Advanced Vector Calculus: Background You’ll Need 3
Create parametric equations for curves and use them to calculate arc length.
Derivatives of Parametric Equations
When working with parametric equations [latex]x = x(t)[/latex] and [latex]y = y(t)[/latex], you can find the slope of the curve without eliminating the parameter.
theorem: derivative of parametric equations
For parametric equations [latex]x = x(t)[/latex] and [latex]y = y(t)[/latex]:
Think of the chain rule: As [latex]t[/latex] changes, both [latex]x[/latex] and [latex]y[/latex] change. The ratio of their rates of change gives us the slope of the curve.
This is particularly useful when eliminating the parameter would result in a complicated expression!
The parametric derivative formula works for any parametric curve—even curves that loop or cross themselves and can’t be written as [latex]y = f(x)[/latex].
Recall that a critical point of a differentiable function [latex]y=f\left(x\right)[/latex] is any point [latex]x={x}_{0}[/latex] such that either [latex]{f}^{\prime }\left({x}_{0}\right)=0[/latex] or [latex]{f}^{\prime }\left({x}_{0}\right)[/latex] does not exist.
The parametric approach gives us the slope at any point along the curve, regardless of how complicated the path might be.
Calculate the derivative [latex]\frac{dy}{dx}[/latex] for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.
This derivative is undefined when [latex]t=0[/latex]. Calculating [latex]x\left(0\right)[/latex] and [latex]y\left(0\right)[/latex] gives [latex]x\left(0\right)={\left(0\right)}^{2}-3=-3[/latex] and [latex]y\left(0\right)=2\left(0\right)-1=-1[/latex], which corresponds to the point [latex]\left(-3,-1\right)[/latex] on the graph. The graph of this curve is a parabola opening to the right, and the point [latex]\left(-3,-1\right)[/latex] is its vertex as shown.
Figure 2. Graph of the parabola described by parametric equations in part a.
To apply the theorem, first calculate [latex]{x}^{\prime }\left(t\right)[/latex] and [latex]{y}^{\prime }\left(t\right)\text{:}[/latex]
which corresponds to the point [latex]\left(3,2\right)[/latex] on the graph. The point [latex]\left(3,2\right)[/latex] is a relative minimum and the point [latex]\left(-1,6\right)[/latex] is a relative maximum, as seen in the following graph.
Figure 3. Graph of the curve described by parametric equations in part b.
To apply the theorem, first calculate [latex]{x}^{\prime }\left(t\right)[/latex] and [latex]{y}^{\prime }\left(t\right)\text{:}[/latex]
This derivative is zero when [latex]\cos{t}=0[/latex] and is undefined when [latex]\sin{t}=0[/latex]. This gives [latex]t=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},\text{and}2\pi[/latex] as critical points for t. Substituting each of these into [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex], we obtain
[latex]t[/latex]
[latex]x\left(t\right)[/latex]
[latex]y\left(t\right)[/latex]
0
5
0
[latex]\frac{\pi }{2}[/latex]
0
5
[latex]\pi[/latex]
−5
0
[latex]\frac{3\pi }{2}[/latex]
0
−5
[latex]2\pi[/latex]
5
0
These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 4). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.
Figure 4. Graph of the curve described by parametric equations in part c.
Watch the following video to see the worked solution to the example above.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Find the equation of the tangent line to the curve defined by the equations
[latex]x\left(t\right)={t}^{2}-3,y\left(t\right)=2t - 1,-3\le t\le 4\text{ when }t=2[/latex].
First find the slope of the tangent line using the theorem, which means calculating [latex]{x}^{\prime }\left(t\right)[/latex] and [latex]{y}^{\prime }\left(t\right)\text{:}[/latex]
When [latex]t=2[/latex], [latex]\frac{dy}{dx}=\frac{1}{2}[/latex], so this is the slope of the tangent line. Calculating [latex]x\left(2\right)[/latex] and [latex]y\left(2\right)[/latex] gives
[latex]x\left(2\right)={\left(2\right)}^{2}-3=1\text{ and }y\left(2\right)=2\left(2\right)-1=3[/latex],
which corresponds to the point [latex]\left(1,3\right)[/latex] on the graph (Figure 6). Now use the point-slope form of the equation of a line to find the equation of the tangent line:
Figure 6. Tangent line to the parabola described by the given parametric equations when [latex]t=2[/latex].
Watch the following video to see the worked solution to the example above.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Arc length measures the actual distance traveled along a curve. If a particle moves from point A to point B along a parametric path, the arc length tells you exactly how far it traveled.
theorem: arc length of a parametric curve
For a plane curve defined by [latex]x = x(t)[/latex], [latex]y = y(t)[/latex] where [latex]t_1 \le t \le t_2[/latex], with [latex]x(t)[/latex] and [latex]y(t)[/latex] differentiable then:
Note that the formula for the arc length of a semicircle is [latex]\pi r[/latex] and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.
Figure 10. The arc length of the semicircle is equal to its radius times [latex]\pi[/latex].
Watch the following video to see the worked solution to the example above.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
