or the speed.

• Parameter Bounds: The limits of integration, $a$ and $b$, must be the bounds for the parameter $t$, not the x or y coordinates of the curve's endpoints.

Step-by-Step Example 1: Basic Application

Problem: Find the length of the curve defined by $x(t)=t^2$ and $y(t)=\frac{2}{3}{t}^3$ for $0\le t\le \sqrt{3}$.

Step 1: Find the derivatives.

Find the derivatives of $x(t)$ and $y(t)$ with respect to $t$.

$$x^{\prime }(t)=\frac{d}{dt}(t^2)=2t$$

$$y^{\prime }(t)=\frac{d}{dt}\left(\frac{2}{3}{t}^3\right)=2t^2$$

Step 2: Square the derivatives and add them.

Square each derivative and find their sum.

$$(x^{\prime }(t){)}^2=(2t{)}^2=4t^2$$

$$(y^{\prime }(t){)}^2=(2t^2{)}^2=4t^4$$

$$(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2=4t^2+4t^4$$

Step 3: Set up the integrand.

Place the sum from Step 2 under a square root and simplify if possible.

$$\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}=\sqrt{4t^2+4t^4}=\sqrt{4t^2(1+t^2)}$$

Since $t\ge 0$ on the interval, we can simplify this to:

$$2t\sqrt{1+t^2}$$

Step 4: Set up and evaluate the definite integral.

Use the simplified integrand and the given bounds for $t$ to set up and solve the integral.

$$L={\int }_0^{\sqrt{3}}2t\sqrt{1+t^2}dt$$

We can use u-substitution. Let $u=1+t^2$. Then $du=2tdt$.

Change the bounds:

• When $t=0$, $u=1+0^2=1$.

• When $t=\sqrt{3}$, $u=1+(\sqrt{3}{)}^2=1+3=4$.

The integral becomes:

$$L={\int }_1^4\sqrt{u}du={\int }_1^4{u}^{1/2}du$$

$$L={\left[\frac{2}{3}{u}^{3/2}\right]}_1^4=\frac{2}{3}(4^{3/2})-\frac{2}{3}(1^{3/2})$$

$$L=\frac{2}{3}(8)-\frac{2}{3}(1)=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$$

The length of the curve is $\frac{14}{3}$.

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

Problem: A particle moves in the xy-plane so that its position at any time $t$, for $0\le t\le 2$, is given by the parametric equations $x(t)=\ln (t^2+1)$ and $y(t)=e^{-t}\sin (t)$. Find the total distance traveled by the particle.

Step 1: Recognize the task.

"Total distance traveled" for a particle moving along a parametric curve is synonymous with arc length. We must use the arc length formula.

Step 2: Find the derivatives.

This requires the chain rule and product rule.

$$x^{\prime }(t)=\frac{d}{dt}(\ln (t^2+1))=\frac{1}{t^2+1}\cdot (2t)=\frac{2t}{t^2+1}$$

$$y^{\prime }(t)=\frac{d}{dt}(e^{-t}\sin (t))=(e^{-t}{)}^{\prime }\cdot \sin (t)+e^{-t}\cdot (\sin (t){)}^{\prime }$$

$$y^{\prime }(t)=-e^{-t}\sin (t)+e^{-t}\cos (t)$$

Step 3: Set up the definite integral.

The integral for the total distance traveled is:

$$L={\int }_0^2\sqrt{{\left(\frac{2t}{t^2+1}\right)}^2+(-e^{-t}\sin (t)+e^{-t}\cos (t){)}^2}dt$$

Step 4: Evaluate using a calculator.

This integral is not practical to solve by hand. This is a calculator-active problem. We use a calculator's numerical integration function to find the value.

$$L\approx 1.514$$

The total distance traveled by the particle from $t=0$ to $t=2$ is approximately $1.514$.

Using Your Calculator

For many arc length problems on the AP Exam, the resulting integral is too difficult to evaluate by hand. You will use your graphing calculator's numerical integration feature.

To calculate $L={\int }_a^b\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}dt$:

1. Find Derivatives: First, find $x^{\prime }(t)$ and $y^{\prime }(t)$ by hand.

2. Enter Functions:

   • In your calculator's function editor (e.g., Y=), enter the expression for $x^{\prime }(t)$ as $Y_1$.

   • Enter the expression for $y^{\prime }(t)$ as $Y_2$.

   • *Note: Use $X$ as the variable in the calculator, e.g., $Y_1=2X/(X^2+1)‘.\ast 3.\ast \ast UseNumericalIntegration:\ast \ast -Gotothemaincalculationscreen.-Accessthenumericalintegrationfunction(e.g.,‘fnInt(‘onaTI-84or‘MATH>9‘).-Setuptheintegralasfollows:‘fnInt(\surd ((Y_1{)}^2+(Y_2{)}^2),X,a,b)‘-Thiscommandtellsthecalculatortointegratetheexpression$\sqrt{(Y_1)^2 + (Y_2)^2}with respect to the variable $X from the lower bound $a$ to the upper bound $b$.

3. Execute: Press ENTER to get the numerical result.

This method avoids typing long expressions into the integrator and reduces the chance of syntax errors.

AP Exam Quick Hit

Common Question Types

• Calculator-Active Distance Traveled: You will be given parametric equations for the position of a particle and asked to find the total distance it travels over a time interval. This is a direct application of the arc length formula using your calculator's numerical integrator.

  • Example: "A particle's velocity components are given by $x^{\prime }(t)=\cos (t^3)$ and $y^{\prime }(t)=\sin (t^2)$. Find the total distance the particle travels from $t=0$ to $t=1$."

• Set Up, Do Not Evaluate: You will be asked to write an integral expression for the arc length of a curve, but not to solve it. This tests your knowledge of the formula itself.

  • Example: "Write, but do not evaluate, an integral expression that gives the length of the curve defined by $x(t)=\sqrt{t}$ and $y(t)=3t-1$ for $1\le t\le 4$."

• Analytically Solvable Arc Length: On the no-calculator section, you may be given a problem where the expression $(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2$ simplifies to a perfect square, making the integral straightforward to solve by hand.

  • Example: "Find the length of the curve given by $x(t)=\frac{1}{3}{t}^3$ and $y(t)=\frac{1}{2}{t}^2$ for $0\le t\le 1$."

Common Mistakes

• Forgetting the Square Root: A very common error is to integrate $(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2$ without the square root. The formula must include the square root.

• Incorrectly Applying the Square Root: Writing $\int \sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}dt$ as $\int (x^{\prime }+y^{\prime })dt$. The square root of a sum is not the sum of the square roots.

• Using Position Instead of Velocity: Integrating $\sqrt{(x(t){)}^2+(y(t){)}^2}$ instead of $\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$. You must use the derivatives (velocity components), not the original functions (position components).

• Using Incorrect Bounds: The limits of integration $a$ and $b$ must be the values for the parameter $t$. Students sometimes mistakenly use the x- or y-coordinates of the curve's endpoints.

• Derivative Errors: Simple mistakes in calculating $x^{\prime }(t)$ or $y^{\prime }(t)$, especially when the chain rule, product rule, or quotient rule is required. Always double-check your derivatives before setting up the integral.