7.2 Calculus of parametric curves

Calculus volume 2 Page 1 / 6

Determine derivatives and equations of tangents for parametric curves.
Find the area under a parametric curve.
Use the equation for 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 parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization 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 just 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

x (t) = 2 t + 3, y (t) = 3 t - 4, −2 \leq t \leq 3 .

The graph of this curve appears in [link] . It is a line segment starting at $(−1, −10)$ and ending at $(9, 5) .$

A straight line from (−1, −10) to (9, 5). The point (−1, −10) is marked t = −2, the point (3, −4) is marked t = 0, and the point (9, 5) is marked t = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t – 4, and −2 ≤ t ≤ 3 — Graph of the line segment described by the given parametric equations.

We can eliminate the parameter by first solving the equation $x (t) = 2 t + 3$ for t :

\begin{matrix} x (t) & = & 2 t + 3 \\ x - 3 & = & 2 t \\ t & = & \frac{x - 3}{2} . \end{matrix}

Substituting this into $y (t),$ we obtain

\begin{matrix} y (t) & = & 3 t - 4 \\ y & = & 3 (\frac{x - 3}{2}) - 4 \\ y & = & \frac{3 x}{2} - \frac{9}{2} - 4 \\ y & = & \frac{3 x}{2} - \frac{17}{2} . \end{matrix}

The slope of this line is given by $\frac{d y}{d x} = \frac{3}{2} .$ Next we calculate $x^{'} (t)$ and $y^{'} (t) .$ This gives $x^{'} (t) = 2$ and $y^{'} (t) = 3 .$ Notice that $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{3}{2} .$ This is no coincidence, as outlined in the following theorem.

Derivative of parametric equations

Consider the plane curve defined by the parametric equations $x = x (t)$ and $y = y (t) .$ Suppose that $x^{'} (t)$ and $y^{'} (t)$ exist, and assume that $x^{'} (t) \neq 0 .$ Then the derivative $\frac{d y}{d x}$ is given by

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{y^{'} (t)}{x^{'} (t)} .

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^{'} (t) = F^{'} (x (t)) x^{'} (t),

F^{'} (x (t)) = \frac{y^{'} (t)}{x^{'} (t)} .

But $F^{'} (x (t)) = \frac{d y}{d x},$ which proves the theorem.

□

[link] 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 $f^{'} (x_{0}) = 0$ or $f^{'} (x_{0})$ does not exist. [link] 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.

Finding the derivative of a parametric curve

Calculate the derivative $\frac{d y}{d x}$ for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.

$x (t) = t^{2} - 3, y (t) = 2 t - 1, −3 \leq t \leq 4$
$x (t) = 2 t + 1, y (t) = t^{3} - 3 t + 4, −2 \leq t \leq 5$
$x (t) = 5 cos t, y (t) = 5 sin t, 0 \leq t \leq 2 π$

To apply [link] , first calculate $x^{'} (t)$ and $y^{'} (t) :$

$\begin{matrix} x^{'} (t) = 2 t \\ y^{'} (t) = 2. \end{matrix}$

Next substitute these into the equation:

$\begin{matrix} \frac{d y}{d x} = \frac{d y / d t}{d x / d t} \\ \frac{d y}{d x} = \frac{2}{2 t} \\ \frac{d y}{d x} = \frac{1}{t} . \end{matrix}$

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.

Graph of the parabola described by parametric equations in part a.
To apply [link] , first calculate $x^{'} (t)$ and $y^{'} (t) :$

$\begin{matrix} x^{'} (t) = 2 \\ y^{'} (t) = 3 t^{2} - 3. \end{matrix}$

Next substitute these into the equation:

$\begin{matrix} \frac{d y}{d x} = \frac{d y / d t}{d x / d t} \\ \frac{d y}{d x} = \frac{3 t^{2} - 3}{2} . \end{matrix}$

This derivative is zero when $t = ±1 .$ When $t = −1$ we have

$x (−1) = 2 (−1) + 1 = −1 and y (−1) = {(−1)}^{3} - 3 (−1) + 4 = −1 + 3 + 4 = 6,$

which corresponds to the point $(−1, 6)$ on the graph. When $t = 1$ we have

$x (1) = 2 (1) + 1 = 3 and y (1) = {(1)}^{3} - 3 (1) + 4 = 1 - 3 + 4 = 2,$

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.

Graph of the curve described by parametric equations in part b.

To apply [link] , first calculate

x^{'} (t)

and

y^{'} (t) :

\begin{matrix} x^{'} (t) = −5 sin t \\ y^{'} (t) = 5 cos t . \end{matrix}

Next substitute these into the equation:

\begin{matrix} \frac{d y}{d x} = \frac{d y / d t}{d x / d t} \\ \frac{d y}{d x} = \frac{5 cos t}{−5 sin t} \\ \frac{d y}{d x} = − cot t . \end{matrix}

This derivative is zero when

cos t = 0

and is undefined when

sin t = 0 .

This gives

t = 0, \frac{π}{2}, π, \frac{3 π}{2}, and 2 π

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
$\frac{π}{2}$	0	5
$π$	−5	0
$\frac{3 π}{2}$	0	−5
$2 π$	5	0

These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations ( [link] ). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.

A circle with radius 5 centered at the origin is graphed with arrow going counterclockwise. The point (5, 0) is marked t = 0, the point (0, 5) is marked t = π/2, the point (−5, 0) is marked t = π, and the point (0, −5) is marked t = 3π/2. On the graph there are also written three equations: x(t) = 5 cos(t), y(t) = 5 sin(t), and 0 ≤ t ≤ 2π. — Graph of the curve described by parametric equations in part c.

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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