- Find derivatives and tangent lines for curves written in parametric form
- Calculate the area underneath a parametric curve
- Find the length of a parametric curve using the arc length formula
- Calculate the surface area when a parametric curve is rotated to create a 3D shape
Arc Length and Speed of Parametric Curves
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 [latex](x(t),y(t))[/latex], 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 just how far that ball has traveled as a function of time.
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 [latex]x[/latex]-axis, the parametric equations for this curve can be written as
[latex]x(t)=140t,y(t)=-16{t}^{2}+2t[/latex]
where [latex]t[/latex] represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex] using [latex]v[/latex] as an independent variable, so as to eliminate any confusion with the parameter [latex]t:[/latex]
[latex]x\left(v\right)=140v,y\left(v\right)=-16{v}^{2}+2v[/latex].
Then we write the arc length formula as follows:
[latex]\begin{array}{cc}\hfill s\left(t\right)& ={\displaystyle\int }_{0}^{t}\sqrt{{\left(\frac{dx}{dv}\right)}^{2}+{\left(\frac{dy}{dv}\right)}^{2}}dv\hfill \\ & ={\displaystyle\int }_{0}^{t}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv.\hfill \end{array}[/latex]
The variable [latex]v[/latex] acts as a dummy variable that disappears after integration, leaving the arc length as a function of time [latex]t[/latex]. To integrate this expression we can use a formula from Appendix A,
[latex]\displaystyle\int \sqrt{{a}^{2}+{u}^{2}}du=\frac{u}{2}\sqrt{{a}^{2}+{u}^{2}}+\frac{{a}^{2}}{2}\text{ln}|u+\sqrt{{a}^{2}+{u}^{2}}|+C[/latex].
We set [latex]a=140[/latex] and [latex]u=-32v+2[/latex]. This gives [latex]du=-32dv[/latex], so [latex]dv=-\frac{1}{32}du[/latex]. Therefore
[latex]\begin{array}{cc}\hfill {\displaystyle\int \sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv}& =-\frac{1}{32}{\displaystyle\int \sqrt{{a}^{2}+{u}^{2}}du}\hfill \\ & =-\frac{1}{32}\left[\begin{array}{c}\frac{\left(-32v+2\right)}{2}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}\hfill \\ +\frac{{140}^{2}}{2}\text{ln}|\left(-32v+2\right)+\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}|\hfill \end{array}\right]+C\hfill \end{array}[/latex]
and
[latex]\begin{array}{cc}\hfill s\left(t\right)& =-\frac{1}{32}\left[\frac{\left(-32t+2\right)}{2}\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}+\frac{{140}^{2}}{2}\text{ln}|\left(-32t+2\right)+\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}|\right]\hfill \\ & +\frac{1}{32}\left[\sqrt{{140}^{2}+{2}^{2}}+\frac{{140}^{2}}{2}\text{ln}|2+\sqrt{{140}^{2}+{2}^{2}}|\right]\hfill \\ & =\left(\frac{t}{2}-\frac{1}{32}\right)\sqrt{1024{t}^{2}-128t+19604}-\frac{1225}{4}\text{ln}|\left(-32t+2\right)+\sqrt{1024{t}^{2}-128t+19604}|\hfill \\ & +\frac{\sqrt{19604}}{32}+\frac{1225}{4}\text{ln}\left(2+\sqrt{19604}\right).\hfill \end{array}[/latex]
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:
[latex]\frac{d}{dx}{\displaystyle\int }_{a}^{x}f\left(u\right)du=f\left(x\right)[/latex].
Therefore:
[latex]\begin{array}{cc}\hfill {s}^{\prime }\left(t\right)& =\frac{d}{dt}\left[s\left(t\right)\right]\hfill \\ & =\frac{d}{dt}\left[{\displaystyle\int }_{0}^{t}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv\right]\hfill \\ & =\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}\hfill \\ & =\sqrt{1024{t}^{2}-128t+19604}\hfill \\ & =2\sqrt{256{t}^{2}-32t+4901}.\hfill \end{array}[/latex]
One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to
[latex]\begin{array}{cc}\hfill s\left(\frac{1}{3}\right)& =\left(\frac{\frac{1}{3}}{2}-\frac{1}{32}\right)\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}\hfill \\ & -\frac{1225}{4}\text{ln}|\left(-32\left(\frac{1}{3}\right)+2\right)+\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}|\hfill \\ & +\frac{\sqrt{19604}}{32}+\frac{1225}{4}\text{ln}\left(2+\sqrt{19604}\right)\hfill \\ & \approx 46.69\text{feet}.\hfill \end{array}[/latex]
This value is just over three quarters of the way to home plate. The speed of the ball is
[latex]{s}^{\prime }\left(\frac{1}{3}\right)=2\sqrt{256{\left(\frac{1}{3}\right)}^{2}-16\left(\frac{1}{3}\right)+4901}\approx 140.34\text{ft/s}[/latex].
This speed translates to approximately 95 mph—a major-league fastball.