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Calculus of Parametric Curves

Calculus of Parametric Curves

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.

Types of Parametric Curves

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!

The Meaning of Parametric Curves in Calculus

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.

Formulas of Parametric Curves in Calculus

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.

Tangent Lines to Parametric Curves

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).

Area Under a Parametric Curve Using Calculus

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:

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

Arc Length of a Parametric Curve

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:

Calculus of Parametric curves arc length between two points StudySmarterFigure 1.- Arc length of a curve between two points - StudySmarter Originals

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.

Surface Area of Parametric Solid of Revolution

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

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

Parametric Curve Examples in Calculus

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

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

Calculus of Parametric Curves tangent lines StudySmarterFigure 2.- The line tangent to the parametric curve at the point \(t=2\)

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:

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


Calculus of Parametric Curves - Key takeaways

  • 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

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

Frequently Asked Questions about Calculus of 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.

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

An example of applying Calculus to a parametric curve is finding the arc length of a parametrized curve over a certain interval.

To find the area under a parametric curve, integrate 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).

Final Calculus of Parametric Curves Quiz

Question

What is the meaning of Calculus of a parametric equation?

Show answer

Answer

You can apply all the Calculus you'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.


Show question

Question

What is the formula for a derivative of a parametric equation?

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Answer

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

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Question

How do we find the equation of a tangent line of a parametric equation?

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Answer

Use the formula for finding the derivative of a parametric equation. This formula produces the slope of the tangent line at a given t. Then, find the corresponding x and y values of the curve in order to write the tangent line in point-slope form.

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Question

What is the formula for the area under a parametric curve?

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Answer

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

$$\int_{a}^{b} y(t) \ x'(t) \,dt$$ 

Show question

Question

What is the formula for the arc length of a parametric curve?

Show answer

Answer

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

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

Show question

Question

What is the formula for surface area of a parametric solid of revolution around the x-axis?

Show answer

Answer

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

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

Show question

Question

What is the formula for surface area of a parametric solid of revolution around the y-axis?

Show answer

Answer

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

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

Show question

Question

How do you find the parametric equation of a curve?

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Answer

Set \(x=f(t)\) and plug in \(f(t)\) to get \(y(t)\). 

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Question

For a parametric curve given by \(x(t)=t^3\) and \(y(t)=t^2\) on the interval \(0<t<1\), which of the following is the correct surface area of the solid around a revolution about the x-axis?  

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Answer

$$S=\frac{\pi(494\sqrt{13}+128)}{1215}$$

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Question

For which of the parametric equations, the curve formed by them has the arc length of \(3\pi\), on the interval \(0 \le t \le \pi\)?

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Answer

\(x=3 \cos t\) and \(y=3 \sin t\) 

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Question

Is it true that for a given curve, the surface area of revolution is the same when revolved around the x-axis and the y-axis?

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Answer

No.

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Question

Eliminate the parameter for the equations \(x=\log_e t\) and \(y=e^t\) by writing \(y\) as a function of \(x\).

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Answer

Solving for \(t\):

$$\begin{aligned} &x=\log_e t \\ \Rightarrow \ &e^x=t \end{aligned}$$ 

Now substituting for \(t\) in \(y=e^t\):

$$y=e^{e^x}$$

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Question

Is the method of arc length an approximating way of finding the length or a precise and exact way?

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Answer

The method using calculus determines the arc length of any curve precisely and exactly, with no approximations. 

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Question

The area under a curve \(y=f(x)\) is given using the classical method of integration. How will this formula differ for parametric equations?

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Answer

For the parametric equation, the area under the curve is given by:

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

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Question

What is the area under the curve whose parametric equations are \(x=t-\sin t\) and \(y=1-\cos t\) on the interval \(0 \le t \le 2\pi\)?

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Answer

\(3\pi\)

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Question

Motion in space refers to motion in ____ dimensions.

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Answer

three.

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Question

The Cartesian plane can be used to model motion in ____ dimensions.

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Answer

two.

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Question

A curve that can be drawn on a plane is known as a ____ curve.

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Answer

plane.

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Question

To plot a trajectory in space, you use three axes, which are known as the \(x-\)axis, the \(y-\)axis, and the ____.

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Answer

\(z-\)axis.

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Question

The function \( z=f(x,y) \) drawn on the three-dimensional space represents a ____.

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Answer

surface.

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Question

In general, the intersection of two planes in space is a(n) ____.

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Answer

curve.

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Question

To describe motion in space you need ____ parametric equation(s).

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Answer

three.

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Question

The derivative of position with respect to time is known as ____.

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Answer

velocity.

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Question

The derivative of velocity with respect to time is known as ____.

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Answer

acceleration.

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Question

Acceleration is the ____ derivative of position with respect to time.

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Answer

second.

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Question

True/False: Speed is a synonym of velocity.

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Answer

False.

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Question

In problems about motion, the word initial usually refers to the instant when \( t \) equals ____.

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Answer

\(0\).

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Question

True/False: A trajectory can be described using parametric equations of one parameter.

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Answer

True.

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Question

True/False: To find the acceleration of an object, you just need to differentiate \(x(t)\).

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Answer

False.

Show question

Question

The position of an object is described by the parametric equations
\[ x(t)=t^2,\]
\[ y(t) = 4,\]
and
\[ z(t)=e^{-t}.\]
Does this movement occur on a plane?

Show answer

Answer

Yes.

Show question

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