Module 1: Parametric Equations and Polar Coordinates
Derivatives of Parametric Equations
Learning Outcomes
Determine derivatives and equations of tangents for parametric curves
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
The graph of this curve appears in Figure 1. It is a line segment starting at [latex]\left(-1,-10\right)[/latex] and ending at [latex]\left(9,5\right)[/latex].
Figure 1. Graph of the line segment described by the given parametric equations.
We can eliminate the parameter by first solving the equation [latex]x\left(t\right)=2t+3[/latex] for t:
The slope of this line is given by [latex]\frac{dy}{dx}=\frac{3}{2}[/latex]. Next we calculate [latex]{x}^{\prime }\left(t\right)[/latex] and [latex]{y}^{\prime }\left(t\right)[/latex]. This gives [latex]{x}^{\prime }\left(t\right)=2[/latex] and [latex]{y}^{\prime }\left(t\right)=3[/latex]. Notice that [latex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}[/latex]. This is no coincidence, as outlined in the following theorem.
theorem: Derivative of Parametric Equations
Consider the plane curve defined by the parametric equations [latex]x=x\left(t\right)[/latex] and [latex]y=y\left(t\right)[/latex]. Suppose that [latex]{x}^{\prime }\left(t\right)[/latex] and [latex]{y}^{\prime }\left(t\right)[/latex] exist, and assume that [latex]{x}^{\prime }\left(t\right)\ne 0[/latex]. Then the derivative [latex]\frac{dy}{dx}[/latex] is given by
This theorem can be proven using the Chain Rule. In particular, assume that the parameter t can be eliminated, yielding a differentiable function [latex]y=F\left(x\right)[/latex]. Then [latex]y\left(t\right)=F\left(x\left(t\right)\right)[/latex]. Differentiating both sides of this equation using the Chain Rule yields
But [latex]{F}^{\prime }\left(x\left(t\right)\right)=\frac{dy}{dx}[/latex], which proves the theorem.
[latex]_\blacksquare[/latex]
The theorem can be used to calculate derivatives of plane curves, as well as critical points. 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 theorem 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 [latex]y=f\left(x\right)[/latex] or not.
Example: Finding the Derivative of a Parametric Curve
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]
This derivative is zero when [latex]t=\pm 1[/latex]. When [latex]t=-1[/latex] we have
[latex]x\left(-1\right)=2\left(-1\right)+1=-1\text{ and }y\left(-1\right)={\left(-1\right)}^{3}-3\left(-1\right)+4=-1+3+4=6[/latex],
which corresponds to the point [latex]\left(-1,6\right)[/latex] on the graph. When [latex]t=1[/latex] we have
[latex]x\left(1\right)=2\left(1\right)+1=3\text{ and }y\left(1\right)={\left(1\right)}^{3}-3\left(1\right)+4=1 - 3+4=2[/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]
[latex]0[/latex]
[latex]5[/latex]
[latex]0[/latex]
[latex]\frac{\pi }{2}[/latex]
[latex]0[/latex]
[latex]5[/latex]
[latex]\pi [/latex]
[latex]-5[/latex]
[latex]0[/latex]
[latex]\frac{3\pi }{2}[/latex]
[latex]0[/latex]
[latex]-5[/latex]
[latex]2\pi [/latex]
[latex]5[/latex]
[latex]0[/latex]
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 Example: Finding the Derivative of a Parametric Curve.
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.
Calculate [latex]{x}^{\prime }\left(t\right)[/latex] and [latex]{y}^{\prime }\left(t\right)[/latex] and use the theorem.
Show Solution
[latex]{x}^{\prime }\left(t\right)=2t - 4[/latex] and [latex]{y}^{\prime }\left(t\right)=6{t}^{2}-6[/latex], so [latex]\frac{dy}{dx}=\frac{6{t}^{2}-6}{2t - 4}=\frac{3{t}^{2}-3}{t - 2}[/latex].
This expression is undefined when [latex]t=2[/latex] and equal to zero when [latex]t=\pm 1[/latex].
Figure 5.
Example: Finding a Tangent Line
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].
Show Solution
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 Example: Finding a Tangent Line.
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}-4t,y\left(t\right)=2{t}^{3}-6t,-2\le t\le 3\text{ when }t=5[/latex].
Hint
Calculate [latex]{x}^{\prime }\left(t\right)[/latex] and [latex]{y}^{\prime }\left(t\right)[/latex] and use the theorem.
Show Solution
The equation of the tangent line is [latex]y=24x+100[/latex].
Try It
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 [latex]y=f\left(x\right)[/latex] is defined to be the derivative of the first derivative; that is,
Since [latex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/latex], we can replace the [latex]y[/latex] on both sides of this equation with [latex]\frac{dy}{dx}[/latex]. This gives us
If we know [latex]\frac{dy}{dx}[/latex] as a function of t, then this formula is straightforward to apply.
Example: Finding a Second Derivative
Calculate the second derivative [latex]\frac{{d}^{2}y}{d{x}^{2}}[/latex] for the plane curve defined by the parametric equations [latex]x\left(t\right)={t}^{2}-3,y\left(t\right)=2t - 1,-3\le t\le 4[/latex].
Show Solution
From the example: Finding the Derivative of a Parametric Curve we know that [latex]\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}[/latex]. Using our above equation, we obtain
Watch the following video to see the worked solution to Example: Finding a Second Derivative.
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.
![A straight line from (−1, −10) to (9, 5). The point (−1, −10) is marked [latex]t[/latex] = −2, the point (3, −4) is marked [latex]t[/latex] = 0, and the point (9, 5) is marked [latex]t[/latex] = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t – 4, and −2 ≤ [latex]t[/latex] ≤ 3](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/4175/2019/04/11234739/CNX_Calc_Figure_11_02_001.jpg)
![A curved line going from (6, −7) through (−3, −1) to (13, 7) with arrow pointing in that order. The point (6, −7) is marked [latex]t[/latex] = −3, the point (−3, −1) is marked [latex]t[/latex] = 0, and the point (13, 7) is marked [latex]t[/latex] = 4. On the graph there are also written three equations: x(t) = t2 − 3, y(t) = 2t − 1, and −3 ≤ [latex]t[/latex] ≤ 4.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/4175/2019/04/11234741/CNX_Calc_Figure_11_02_002.jpg)
![A vaguely sinusoidal curve going from (−3, 2) through (−1, 6) and (3, 2) to (5, 6). The point (−3, 2) is marked [latex]t[/latex] = −2, the point (−1, 6) is marked [latex]t[/latex] = −1, the point (3, 2) is marked [latex]t[/latex] = 1, and the point (5, 6) is marked [latex]t[/latex] = 2. On the graph there are also written three equations: x(t) = 2t + 1, y(t) = t3 – 3t + 4, and −2 ≤ [latex]t[/latex] ≤ 2.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/4175/2019/04/11234744/CNX_Calc_Figure_11_02_003.jpg)
![A circle with radius 5 centered at the origin is graphed with arrow going counterclockwise. The point (5, 0) is marked [latex]t[/latex] = 0, the point (0, 5) is marked [latex]t[/latex] = π/2, the point (−5, 0) is marked [latex]t[/latex] = π, and the point (0, −5) is marked [latex]t[/latex] = 3π/2. On the graph there are also written three equations: x(t) = 5 cos(t), y(t) = 5 sin(t), and 0 ≤ [latex]t[/latex] ≤ 2π.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/4175/2019/04/11234746/CNX_Calc_Figure_11_02_004.jpg)
![A curve going from (12, −4) through the origin and (−4, 0) to (−3, 36) with arrows in that order. The point (12, −4) is marked [latex]t[/latex] = −2 and the point (−3, 36) is marked [latex]t[/latex] = 3. On the graph there are also written three equations: x(t) = t2 – 4t, y(t) = 2t3 – 6t, and −2 ≤ [latex]t[/latex] ≤ 3.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/4175/2019/04/11234751/CNX_Calc_Figure_11_02_005.jpg)
![A curved line going from (6, −7) through (−3, −1) to (13, 7) with arrow pointing in that order. The point (6, −7) is marked [latex]t[/latex] = −3, the point (−3, −1) is marked [latex]t[/latex] = 0, and the point (13, 7) is marked [latex]t[/latex] = 4. On the graph there are also written three equations: x(t) = t2 − 3, y(t) = 2t − 1, and −3 ≤ [latex]t[/latex] ≤ 4. At the point (1, 3), which is marked [latex]t[/latex] = 2, there is a tangent line with equation y = x/2 + 5/2.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/4175/2019/04/11234753/CNX_Calc_Figure_11_02_006.jpg)
