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Calculus of Parametric Curves

- Calculus
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Now that you've learned how to represent curves as parametric equations, we can apply Calculus to parametric curves!

Suppose you wanted to model the way a football flies through the air after leaving a quarterback's hand. Since the ball moves in both the vertical \(y\) direction and the horizontal \(x\) direction, it would be useful to model the football's position with parametric equations. Just as we used Calculus to find a particle in motion's speed or acceleration, we can use Calculus to find the motion of an object that follows a parametric curve.

Parametric curves can be used to describe curves as surfaces. They're particularly useful in graphic curves that are not functions, such as the unit circle. So, parametric curves can be used to describe a large variety of curves including parabolas, circles, and ellipses. We can also parametrize more complex curves, like the Lissajous curve (used to describe harmonic motion), kites, rounded triangles, and many other funky shapes!

We can apply all the Calculus we've learned up to this point to parametric curves. The most important Calculus applications to parametric curves are finding a tangent line, finding the area under a parametric curve, calculating the arc length of a parametric curve, and finding the surface area to a parametric curve volume.

There are a few important formulas for applying Calculus to parametric curves. These formulas cover some of the most important aspects of **Parametric Curves. **

There is stuff that may first seem far from calculus, but calculus proves to be an indispensable tool in our study of **Parametric Curves. **

The formulas we will take a look at, are:

Tangent Lines to Parametric Curves

Area under a Parametric Curve

Arc length of a Parametric Curve

Surface Area of a Parametric solid

In the next sections, you will look at each one, along with an example of each.

To find the equation for a tangent line, we need the derivative of the parametric equations.

Consider a curve defined by the parametric equations \(x(t)\) and \(y(t)\). Assume that \(x'(t)\) and \(y'(t)\) exist and \(x'(t) \neq 0\). Then the derivative \(\frac{dy}{dx}\) is

\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'(t)}{x'(t)}.\]

Remember that the parameter need not be \(t\), it can be anything.

Find the derivative of \(y\) w.r.t. \(x\) for the curve given by the equations \(y=e^{2t}\) and \(x=t^3-t\) where \(t \neq 0\) at the point where \(t=2\).

**Solution:**

Using the formula that we saw above:

$$\begin{aligned} \frac{dy}{dx}&=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ & =\frac{\frac{d(e^{2t})}{dt}}{\frac{d(t^3-t)}{dt}} \\ &=\frac{2e^{2t}}{3t^2-1}. \end{aligned}$$

Evaluating it at \(t=2\):

$$\frac{dy}{dx}=\frac{2e^{4}}{11}.$$

This is an example of how we can find the derivative of a dependent variable w.r.t. the independent variable, for a pair of parametric equations, using this we can find the equation of the tangent line to the curve (an example of this is at the end).

You already know how to calculate the area under a curve, whose equation is given by \(y=f(x)\). The calculus is quite straightforward for that. But what if we are given parametric equations, you need a different formula in this case.

Consider the following theorem for the area under a parametric curve.

Consider a non-self intersecting parametric curve defined by \(x(t)\) and \(y(t)\) for \(a \le t \le b\). Assume that \(x(t)\) is differentiable. The area under this specified curve is:

$$A=\int_a^{b} y(t)x'(t)dt.$$

Let's go over the theorem for the arc length of a parametric curve.

Consider the curve defined by \(x(t)\) and \(y(t)\) for \(a \le t \le b\). Assume that \(x(t)\) and \(y(t)\) are differentiable. We want to find the **Arc Length **of the curve between that interval. But what is arc length?

Suppose you have a curve, and you want to find the length of the curve between a given interval, the matter is easy for a straight line or even for a circle. But what if the curve is any random function, then we need the help of calculus to find the length of any random curve.

To visualize what exactly is meant by **Arc Length**, take a look at the figure below:

Then, the arc length of the curve is given by

$$S=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt.$$

Find the arc length of the parametric curve given by \(x=3t^2\) and \(y=2t^3\) for the interval \(1<t<3\).

**Solution:**

Firstly, we will calculate the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):

$$\frac{dx}{dt}=\frac{d(3t^2)}{dt}=6t,$$

And for \(\frac{dy}{dt}\):

$$\frac{dy}{dt}=\frac{d(2t^3)}{dt}=6t^2.$$

Now substituting these in the formula of Arc length:

$$\begin{aligned} S &=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt \\ &=\int_1^{3} \sqrt{(6t)^2+(6t^2)^2} \,dt \\ &=6 \int_1^{3} t \sqrt{1+t^2} \,dt. \end{aligned}$$

The problem is now of pure integration, here we shall use integration by substitution.

Substituting for \(1+t^2=u^2\), we get \( t \,dt=u \,du\). For the integral limits, \(1<t<3\) becomes \(\sqrt{2}<u<\sqrt{10}\).

Our integral now becomes:

$$ \begin{aligned} S&=6 \int_\sqrt{2}^{\sqrt{10}} u^2 \,du \\ &= 2u^3 \Biggr|_{\sqrt{2}}^{\sqrt{10}} \\ &=2(10^{3/2}-2^{3/2}). \end{aligned}$$

Hence, the arc length of the curve is \(2(10^{3/2}-2^{3/2})\) units.

There is one last formula you need to go through in this section. This formula involves finding the surface area of a parametric curve that is revolves around the \(x\)-axis.

Again, let's consider the curve defined by \(x(t)\) and \(y(t)\) for \(a \le t \le b\). Assume that \(x(t)\) and \(y(t)\) are differentiable. This curve is revolved around the \(x\)-axis to create a Solid of Revolution. The surface area of the solid is defined by

$$S=2\pi \int_a^{b} y(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}} \,dt$$

If this curve is revolved around the \(y\)-axis to create a Solid of Revolution, the surface area of the solid is defined by

\[S=2\pi \int_a^{b} x(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}}dt\]

Find the derivative \(\frac{dy}{dx}\) of the parametric curve defined by \(x(t)=t^{3}-4\) and \(y(t)=3t\) for \(t \neq 0\). Then, find the equation of the line tangent to the curve at the point \(t=2\).

**Solution:**

$$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{3}{3t^{2}}=\frac{1}{t^{2}}.$$

Plugging in \(t=2\) to \(x(t)\) and \(y(t)\).

\[x(2)=2^{3}-4=4,\]

\[y(2)=3(2)=6.\]

Plugging in \(t=2\) to \(\frac{dy}{dx}\) to find the slope of the tangent line at the point \((4, 6)\).

\[\frac{dy}{dx}=\frac{1}{2^{2}}=\frac{1}{4}.\]

Plugging in these values to point-slope form

\[y-6=\frac{1}{4}(x-4).\]

Consider the curve defined by \(x(t)=3cost\) and \(y(t)=3sint\). Find the area under the curve on the interval \(0 \le t \le \pi\).

**Solution:**

Based on the above theorem, you need to differentiate \(x(t)\) before integrating.

$$x'(t)=-3sint.$$

All you have to do is plug in \(x'(t)\) and \(y(t)\) and evaluate the integral over the interval \(0 \le t \le \pi\): \(\int_a^{b} y(t)x'(t)dt\)

$$ \begin{aligned} \int_0^{\pi} (3sint)(-3sint) \,dt &=\int_0^{\pi} -9\sin^{2} \,dt \\ &=-9\int_0^{\pi} \frac{1-\cos(2t)}{2} \,dt \\ &=\frac{-9}{2} \int_0^{\pi} 1-\cos(2t) \,dt \\ &=\frac{-9}{2}[(\pi - \frac{\sin(2\pi)}{2})-(0 - \frac{\sin(2(0))}{2}) \\ &=\frac{-9\pi}{2}. \end{aligned}$$

Find the arc length of the parametric curve defined by \(x(t)=cost\) and \(y(t)=sint\) on the interval \(0 \le t \le 2\pi\).

**Solution:**

First, let's differentiate \(x(t)\) and \(y(t)\).

$$ \begin{align} x'(t)=-\sin t, \\ y'(t)=\cos t. \end{align}$$

Recall that the arc length is given as follows:

$$s=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt.$$

Now plugging our data into this formula:

$$ \begin{aligned} s &=\int_0^{2\pi} \sqrt{(\cos t)^{2} + (\sin t)^{2}} \,dt \\ s&=\int_0^{2\pi} \sqrt{1} \,dt \\ s&=2\pi - 0 \\ s &=2\pi. \end{aligned}$$

Essentially, what this tells us is the circumference of the unit circle is \(2\pi\).

Find the surface area of the curve described by \(x(t)=t^{2}\) and \(y(t)=t\) between \(0 \le t \le 3\) revolved around the \(x\)-axis.

**Solution:**

First, let's differentiate \(x(t)\) and \(y(t)\).

$$ \begin{align} x'(t)=2t, \\ y'(t)=1. \end{align}$$

Recall that the surface area of revolution is given by:

$$S=2\pi \int_a^{b} y(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}} \,dt.$$

Substituting the given data in the equation:

$$ \begin{aligned} S &=2\pi \int_0^{3} t \sqrt{(2t)^{2} + (1)^{2}} \,dt \\ &=2\pi \int_0^{3} t \sqrt{4t^{2}+1} \,dt. \end{aligned}$$

To integrate, we need to use \(u\)-substitution.

Let \(u=4t^{2}+1\) and \(\,du=8t \,dt\). For the integral limits, \(0 \le t \le 3\) becomes \(1 \le u \le 37\).

Your integral now becomes:

$$ \begin{aligned} S &=2\pi \int_0^{3} \frac{8}{8}t \sqrt{4t^{2}+1} \,dt \\ &=\frac{\pi}{4} \int_1^{37} \sqrt{u} \,du \\ S &=\frac{\pi}{4} \left( \frac{2}{3}u^{3/2} \right) \Biggr|_{1}^{37} \end{aligned} $$

Now substituting for the limits, you get:

$$ \begin{gathered} S=\frac{\pi}{6}(37^ \frac {3}{2} -1)\\ S \approx 117.3\, \text{square units}. \end{gathered}$$

- We can apply all the Calculus we've learned up to this point to parametric curves
- Then the derivative \(\frac{dy}{dx}\) of a curve \(x(t)\) and \(y(t)\) is
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'(t)}{x'(t)}.\]

The area under the curve \(x(t)\) and \(y(t)\) over the interval \(a \le t \le b\) is

\[\int_a^{b} y(t)x'(t)dt.\]

The arc length of the curve \(x(t)\) and \(y(t)\) over the interval \(a \le t \le b\) is given by

$$S=\int_a^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \,dt.$$

The surface area of the solid obtained by revolving the curve \(x(t)\) and \(y(t)\) around the \(x\)-axis is defined by

\[S=2\pi \int_a^{b} y(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}}dt.\]

- The surface area of the solid obtained by revolving the curve \(x(t)\) and \(y(t)\) around the \(y\)-axis is defined by
\[S=2\pi \int_a^{b} x(t) \sqrt{(x'(t))^{2} + (y'(t))^{2}}dt.\]

- Then the derivative \(\frac{dy}{dx}\) of a curve \(x(t)\) and \(y(t)\) is

Set *x(t)* = *t* and plug in *t* to *f(x)* to derive *y(t)*.

*y(t)**x'(t)* over the given interval with respect to *t*.

Set *x(t)* = *t* and plug in *t* to *f(x)* to derive *y(t)*.

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