The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY) - AP Calculus BC Study Guide

The Core Idea: The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC ONLY)

This topic introduces a method for calculating the precise length of a smooth curve in a two-dimensional plane using definite integrals. The fundamental concept is that we can approximate the length of a curve by summing the lengths of many small, straight line segments that lie along the curve. By taking the limit as the number of segments approaches infinity (and their individual lengths approach zero), this sum becomes a definite integral, giving us the exact length.

This geometric concept of "arc length" is directly analogous to the physical concept of "distance traveled." If a particle moves along a path described by a set of equations, the total distance it covers is simply the length of the path it traces. Therefore, the same integral formula used to find the length of a curve is also used to find the total distance a particle travels over a given time interval. This topic provides the tools to solve for this length or distance for curves defined either as functions of $x$ or by a set of parametric equations.

Key Formulas

The specific formula for arc length depends on how the curve is defined. Both formulas are applications of a definite integral.

1. Arc Length for a Function $y=f(x)$

The length, $L$, of a smooth, planar curve defined by the function $y=f(x)$ from $x=a$ to $x=b$ is given by the integral:

$$L={\int }_a^b\sqrt{1+(f^{\prime }(x){)}^2}dx$$

2. Arc Length for a Parametrically Defined Curve

The length, $L$, of a smooth, planar curve defined by the parametric equations $x=g(t)$ and $y=h(t)$ from $t=a$ to $t=b$ is given by the integral:

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

This can also be written using Leibniz notation as:

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

3. Distance Traveled by a Particle

The total distance traveled by a particle moving along a curve from time $t=a$ to $t=b$ is equal to the arc length of the curve over that interval. If the particle's position is given by $(x(t),y(t))$, the distance traveled is found using the parametric arc length formula.

Understanding the Connection: Arc Length and Distance Traveled

The Essential Knowledge for this topic establishes a direct and crucial link between a geometric property and a physical one. The formula for the length of a parametrically defined curve, $\int [a,b]\surd ((g^{\prime }(t){)}^2+(h^{\prime }(t){)}^2)dt$, is precisely the same formula used to calculate the total distance a particle travels along that curve.

This is because the expression inside the square root, $\surd ((g^{\prime }(t){)}^2+(h^{\prime }(t){)}^2)$, represents the speed of the particle at time $t$. The derivatives $g^{\prime }(t)=dx/dt$ and $h^{\prime }(t)=dy/dt$ are the components of the particle's velocity vector. The speed is the magnitude of this velocity vector, found using the Pythagorean theorem. The definite integral of speed with respect to time, $\int speeddt$, accumulates the instantaneous rates of travel over the interval $[a,b]$ to yield the total distance covered, which is by definition the length of the path.

Therefore, calculating "arc length" and calculating "total distance traveled" for a parametrically defined path are identical mathematical processes.

Core Concepts & Rules

• To find the length of a curve given by $y=f(x)$ on the interval $[a,b]$, you must first find the derivative $f^{\prime }(x)$, square it, add 1, take the square root of the entire expression, and then integrate that result from $a$ to $b$.

• To find the length of a curve given parametrically by $x=g(t)$ and $y=h(t)$ on the interval $[a,b]$, you must first find the derivatives $g^{\prime }(t)$ and $h^{\prime }(t)$. Then, square each derivative, add the results, take the square root of the sum, and integrate that final expression from $a$ to $b$.

• The calculation for the total distance a particle travels along a curve between two points in time is identical to the calculation for the arc length of its path between those same two points.

Step-by-Step Example 1: Arc Length of a Function $y=f(x)$

Problem: Find the exact length of the curve $y=\frac{1}{3}(x^2+2{)}^{3/2}$ from $x=0$ to $x=3$.

Step 1: Find the derivative $f^{\prime }(x)$

The function is $f(x)=\frac{1}{3}(x^2+2{)}^{3/2}$. We use the chain rule.

$$f^{\prime }(x)=\frac{1}{3}\cdot \frac{3}{2}(x^2+2{)}^{1/2}\cdot (2x)$$

$$f^{\prime }(x)=x\sqrt{x^2+2}$$

Step 2: Square the derivative, $(f^{\prime }(x){)}^2$

$$(f^{\prime }(x){)}^2={\left(x\sqrt{x^2+2}\right)}^2=x^2(x^2+2)=x^4+2x^2$$

Step 3: Set up the integrand $\surd (1+(f^{\prime }(x){)}^2)$

$$1+(f^{\prime }(x){)}^2=1+x^4+2x^2$$

Rearranging the terms gives a perfect square trinomial:

$$x^4+2x^2+1=(x^2+1{)}^2$$

Now, take the square root:

$$\sqrt{1+(f^{\prime }(x){)}^2}=\sqrt{(x^2+1{)}^2}=x^2+1$$

Step 4: Set up and evaluate the definite integral

The arc length $L$ is the integral of the expression from Step 3 over the interval $[0,3]$.

$$L={\int }_0^3(x^2+1)dx$$

$$L={\left[\frac{x^3}{3}+x\right]}_0^3$$

$$L=\left(\frac{3^3}{3}+3\right)-\left(\frac{0^3}{3}+0\right)$$

$$L=\left(\frac{27}{3}+3\right)-0=(9+3)=12$$

The arc length of the curve is 12 units.

Step-by-Step Example 2: Distance Traveled by a Particle (Parametric)

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

Step 1: Find the derivatives $x^{\prime }(t)$ and $y^{\prime }(t)$

We use the product rule for both derivatives.

For $x(t)=e^t\sin (t)$:

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

For $y(t)=e^t\cos (t)$:

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

Step 2: Square each derivative

$$(x^{\prime }(t){)}^2=(e^t(\sin (t)+\cos (t)){)}^2=e^{2t}({\sin }^2(t)+2\sin (t)\cos (t)+{\cos }^2(t))$$

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

Step 3: Add the squared derivatives and simplify

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

Factor out $e^{2t}$:

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

Using the identity $si{n}^2(t)+co{s}^2(t)=1$:

$$=e^{2t}[1+2\sin (t)\cos (t)+1-2\sin (t)\cos (t)]=e^{2t}[2]$$

Step 4: Set up the integrand $\surd ((x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2)$

$$\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}=\sqrt{2e^{2t}}=\sqrt{2}\cdot \sqrt{e^{2t}}=\sqrt{2}{e}^t$$

Step 5: Set up and evaluate the definite integral

The total distance traveled is the integral of the expression from Step 4 over the interval $[0,\pi ]$.

$$L={\int }_0^{\pi }\sqrt{2}{e}^{t}dt$$

$$L=\sqrt{2}{\int }_0^{\pi }{e}^{t}dt=\sqrt{2}{\left[e^t\right]}_0^{\pi }$$

$$L=\sqrt{2}(e^{\pi }-e^0)=\sqrt{2}(e^{\pi }-1)$$

The total distance traveled by the particle is $\surd 2(e^{\pi }-1)$.

Using Your Calculator

For many arc length problems, the resulting integral is difficult or impossible to evaluate by hand. In these cases, a graphing calculator is used to find a numerical approximation.

To find the arc length of $y=f(x)$ from $x=a$ to $x=b$:

1. Calculate the derivative $f^{\prime }(x)$ by hand.

2. On your calculator, use the numerical integration function (e.g., fnInt on a TI-84 or $\int$ on the math menu).

3. Enter the expression for the integral. The syntax will look similar to this:

   fnInt(√(1 + (f'(x))^2), X, a, b)

   For example, to find the length of $y=x^3$ from $x=0$ to $x=2$, you would first find $y^{\prime }=3x^2$. Then you would enter:

   fnInt(√(1 + (3X^2)^2), X, 0, 2)

To find the distance traveled for a parametric curve $(x(t),y(t))$ from $t=a$ to $t=b$:

1. Calculate the derivatives $x^{\prime }(t)$ and $y^{\prime }(t)$ by hand.

2. Use the numerical integration function.

3. Enter the expression for the parametric arc length integral. The syntax will look similar to this:

   fnInt(√((x'(t))^2 + (y'(t))^2), T, a, b)

   For example, if a particle's velocity components are $x^{\prime }(t)=cos(t^2)$ and $y^{\prime }(t)=t$, to find the distance traveled from $t=1$ to $t=3$, you would enter:

   fnInt(√((cos(T^2))^2 + (T)^2), T, 1, 3)

AP Exam Quick Hit

Common Question Types

• Set up, but do not evaluate, an integral: This tests your knowledge of the correct formula.

  • Example: "Write an integral expression that gives the length of the curve $y=\arctan (x)$ from $x=0$ to $x=1$." (Answer: $\int [0,1]\surd (1+(1/(1+x^2){)}^2)dx$)

• Calculator-active distance traveled: This is a very common free-response question part. You are given $x^{\prime }(t)$ and $y^{\prime }(t)$ (or $x(t)$ and $y(t)$) and asked to find the total distance traveled over an interval.

  • Example: "A particle's velocity is given by $v(t)=<1-t^2,2^t>$. Find the total distance the particle travels from $t=0$ to $t=2$."

• By-hand calculation with algebraic simplification: This appears in the multiple-choice section, requiring you to set up the integral and evaluate it after simplifying the integrand, often by finding a hidden perfect square.

  • Example: "Find the length of the curve defined by $x(t)=8t^{3/2}$ and $y(t)=3-6t$ for $1\le t\le 4$."

Common Mistakes

• Forgetting to square the derivative(s): A very common error is to write $\surd (1+f^{\prime }(x))$ or $\surd ((x^{\prime }(t))+(y^{\prime }(t)))$. The derivatives must be squared before they are added.

• Incorrectly simplifying the radical: Students often make the algebraic error of assuming $\surd (A+B)=\surd A+\surd B$. For example, writing $\surd (1+(f^{\prime }(x){)}^2)=1+f^{\prime }(x)$. The square root cannot be distributed over addition.

• Using position instead of velocity for distance traveled: For parametric curves, students might integrate $\surd ((x(t){)}^2+(y(t){)}^2)dt$ instead of the correct $\surd ((x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2)dt$. Remember, distance is the integral of speed, and speed is the magnitude of the velocity vector.

• Mixing up the $y=f(x)$ and parametric formulas: Applying the $\surd (1+...)$ formula to a parametric problem or vice-versa.

• Forgetting the chain rule: When finding $f^{\prime }(x)$ or the parametric derivatives, complex functions require careful application of the chain rule. An incorrect derivative will lead to an incorrect integral setup.