Calculus and parametric curves

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We discuss derivatives of parametrically defined curves.

Derivatives

Recall from differential calculus that the tangent line provides the best linear approximation to a curve at a given point. Thus, we are often interested in calculating the tangent line. If the curve can be expressed as a function of either $x$ or $y$ then the slope of the tangent line is obtained by taking the derivative at the given point. If the curve has a Cartesian description $F(x,y)=0$ , then even if we can’t express it as a function, we can still use implicit differentiation to find the slope of the tangent line.

Suppose we have a curve given parametrically. Remember that for some parametric curves would be difficult or impossible to find Cartesian forms. We would like to be able to find the slope of the tangent line directly from the parametric description without having to convert to a Cartesian form.

Suppose we have a curve $C$ that is traced out by the parametric equations

$\begin {cases} x=x(t) \\ y=y(t) \end {cases}$

In order to find the slope of the tangent line we need to compute the rate of change $\frac {dy}{dx}$ of $y$ with respect to $x$ ,

We use the chain rule to write $\frac {dy}{dx} = \frac {dy}{dt} \frac {dt}{dx}$

Recall from the Inverse Function Theorem, we have $\frac {dt}{dx}=\frac {1}{\frac {dx}{dt}}$ . This means we can rewrite $\frac {dy}{dx}$ as

$\frac {dy}{dx} = \frac {dy / dt }{dx/ dt}$

Notice that this formula allows us to calculate $\frac {dy}{dx}$ directly from our parametric description of $C$ .

Let a curve $C$ be parametrized by

$\begin {cases} x&=x(t) \\ y&=y(t) \end {cases}$ for $t$ in an interval $I$ .

Suppose that $x$ and $y$ are differentiable functions on $I$ and let $t_{0}$ be a point in $I$ . The tangent line to $C$ when $t=t_0$ is the line through $\big (x(t_0),y(t_0)\big )\text { with slope }m=\frac {y'(t_0)}{x'(t_0)},$ provided $x'(t_0)\neq 0$ .

The normal line to $C$ at $t=t_0$ is the line through $\big (x(t_0),y(t_0)\big )\text { with slope }m=\frac {-x'(t_0)}{y'(t_0)},$ provided $y'(t_0)\neq 0$ . The normal line is perpendicular to the tangent line.

The definition leaves two special cases to consider. When the tangent line is horizontal, the normal line is undefined by the above definition as $y'(t_0)=0$ . Likewise, when the normal line is horizontal, the tangent line is undefined. It seems reasonable that these lines be defined (one can draw a line tangent to the “right side” of a circle, for instance), so we add the following to the above definition.

If the tangent line at $t=t_0$ has a slope of $0$ , the normal line to $C$ at $t=t_0$ is the vertical line $x=x(t_0)$ .
If the normal line at $t=t_0$ has a slope of $0$ , the tangent line to $C$ at $t=t_0$ is the line $x=x(t_0)$ .

Let’s look at some examples.

Consider the curve $C$ given by the parametrization $\begin {cases} x(t)&=5t^2-6t+4 \\ y(t)&=t^2+6t-1 \end {cases}$ where $t$ ranges over all of $\mathbb {R}$ .

Find the equations of the tangent and normal lines to $C$ at $t=3$ .

We start by computing $x'(t) = \answer [given]{10t-6}$ and $y'(t) =\answer [given]{2t+6}.$ Thus $\frac {dy}{dx} = \answer [given]{\frac {2t+6}{10t-6}}.$ Make note of something that might seem unusual: $\frac {dy}{dx}$ is a function of $t$ , not $x$ . Just as points on the curve are found in terms of $t$ , so are the slopes of the tangent lines.

The point on $C$ at $t=3$ is $(\answer [given]{31},\answer [given]{26})$ . The slope of the tangent line is $m=\answer [given]{1/2}$ and the slope of the normal line is $m=\answer [given]{-2}$ . Thus,

the equation of the tangent line is $y=\answer [given]{\frac {(x-31)}{2}+26}$ (shown below in orange), and
the equation of the normal line is $y=\answer [given]{-2(x-31)+26}$ (shown below in green).

Find where the circle, defined by $x=\cos ( t)$ and $y=\sin ( t)$ on $[0,2\pi )$ , has vertical and horizontal tangent lines.

We compute the derivative $\frac {dy}{dx} = \frac {y'(t)}{x'(t)} = \answer [given]{-\frac {\cos t}{\sin t}}.$ The derivative is $0$ when $\cos ( t)= \answer [given]{0}$ . Note that we only need the values of $t$ in the interval $[0, 2\pi )$ that satisfy $\cos (t)$ . This happens when $t=\pi /2$ , and $t= 3\pi /2$ . These two $t$ values correspond to the points $(0,1)$ and $(0,-1)$ on the circle.

The normal line is horizontal (and hence, the tangent line is vertical) when $\sin t=\answer [given]{0}$ . This happens when $t= 0$ and $t=\pi$ corresponding to the points $(-1,0)$ and $(0,1)$ on the circle. These results should make intuitive sense.

Consider the parametric equations

$\begin {cases} x(t)&=\sin (2t) \\ y(t)&=\cos (3t) \end {cases}$

Below is the curve traced out by the above parametric curves as $t$ varies over $[0, 2\pi )$ .

Notice that the origin belongs to the curve. What is the tangent line to the curve when at the origin? Phrased in this manner, this question does not make sense. If we zoom in on the origin, the curve does not begin to look more and more like a straight line. In fact, it will look like two lines crossing no matter how far we zoom in. Thus it seems that the curve is not differentiable at the origin. However we are only focusing on the curve itself rather than how the curve is traced out. Let’s use the parametrization to investigate how the curve is traced out.

The $t$ values for which the curve passes through the origin are obtained by find the $t$ values that simulaneously solve $\cos (3t)=0$ and $\sin (2t)=0$ . This gives us $t=\pi /2$ and $t=9\pi /6$ .

See below how the curve is traced out as $t$ varies. Use the slider for $a$ to traced out the curve.

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What is the tangent line to the curve when $t=\frac {\pi }{2}$ ? What about when $t=\frac {9\pi }{6}$ . Look at the interactive figure below. Think of the parameter $t$ as time. Then the figure below shows a point corresponding to the time as well as some points corresponding to both future and past time values close to the given time. We see that if we look our curve for times close to $\frac {\pi }{2}$ we see that the curve looks approximately like a line. Similarly, if we look at our curve for times close to $\frac {9\pi }{6}$ then again we see that the curve looks approximately linear if we zoom in on the origin.

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Now we can find the tangent line to the curve at the two different times it passes through the origin. Apply the formula

$\frac {dy}{dx} = \frac { y'(t)}{x'(t)}$

we get $\frac {dy}{dx}=\frac {\answer [given]{-3\sin (3t)}}{\answer [given]{2\cos (2t)}}$

Then calculating the $\frac {dy}{dx}$ when $t=\frac {\pi }{2}$ gives us $\answer [given]{ \frac {-3}{2} }$ . Thus the tangent line to the curve when $t=\frac {\pi }{2}$ is $y-\answer [given]{ 0}=\answer [given]{\frac {-3}{2}}(x-\answer [given]{0})$ (shown in orange below).

Similarly the tangent line to the curve when $t=\frac {9\pi }{6}$ is $y-\answer [given]{0}=\answer [given]{\frac {3}{2} }(x-\answer {0})$ (shown in green below).

Finally here is a visualization that shows how the tangent line to the curve varies as we vary $t$ .

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