10.3: Calculus of Parametric Curves

Learning Objectives

• Determine derivatives and equations of tangents for parametric curves.
• Find the area under a parametric curve.
• Use the equation for the arc length of a parametric curve.
• Apply the formula for surface area to a volume generated by a parametric curve.

Now that we have introduced the concept of a parametrized curve, our next step is to learn how to work with this concept in the context of Calculus. For example, if we know a parametrization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve?

Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher's hand. If the position of the baseball is represented by the plane curve \((x(t),y(t))\), then we should be able to use Calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate how far that ball has traveled as a function of time.

Derivatives of Parametric Equations

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations\[\begin{align} x(t) & = 2t+3 \label{eq1} \\[6pt]
y(t) & = 3t−4 \label{eq2} \\[6pt]
\end{align} \]within \(−2 \leq t \leq 3\).

The graph of this curve appears in Figure \(\PageIndex{1}\). It is a line segment starting at \((−1,−10)\) and ending at \((9,5)\).

Figure \(\PageIndex{1}\): Graph of the line segment described by the given parametric equations.

We can eliminate the parameter by first solving Equation \ref{eq1} for \(t\):\[ \begin{array}{crcl}
& x(t) & = & 2t+3 \\[6pt]
\implies & x−3 & = & 2t \\[6pt]
\implies & t & = & \dfrac{x−3}{2}. \\[6pt]
\end{array} \nonumber \]Substituting this into \(y(t)\) (Equation \ref{eq2}), we obtain\[ \begin{array}{crcl}
& y(t) & = & 3t−4 \\[6pt]
\implies & y & = & 3\left(\dfrac{x−3}{2}\right)−4 \\[6pt]
\implies & y & = & \dfrac{3x}{2}−\dfrac{9}{2}−4 \\[6pt]
\implies & y & = & \dfrac{3x}{2}−\dfrac{17}{2}. \\[6pt]
\end{array} \nonumber \]The slope of this line is given by \(\frac{dy}{dx}=\frac{3}{2}\). Next we calculate \(x^{\prime}(t)\) and \(y^{\prime}(t)\). This gives \(x^{\prime}(t)=2\) and \(y^{\prime}(t)=3\). Notice that\[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3}{2}. \nonumber \]This is no coincidence, as outlined in the following theorem.

Theorem: Derivative of Parametric Equations

Consider the plane curve defined by the parametric equations \(x=x(t)\) and \(y=y(t)\). Suppose that \(x^{\prime}(t)\) and \(y^{\prime}(t)\) exist, and assume that \(x^{\prime}(t) \neq 0\). Then the derivative \(\frac{dy}{dx}\) is given by\[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{y^{\prime}(t)}{x^{\prime}(t)}. \label{paraD} \]

Proof

This theorem can be proven using the Chain Rule. In particular, assume that the parameter \(t\) can be eliminated, yielding a differentiable function \(y=F(x)\). Then \(y(t)=F(x(t))\). Differentiating both sides of this equation using the Chain Rule yields\[y^{\prime}(t) = F^{\prime} \left( x(t) \right) x^{\prime}(t), \nonumber \]so\[F^{\prime} \left( x(t) \right) =\dfrac{y^{\prime}(t)}{x^{\prime}(t)}. \nonumber \]But \(F^{\prime} \left( x(t) \right) = \frac{dy}{dx}\), which proves the theorem.

Equation \ref{paraD} can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function \(y=f(x)\) is any point \(x=x_0\) such that either \(f^{\prime}(x_0)=0\) or \(f^{\prime}(x_0)\) does not exist. Equation \ref{paraD} gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function \(y=f(x)\) or not.

Example \(\PageIndex{1}\): Finding the Derivative of a Parametric Curve

Calculate the derivative \(\frac{dy}{dx}\) for each of the following parametrically-defined plane curves, and locate any critical points on their respective graphs.

1. \(x(t)=t^2−3, \quad y(t)=2t−1, \quad\text{for }−3 \leq t \leq 4\)
2. \(x(t)=2t+1, \quad y(t)=t^3−3t+4, \quad\text{for }−2 \leq t \leq 2\)
3. \(x(t)=5\cos t, \quad y(t)=5\sin t, \quad\text{for }0 \leq t \leq 2 \pi \)

Solutions

1. To apply Equation \ref{paraD}, first calculate \(x^{\prime}(t)\) and \(y^{\prime}(t)\):\[\begin{array}{rcl}
   x^{\prime}(t) & = & 2t \\[6pt]
   y^{\prime}(t) & = & 2. \\[6pt]
   \end{array} \nonumber \]Next substitute these into the equation:\[ \begin{array}{crcl}
   & \dfrac{dy}{dx} & = & \dfrac{dy/dt}{dx/dt} \\[6pt]
   \implies & \dfrac{dy}{dx} & = & \dfrac{2}{2t} \\[6pt]
   \implies & \dfrac{dy}{dx} & = & \dfrac{1}{t}. \\[6pt]
   \end{array} \nonumber \]This derivative is undefined when \(t=0\). Calculating \(x(0)\) and \(y(0)\) gives \(x(0)=(0)^2−3=−3\) and \(y(0)=2(0)−1=−1\), which corresponds to the point \((−3,−1)\) on the graph. The graph of this curve is a parabola opening to the right, and the point \((−3,−1)\) is its vertex, as shown.

   
   Figure \(\PageIndex{2}\): Graph of the parabola described by parametric equations in part a.

2. To apply Equation \ref{paraD}, first calculate \(x^{\prime}(t)\) and \(y^{\prime}(t)\):\[ \begin{array}{rcl}
   x^{\prime}(t) & = & 2 \\[6pt]
   y^{\prime}(t) & = & 3t^2−3. \\[6pt]
   \end{array} \nonumber \]Next substitute these into the equation:\[ \begin{array}{rcl}
   \dfrac{dy}{dx} & = & \dfrac{dy/dt}{dx/dt} \\[6pt]
   \dfrac{dy}{dx} & = & \dfrac{3t^2−3}{2}. \\[6pt]
   \end{array} \nonumber \]This derivative is zero when \(t= \pm 1\). When \(t=−1\) we have\[x(−1)=2(−1)+1=−1 \text{ and } y(−1)=(−1)^3−3(−1)+4=−1+3+4=6, \nonumber \]which corresponds to the point \((−1,6)\) on the graph. When \(t=1\) we have\[ x(1)=2(1)+1=3 \text{ and } y(1)=(1)^3−3(1)+4=1−3+4=2, \nonumber \]which corresponds to the point \((3,2)\) on the graph. The point \((3,2)\) is a relative minimum, and the point \((−1,6)\) is a relative maximum, as seen in the following graph.

   
   Figure \(\PageIndex{3}\): Graph of the curve described by parametric equations in part b.

3. To apply Equation \ref{paraD}, first calculate \(x^{\prime}(t)\) and \(y^{\prime}(t)\):\[ \begin{array}{rcl}
   x^{\prime}(t) & = & −5\sin t \\[6pt]
   y^{\prime}(t) & = & 5\cos t. \\[6pt]
   \end{array} \nonumber \]Next substitute these into the equation:\[ \begin{array}{crcl}
   & \dfrac{dy}{dx} & = & \dfrac{dy/dt}{dx/dt} \\[6pt]
   \implies & \dfrac{dy}{dx} & = & \dfrac{5\cos t}{−5\sin t} \\[6pt]
   \implies & \dfrac{dy}{dx} & = & −\cot t. \\[6pt]
   \end{array} \nonumber \]This derivative is zero when \(\cos t=0\) and is undefined when \(\sin t=0\). This gives \(t=0,\frac{ \pi }{2}, \pi ,\frac{3 \pi }{2}\), and \(2 \pi \) as critical points for \( t \). Substituting each of these into \(x(t)\) and \(y(t)\), we obtain

   \(t\)	\(x(t)\)	\(y(t)\)
   0	5	0
   \(\frac{ \pi }{2}\)	0	5
   \( \pi \)	−5	0
   \(\frac{3 \pi }{2}\)	0	−5
   \(2 \pi \)	5	0

   These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure \(\PageIndex{4}\)). The derivative is undefined on the left and right edges of the circle. On the top and bottom, the derivative equals zero.

   
   Figure \(\PageIndex{4}\): Graph of the curve described by parametric equations in part c.

Checkpoint \(\PageIndex{1}\)

Calculate the derivative \(dy/dx\) for the plane curve defined by the equations\[x(t)=t^2−4t, \quad y(t)=2t^3−6t, \quad\text{for }−2 \leq t \leq 3 \nonumber \]and locate any critical points on its graph.

Answer

\(x^{\prime}(t)=2t−4\) and \(y^{\prime}(t)=6t^2−6\), so \(\frac{dy}{dx}=\frac{6t^2−6}{2t−4}=\frac{3t^2−3}{t−2}\). This expression is undefined when \(t=2\) and equal to zero when \(t= \pm 1\).

Example \(\PageIndex{2}\): Finding a Tangent Line

Find the equation of the tangent line to the curve defined by the equations\[x(t)=t^2−3, \quad y(t)=2t−1, \quad\text{for }−3 \leq t \leq 4 \nonumber \]when \(t=2\).

Solution

First find the slope of the tangent line using Equation \ref{paraD}, which means calculating \(x^{\prime}(t)\) and \(y^{\prime}(t)\):\[ \begin{array}{rcl}
x^{\prime}(t) & = & 2t \\[6pt]
y^{\prime}(t) & = & 2. \\[6pt]
\end{array} \nonumber \]Next substitute these into the equation:\[ \begin{array}{crcl}
& \dfrac{dy}{dx} & = & \dfrac{dy/dt}{dx/dt} \\[6pt]
\implies & \dfrac{dy}{dx} & = & \dfrac{2}{2t} \\[6pt]
\implies & \dfrac{dy}{dx} & = & \dfrac{1}{t}. \\[6pt]
\end{array} \nonumber \]When \(t=2, \frac{dy}{dx}=\frac{1}{2}\), so this is the slope of the tangent line. Calculating \(x(2)\) and \(y(2)\) gives\[x(2)=(2)^2−3=1 \text{ and } y(2)=2(2)−1=3, \nonumber \]which corresponds to the point \((1,3)\) on the graph (Figure \(\PageIndex{5}\)). Now use the point-slope form of the equation of a line to find the equation of the tangent line:\[ \begin{array}{crcl}
& y−y_0 & = & m(x−x_0) \\[6pt]
\implies & y−3 & = & \dfrac{1}{2}(x−1) \\[6pt]
\implies & y−3 & = & \dfrac{1}{2}x−\dfrac{1}{2} \\[6pt]
\implies & y & = & \dfrac{1}{2}x+\dfrac{5}{2}. \\[6pt]
\end{array} \nonumber \]

Figure \(\PageIndex{5}\): Tangent line to the parabola described by the given parametric equations when \(t=2\).

Checkpoint \(\PageIndex{2}\)

Find the equation of the tangent line to the curve defined by the equations\[x(t)=t^2−4t, \quad y(t)=2t^3−6t, \quad \text{for }−2 \leq t \leq 6 \nonumber \]when \(t=5\).

Answer

The equation of the tangent line is \(y=24x+100\).

Second-Order Derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function \(y=f(x)\) is defined to be the derivative of the first derivative; that is,\[\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left[\dfrac{dy}{dx}\right]. \label{eqD2} \]Since\[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}, \nonumber \]we can replace the \(y\) on both sides of Equation \ref{eqD2} with \(\frac{dy}{dx}\). This gives us\[\dfrac{d^2y}{dx^2}=\dfrac{d}{dx} \left(\dfrac{dy}{dx} \right)=\dfrac{(d/dt)(dy/dx)}{dx/dt}.\label{paraD2} \]If we know \(dy/dx\) as a function of \(t\), then this formula is straightforward to apply

Example \(\PageIndex{3}\): Finding a Second Derivative

Calculate the second derivative \(d^2y/dx^2\) for the plane curve defined by the parametric equations \(x(t)=t^2−3, \quad y(t)=2t−1, \quad\text{for }−3 \leq t \leq 4\).

Solution

From Example \(\PageIndex{1}\) we know that \(\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}\). Using Equation \ref{paraD2}, we obtain\[ \dfrac{d^2y}{dx^2}=\dfrac{(d/dt)(dy/dx)}{dx/dt}=\dfrac{(d/dt)(1/t)}{2t}=\dfrac{−t^{−2}}{2t}=−\dfrac{1}{2t^3}. \nonumber \]

Checkpoint \(\PageIndex{3}\)

Calculate the second derivative \(d^2y/dx^2\) for the plane curve defined by the equations\[ x(t)=t^2−4t, \quad y(t)=2t^3−6t, \quad\text{for }−2 \leq t \leq 3 \nonumber \]and locate any critical points on its graph.

Answer

\(\frac{d^2y}{dx^2}=\frac{3t^2−12t+3}{2(t−2)^3}\). Critical points \((5,4),\, (−3,−4)\),and \((−4,6)\).

Integrals Involving Parametric Equations

Now that we have seen how to calculate the derivative of a plane curve, we naturally ask ourselves how to find the area under a curve defined parametrically. Recall the cycloid defined by these parametric equations\[ \begin{array}{rcl}
x(t) & = & t−\sin t \\[6pt]
y(t) & = & 1−\cos t. \\[6pt]
\end{array}\]Suppose we want to find the area of the shaded region in the following graph.

Figure \(\PageIndex{6}\): Graph of a cycloid with the arch over \([0,2 \pi ]\) highlighted.

To derive a formula for the area under the curve defined by the functions\[\begin{array}{rcl}
x & = & x(t) \\[6pt]
y & = & y(t) \\[6pt]
\end{array}\]where \(a \leq t \leq b\), we assume that \(x(t)\) is differentiable and start with an equal partition of the interval \(a \leq t \leq b\). Suppose \(t_0=a<t_1<t_2< \cdots <t_n=b\) and consider the following graph.

Figure \(\PageIndex{7}\): Approximating the area under a parametrically defined curve.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is \(y(x(\bar{t_i}))\) for some value \(\bar{t_i}\) in the \(i^\text{th}\) subinterval, and the width can be calculated as \(x(t_i)−x(t_{i−1})\). Thus the area of the \(i^\text{th}\) rectangle is given by\[A_i=y(x(\bar{t_i}))(x(t_i)−x(t_{i−1})). \nonumber \]Then a Riemann sum for the area is\[A_n=\sum_{i=1}^ny(x(\bar{t_i}))(x(t_i)−x(t_{i−1})).\nonumber \]Multiplying and dividing each area by \(t_i−t_{i−1}\) gives\[ \begin{array}{rcl}
A_n & = & \displaystyle \sum_{i=1}^ny(x(\bar{t_i})) \left(\dfrac{x(t_i)−x(t_{i−1})}{t_i−t_{i−1}}\right)(t_i−t_{i−1}) \\[6pt]
& = & \displaystyle \sum_{i=1}^ny(x(\bar{t_i})) \left(\dfrac{x(t_i)−x(t_{i−1})}{ \Delta t}\right) \Delta t. \\[6pt]
\end{array} \nonumber \]Taking the limit as \(n\) approaches infinity gives\[A=\lim_{n \to \infty }A_n= \int ^b_ay(t)x^{\prime}(t)\,dt. \nonumber \]This leads to the following theorem.

Theorem: Area under a Parametric Curve

Consider the non-self-intersecting plane curve defined by the parametric equations\[x=x(t),\quad y=y(t),\quad \text{for }a \leq x \leq b \nonumber \]and assume that \(x(t)\) is differentiable. The area under this curve is given by\[A= \int^{t_b}_{t_a} y(t)x^{\prime}(t)\,dt,\label{ParaArea} \]where \( t_a \) is the time-value associated with \( x = a \) and \( t_b \) is the time-value associated with \( x=b \).

If \( x(t) \) is invertible, then \( t_a = x^{-1}(a) \) and \( t_b = x^{-1}(b) \).

As with our previous work, we require \( \Delta x \) from our partition of the interval \( [a,b] \) to be positive. That is, when building the approximating rectangles from the Riemann sums, we get \( x_0 = a \), \( x_1 = a + i \Delta x \), \( \ldots \), \( x_n = b \). These values imply the computed width of each rectangle is positive. This creates a need to be careful when computing areas under parametric curves. It is often the case that \( a \) is not associated with the smallest value of \( t \) and \( b \) is not associated with the largest value of \( t \). Therefore, it is best to initially think of the area integral for parametric curves as going from \( x = a \) to \( x=b \).

For example, when finding the area under the parametric curve defined by\[ x(t) = e^{-t}, \quad y(t) = t^2 + 3t + 4, \nonumber \]where \( 0 \leq t \leq 1 \), we start by thinking in terms of how we normally find areas under curves:\[ A = \int f(x) \, dx. \nonumber \]Of course, this is an indefinite integral, but it is a placeholder for our thought-process. We know that \( f(x) \) is a representation of the \( y \)-values for the curve and, in this case, \( y(t) = t^2 + 3t + 4 \). Moreover, \( dx = x^{\prime}(t) \, dt = -e^{-t} \, dt \). Thus, our "placeholder" integral becomes\[ A = \int f(x) \, dx = \int y(t) x^{\prime}(t) \, dt = \int (t^2 + 3t + 4) (- e^{-t} ) \, dt = -\int (t^2 + 3t + 4) e^{-t} \, dt. \nonumber \]The only thing we have left to consider are the limits of integration.

Assuming the lower limit is \( t = 0 \) and the upper is \( t = 1 \) would be incorrect. Why? Since \( x(0) = 1 \gt e^{-1} = x(1) \), our geometric derivation of the Riemann sum (from Calculus I) tells us that the lower limit is the time value corresponding to \( x = e^{-1} \) and the upper limit is the time value corresponding to \( x = 1 \). Thus, the true integral representing the area under the curve from \( t = 0 \) to \( t = 1 \) is\[ -\int_{t = 1}^{t = 0} (t^2 + 3t + 4) e^{-t} \, dt. \nonumber \]The following Cautionary statement summarizes our concerns.

Caution: Areas Are Computed from \( x_{\text{low}} \) to \( x_{\text{high}} \)

When computing the area under a curve (parametric or not), we integrate from \( x = a \) to \( x = b \), where \( a \leq x \leq b \). Any other method will result in an incorrect area.

Example \(\PageIndex{4}\): Finding the Area under a Parametric Curve

Find the area under the curve of the cycloid defined by the equations\[x(t)=t−\sin t, \quad y(t)=1−\cos t, \quad \text{for }0 \leq t \leq 2 \pi . \nonumber \]

Solution

We start by making sure we are integrating in the proper direction.\[ x(0) = 0 \quad \text{and} \quad x(2\pi) = 2\pi. \nonumber \]This seems convincing that our time-values are "flowing" from a lower \( x \)-value to a higher one, but just to be sure the parametric curve doesn't "double-back" on itself, let's compute \( x^{\prime}(t) \) to see if it is always increasing (we have to compute \( x^{\prime} \) in any case, so why not get it out of the way right now?).\[ x^{\prime}(t) = 1 - \cos t, \nonumber \]which is non-negative for all \( t \). Thus, our \( x \)-values are never decreasing with respect to \( t \). Using Equation \ref{ParaArea}, we have\[\begin{array}{rcl}
A & = & \displaystyle \int ^{t_b}_{t_a} y(t)x^{\prime}(t)\,dt \\[6pt]
& = & \displaystyle \int ^{2 \pi }_0(1−\cos t)(1−\cos t)\,dt \\[6pt]
& = & \displaystyle \int ^{2 \pi }_0(1−2\cos t+\cos^2t)\,dt \\[6pt]
& = & \displaystyle \int ^{2 \pi }_0\left(1−2\cos t+\dfrac{1+\cos (2t)}{2}\right)\,dt \\[6pt]
& = & \displaystyle \int ^{2 \pi }_0\left(\dfrac{3}{2}−2\cos t+\dfrac{\cos (2t)}{2}\right)\,dt \\[6pt]
& = & \dfrac{3t}{2}−2\sin t+\dfrac{\sin (2t)}{4}\bigg|^{2 \pi }_0 \\[6pt]
& = & 3 \pi \\[6pt]
\end{array} \nonumber \]

Checkpoint \(\PageIndex{4}\)

Find the area under the curve of the hypocycloid defined by the equations\[x(t)=3\cos t+\cos(3t), \quad y(t)=3\sin t−\sin(3t), \quad \text{for }0 \leq t \leq \pi . \nonumber \]

Answer

\(A=3 \pi \) (Note that this is one of those cases where the limits of integration go from high time-value to low. This is because \(x(t)\) is a decreasing function over the interval \([0, \pi ];\) that is, the curve is traced from right to left.)

Arc Length of a Parametric Curve

Besides finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, the arc length is the same as the distance between the endpoints. If a particle travels from point \(A\) to point \(B\) along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments, as shown in the following graph.

Figure \(\PageIndex{7}\): Approximation of a curve by line segments.

Given a plane curve defined by the functions \(x=x(t)\), \(y=y(t)\), for \(a \leq t \leq b\), we start by partitioning the interval \([a,b]\) into \(n\) equal subintervals: \(t_0=a<t_1<t_2< \cdots <t_n=b\). The width of each subinterval is given by \( \Delta t=(b−a)/n\). We can calculate the length of each line segment:\[ \begin{array}{rcl}
d_1 & = & \sqrt{(x(t_1)−x(t_0))^2+(y(t_1)−y(t_0))^2} \\[6pt]
d_2 & = & \sqrt{(x(t_2)−x(t_1))^2+(y(t_2)−y(t_1))^2} \\[6pt]
\vdots & \vdots & \vdots \\[6pt]
\end{array} \nonumber \]Then add these up. We let \(s\) denote the exact arc length and \(s_n\) denote the approximation by \(n\) line segments:\[s \approx \sum_{k=1}^ns_k=\sum_{k=1}^n\sqrt{(x(t_k)−x(t_{k−1}))^2+(y(t_k)−y(t_{k−1}))^2}. \label{arc5} \]If we assume that \(x(t)\) and \(y(t)\) are differentiable functions of \(t\), then the Mean Value Theorem applies, so in each subinterval \([t_{k−1},t_k]\) there exist \(\hat{t_k}\) and \(\tilde{t_k}\) such that\[ \begin{array}{rclcl}
x(t_k)−x(t_{k−1}) & = & x^{\prime}(\hat{t_k})(t_k−t_{k−1}) & = & x^{\prime}(\hat{t_k})\, \Delta t \\[6pt]
y(t_k)−y(t_{k−1}) & = & y^{\prime}(\tilde{t_k})(t_k−t_{k−1}) & = & y^{\prime}(\tilde{t_k})\, \Delta t. \\[6pt]
\end{array} \nonumber \]Therefore Equation \ref{arc5} becomes\[\begin {array}{rcl}
s & \approx & \displaystyle \sum_{k=1}^ns_k \\[6pt]
& = & \displaystyle \sum_{k=1}^n\sqrt{(x^{\prime}(\hat{t_k}) \Delta t)^2+(y^{\prime}(\tilde{t_k}) \Delta t)^2} \\[6pt]
& = & \displaystyle \sum_{k=1}^n\sqrt{(x^{\prime}(\hat{t_k}))^2( \Delta t)^2+(y^{\prime}(\tilde{t_k}))^2( \Delta t)^2} \\[6pt]
& = & \displaystyle \sum_{k=1}^n\sqrt{(x^{\prime}(\hat{t_k}))^2+(y^{\prime}(\tilde{t_k}))^2} \Delta t. \\[6pt]
\end{array} \nonumber \]This is a Riemann sum that approximates the arc length over a partition of the interval \([a,b]\). If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives\[ \begin{array}{rcl}
s & = & \displaystyle \lim_{n \to \infty }\sum_{k=1}^ns_k \\[6pt]
& = & \displaystyle \lim_{n \to \infty }\sum_{k=1}^n\sqrt{(x^{\prime}(\hat{t_k}))^2+(y^{\prime}(\tilde{t_k}))^2} \Delta t \\[6pt]
& = & \displaystyle \int ^b_a\sqrt{(x^{\prime}(t))^2+(y^{\prime}(t))^2}\,dt . \\[6pt]
\end{array} \nonumber \]When taking the limit, the values of \(\hat{t_k}\) and \(\tilde{t_k}\) are both contained within the same ever-shrinking interval of width \( \Delta t\), so they must converge to the same value.

We can summarize this method in the following theorem.

Theorem: Arc Length of a Parametric Curve

Consider the plane curve defined by the parametric equations\[x=x(t), \quad y=y(t), \quad \text{for }t_1 \leq t \leq t_2 \nonumber \]and assume that \(x(t)\) and \(y(t)\) are differentiable functions of \(t\). Then the arc length of this curve is given by\[s= \int ^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt. \label{arcP} \]

Since the arc length formula involves a square root (which is always non-negative), the value of \( dt \) must also be non-negative for the length of an arc to be non-negative. Hence, our integrals for arc length will always go from a lower value of \( t \) to a higher value of \( t \). This is summarized in the following Note.

Note: Arc Lengths Are Computed from \( t_{\text{low}} \) to \( t_{\text{high}} \)

Unlike areas under parametric curves, arc lengths of parametric curves are strictly computed from the lowest time value to the highest.

At this point, a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function \(y=F(x)\). Then \(y(t)=F(x(t))\) and the Chain Rule gives\[y^{\prime}(t)=F^{\prime} \left( x(t) \right) x^{\prime}(t). \nonumber \]Substituting this into Equation \ref{arcP} gives\[\begin {array}{rcl}
s & = & \displaystyle \int ^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(F^{\prime}(x)\dfrac{dx}{dt}\right)^2}\,dt \\[6pt]
& = & \displaystyle \int ^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2(1+\left(F^{\prime}(x)\right)^2)}\,dt \\[6pt]
& = & \displaystyle \int ^{t_2}_{t_1}x^{\prime}(t)\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dt. \\[6pt]
\end{array} \nonumber \]Here we have assumed that \(x^{\prime}(t)>0\), which is a reasonable assumption. The Chain Rule gives \(dx=x^{\prime}(t)\,dt\), and letting \(a=x(t_1)\) and \(b=x(t_2)\) we obtain the formula\[s= \int ^b_a\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\, dx, \nonumber \]which is the formula for arc length obtained in the Introduction to the Applications of Integration.

Example \(\PageIndex{5}\): Finding the Arc Length of a Parametric Curve

Find the arc length of the semicircle defined by the equations\[x(t)=3\cos t, \quad y(t)=3\sin t, \quad \text{for }0 \leq t \leq \pi . \nonumber \]

Solution

The values \(t=0\) to \(t= \pi \) trace out the blue curve in Figure \(\PageIndex{8}\). To determine its length, use Equation \ref{arcP}:\[\begin{array}{rcl}
s & = & \displaystyle \int^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\\[6pt]
& = & \displaystyle \int ^ \pi _0\sqrt{(−3\sin t)^2+(3\cos t)^2}\,dt\\[6pt]
& = & \displaystyle \int ^ \pi _0\sqrt{9\sin^2t+9\cos^2t}\,dt\\[6pt]
& = & \displaystyle \int ^ \pi _0\sqrt{9(\sin^2t+\cos^2t)}\,dt\\[6pt]
& = & \displaystyle \int ^ \pi _03\,dt \\[6pt]
& = & 3t\Big|^ \pi _0\\[6pt]
& = & 3 \pi \text{ units}. \\[6pt]
\end{array} \nonumber \]Note that the formula for the arc length of a semicircle is \( \pi r\), and the radius of this circle is \(3\). This is a great example of using Calculus to derive a known formula for a geometric quantity.

Figure \(\PageIndex{8}\): The arc length of the semicircle is equal to its radius times \( \pi \).

Checkpoint \(\PageIndex{5}\)

Find the arc length of the curve defined by the equations\[x(t)=3t^2, \quad y(t)=2t^3, \quad \text{for }1 \leq t \leq 3. \nonumber \]

Answer

\(s=2(10^{3/2}−2^{3/2}) \approx 57.589\) units

We now return to the problem at the beginning of the section about a baseball leaving a pitcher's hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher's hand is at the origin, and the ball travels left to right in the direction of the positive \(x\)-axis, the parametric equations for this curve can be written as\[x(t)=140t, \quad y(t)=−16t^2+2t \nonumber \]where \(t\) represents time. We first calculate the distance the ball travels as a function of time. The arc length represents this distance. We can modify the arc length formula slightly. First, rewrite the functions \(x(t)\) and \(y(t)\) using \( v \) as an independent variable to eliminate any confusion with the parameter \(t\):\[x(v)=140v, \quad y(v)=−16v^2+2v. \nonumber \]Then we write the arc length formula as follows:\[\begin{array}{rcl}
s(t) & = & \displaystyle \int ^t_0\sqrt{(\dfrac{dx}{dv})^2+(\dfrac{dy}{dv})^2}\,dv\\[6pt]
& = & \displaystyle \int ^t_0\sqrt{140^2+(−32v+2)^2}\,dv \\[6pt]
\end{array} \nonumber \]The variable \(v\) acts as a dummy variable that disappears after integration, leaving the arc length as a function of time \(t\). Using our previous integration techniques,\[ \int \sqrt{a^2+u^2}\,du=\dfrac{u}{2}\sqrt{a^2+u^2}+\dfrac{a^2}{2}\ln |u+\sqrt{a^2+u^2}|+C. \nonumber \]We set \(a=140\) and \(u=−32v+2\). This gives \(du=−32\,dv\), so \(dv=−\frac{1}{32}\,du\). Therefore,\[\begin{array}{rcl}
\displaystyle \int \sqrt{140^2+(−32v+2)^2}\,dv & = & −\dfrac{1}{32} \displaystyle \int \sqrt{a^2+u^2}\,du\\[6pt]
& = & −\dfrac{1}{32}\left[\dfrac{(−32v+2)}{2}\sqrt{140^2+(−32v+2)^2}+\dfrac{140^2}{2}\ln |(−32v+2)+\sqrt{140^2+(−32v+2)^2}|+C\right] \\[6pt]
\end{array} \nonumber \]and\[\begin{array}{rcl}
s(t) & = & -\dfrac{1}{32}\left[\dfrac{(-32t+2)}{2}\sqrt{140^2+(-32t+2)^2}+\dfrac{140^2}{2}\ln \left |(-32t+2)+\sqrt{140^2+(-32t+2)^2}\right |\right] +\dfrac{1}{32}\left[\sqrt{140^2+2^2}+\dfrac{140^2}{2}\ln \left |2+\sqrt{140^2+2^2}\right |\right] \\[6pt]
& = & \left(\dfrac{t}{2}-\dfrac{1}{32}\right)\sqrt{1024t^2-128t+19604}-\dfrac{1225}{4}\ln \left |(-32t+2)+\sqrt{1024t^2-128t+19604}\right |+\dfrac{\sqrt{19604}}{32}+\dfrac{1225}{4}\ln (2+\sqrt{19604}). \\[6pt]
\end{array} \nonumber \]This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to \(t\). While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:\[\dfrac{d}{dx} \int ^x_af(u)\,du=f(x). \nonumber \]Therefore\[ \begin{array}{rcl}
s^{\prime}(t) & = & \dfrac{d}{dt}\Big[s(t)\Big]\\[6pt]
& = & \dfrac{d}{dt}\left[ \displaystyle \int ^t_0\sqrt{140^2+(−32v+2)^2}\,dv\right]\\[6pt]
& = & \sqrt{140^2+(−32t+2)^2}\\[6pt]
& = & \sqrt{1024t^2−128t+19604}\\[6pt]
& = & 2\sqrt{256t^2−32t+4901}. \\[6pt]
\end{array} \nonumber \] One-third of a second after the ball leaves the pitcher's hand, the distance it travels is equal to\[\begin{array}{rcl}
s\left(\dfrac{1}{3}\right) & = & \left(\dfrac{1/3}{2}−\dfrac{1}{32}\right)\sqrt{1024\left(\dfrac{1}{3}\right)^2−128\left(\dfrac{1}{3}\right)+19604}\\[6pt]
& & −\dfrac{1225}{4}\ln \Bigg|\left(−32\left(\dfrac{1}{3}\right)+2\right)+\sqrt{1024\left(\dfrac{1}{3}\right)^2−128\left(\dfrac{1}{3}\right)+19604}\Bigg|\\[6pt]
& & +\dfrac{\sqrt{19604}}{32}+\dfrac{1225}{4}\ln(2+\sqrt{19604})\\[6pt]
& \approx & 46.69 \text{ feet}. \\[6pt]
\end{array} \nonumber \]This value is just over three-quarters of the way to home plate. The speed of the ball is\[s^{\prime}\left(\dfrac{1}{3}\right)=2\sqrt{256\left(\dfrac{1}{3}\right)^2−32\left(\dfrac{1}{3}\right)+4901} \approx 140.27 \text{ ft/s}. \nonumber \]This speed translates to approximately \(95\) mph—a major-league fastball.

Surface Area Generated by a Parametric Curve

Recall the problem of finding the surface area of a volume of revolution. Earlier in the text, we derived a formula for finding the surface area of a volume generated by a function \(y=f(x)\) from \(x=a\) to \(x=b\), revolved around the \(x\)-axis:\[S=2 \pi \int ^b_af(x)\sqrt{1+(f^{\prime}(x))^2}\, dx. \nonumber \]We now consider a volume of revolution generated by revolving a parametrically defined curve \(x=x(t)\), \(y=y(t)\), for \(a \leq t \leq b\) around the \(x\)-axis as shown in Figure \(\PageIndex{9}\).

Figure \(\PageIndex{9}\): A surface of revolution generated by a parametrically defined curve.

The analogous formula for a parametrically defined curve is\[S=2 \pi \int ^b_ay(t)\sqrt{(x^{\prime}(t))^2+(y^{\prime}(t))^2}\,dt \label{ParSurface} \]provided that \(y(t)\) is not negative on \([a,b]\).

Example \(\PageIndex{6}\): Finding Surface Area

Find the surface area of a sphere of radius \(r\) centered at the origin.

Solution

We start with the curve defined by the equations\[x(t)=r\cos t, \quad y(t)=r\sin t, \quad \text{for }0 \leq t \leq \pi . \nonumber \]This generates an upper semicircle of radius \(r\) centered at the origin, as shown in the following graph.

Figure \(\PageIndex{10}\): A semicircle generated by parametric equations.

When this curve is revolved around the \(x\)-axis, it generates a sphere of radius \(r\). To calculate the surface area of the sphere, we use Equation \ref{ParSurface}:\[\begin{array}{rcl}
S & = & 2 \pi \displaystyle \int ^b_ay(t)\sqrt{(x^{\prime}(t))^2+(y^{\prime}(t))^2}\,dt\\[6pt]
& = & 2 \pi \displaystyle \int ^ \pi _0r\sin t\sqrt{(−r\sin t)^2+(r\cos t)^2}\,dt\\[6pt]
& = & 2 \pi \displaystyle \int ^ \pi _0r\sin t\sqrt{r^2\sin^2t+r^2\cos^2t}\,dt\\[6pt]
& = & 2 \pi \displaystyle \int ^ \pi _0r\sin t\sqrt{r^2(\sin^2t+\cos^2t)}\,dt\\[6pt]
& = & 2 \pi \displaystyle \int ^ \pi _0r^2\sin t\,dt\\[6pt]
& = & 2 \pi r^2\left(−\cos t\Big|^ \pi _0\right)\\[6pt]
& = & 2 \pi r^2(−\cos \pi +\cos 0)\\[6pt]
& = & 4 \pi r^2 \text{ units}^2. \\[6pt]
\end{array} \nonumber \]This is, in fact, the formula for the surface area of a sphere.

Checkpoint \(\PageIndex{6}\)

Find the surface area generated when the plane curve defined by the equations\[x(t)=t^3, \quad y(t)=t^2, \quad \text{for }0 \leq t \leq 1 \nonumber \]is revolved around the \(x\)-axis.

Answer

\(A=\frac{ \pi (494\sqrt{13}+128)}{1215} \text{ units}^2\)

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Key Concepts

• The derivative of the parametrically defined curve \(x=x(t)\) and \(y=y(t)\) can be calculated using the formula \(\frac{dy}{dx}=\frac{y^{\prime}(t)}{x^{\prime}(t)}\). Using the derivative, we can find the equation of a tangent line to a parametric curve.
• The area between a parametric curve and the \(x\)-axis can be determined by using the formula \(\displaystyle A= \int ^{t_2}_{t_1}y(t)x^{\prime}(t)\,dt\).
• The arc length of a parametric curve can be calculated by using the formula \[s= \int ^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt. \nonumber \]
• The surface area of a volume of revolution revolved around the \(x\)-axis is given by \[S=2 \pi \int ^b_ay(t)\sqrt{(x^{\prime}(t))^2+(y^{\prime}(t))^2}\,dt. \nonumber \]
• If the curve is revolved around the \(y\)-axis, then the formula is \[S=2 \pi \int ^b_a x(t)\sqrt{(x^{\prime}(t))^2+(y^{\prime}(t))^2}\,dt. \nonumber \]

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Key Equations

• Derivative of parametric equations

\[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{y^{\prime}(t)}{x^{\prime}(t)} \nonumber \]

• Second-order derivative of parametric equations

\[\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{(d/dt)(dy/dx)}{dx/dt} \nonumber \]

• Area under a parametric curve

\[A= \int ^b_ay(t)x^{\prime}(t)\,dt \nonumber \]

• Arc length of a parametric curve

\[s= \int ^{t_2}_{t_1}\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \nonumber \]

• Surface area generated by a parametric curve

\[S=2 \pi \int ^b_ay(t)\sqrt{(x^{\prime}(t))^2+(y^{\prime}(t))^2}\,dt \nonumber \]