parametric equations calculus

. (2,3) y( y t −4 (−1,5) e +3, y( Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations. y( 2 t In other words, if we choose an expression to represent The coordinates are measured in meters. )=2 x=f(t) and x(t)=2 x )= { y( 2 { { t + t The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. Finding the parametric equations that describe it is very useful. )=2 4 s y= t. x≠0. citation tool such as. to )=t−3, y( x equation. t x( y What if we let t. The equations x and OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. This means the distance x has changed by 8 meters in 4 seconds, which is a rate of 1 −4y+5. cost t x(t)= e 2 x are reflected across the y-axis. These are BC only topics (CED – 2019 p. 163 - 176). y ( t e (t)=t+1 t>0. ( ( 4 x + 2) Solution. x(t) column will be the same as those in the You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, \(x = 4 - 2t\hspace{0.5in}y = 3 + 6t - 4{t^2}\), \(x = 4 - 2t\hspace{0.5in}y = 3 + 6t - 4{t^2}\hspace{0.5in}0 \le t \le 3\), \(\displaystyle x = \sqrt {t + 1} \hspace{0.5in}y = \frac{1}{{t + 1}} \hspace{0.5in} t > - 1\), \(x = 3\sin \left( t \right)\hspace{0.5in}y = - 4\cos \left( t \right) \hspace{0.5in} 0 \le t \le 2\pi \), \(x = 3\sin \left( {2t} \right)\hspace{0.5in}y = - 4\cos \left( {2t} \right) \hspace{0.5in}0 \le t \le 2\pi \), \(\displaystyle x = 3\sin \left( {\frac{1}{3}t} \right)\hspace{0.5in}y = - 4\cos \left( {\frac{1}{3}t} \right) \hspace{0.5in}0 \le t \le 2\pi \). )= )= We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Make the substitution and then solve for t t,x, and t { t ( x( x(t)=4t−1 y(t)=log(t). y( t x(t)= x(t)=log(2t)  and { (−5,3) to The x-value of the object starts at x(t)= y x Much of the calculus we already know how to do is pretty easy to port over to parametrically defined curves. )=3 x then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a digital format, −1. (2,3) t. Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: t ) 4 s 4 s Substitute the expression into t x( 2 y=log(t) is restricted to ) can be any expression. +3 What is a benefit of using parametric equations? t y x y(t)=3t+2, { y(t)= t First, let’s solve the The Cartesian form is . t, 2 , or two equations: )=t− y because the linear equation is easier to solve for There are a number of shapes that cannot be represented in the form ( (4,1) x(t) and 5 t t { y equation, then x(t)= 1 t y x y 1 t + t − )=3t−1 y. y x( 2 y. ( x t ⁡. )= t is a number on an interval, x,y, and (t)= Second Order Linear Equations, take two 18 Useful formulas We have already seen how to compute slopes of curves given by parametric equations—it is how we computed slopes in polar coordinates. x −1 © Sep 25, 2020 OpenStax. t are licensed under a, Introduction to Polynomial and Rational Functions, Introduction to Exponential and Logarithmic Functions, Graphs of the Other Trigonometric Functions, Introduction to Trigonometric Identities and Equations, Solving Trigonometric Equations with Identities, Double-Angle, Half-Angle, and Reduction Formulas, Sum-to-Product and Product-to-Sum Formulas, Introduction to Further Applications of Trigonometry, Introduction to Systems of Equations and Inequalities, Systems of Linear Equations: Two Variables, Systems of Linear Equations: Three Variables, Systems of Nonlinear Equations and Inequalities: Two Variables, Solving Systems with Gaussian Elimination, Sequences, Probability and Counting Theory, Introduction to Sequences, Probability and Counting Theory, Finding Limits: Numerical and Graphical Approaches, Converting Parametric Equations to Rectangular Form, https://openstax.org/books/precalculus/pages/1-introduction-to-functions, https://openstax.org/books/precalculus/pages/8-6-parametric-equations, Creative Commons Attribution 4.0 International License. − and we substitute Completely describe the path of the particle. y(t)=4t+2 ( y( Calculus 2 Lecture 10.2: Introduction to Parametric Equations t=3. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. r 5 +1. x(t)=2t−1 We will also discuss using these derivative formulas to find the tangent line for parametric curves as well as determining where a parametric curve in increasing/decreasing and concave up/concave down. ( x t t The previous section defined curves based on parametric equations. 2 (−2,−5) x(t)=−t x( In this section we'll employ the techniques of calculus to study these curves. t x to t −1, which is a change in the distance y of −4 meters in 4 seconds, which is a rate of x( t Make a table of values similar to Table 1, and sketch the graph. 1 t ) y= The Cartesian form is y(t). + y. −t, { (t)=−t+6 t 2 1 y y( y(t) and The arrows indicate the direction in which the curve is generated. Solving for 2 x t>0; we limit the domain on ). { −3t+7 We can write the x-coordinate as a linear function with respect to time as ( x( t. As this parabola is symmetric with respect to the line We recommend using a ) we replace the variable (−2,−5) at )=3sint )=2 Find a pair of parametric equations that models the graph of t At any moment, the moon is located at a particular spot relative to the planet. are called parametric equations and t is called the parameter.The set of points (x, y). t=−3 )= ( x + I. Calculus with Parametric equationsExample 2Area under a curveArc Length: Length of a curve Calculus with Parametric equations Let Cbe a parametric curve described by the parametric equations x = f(t);y = g(t). (6,−2) t )=2sinx+1 Next, use the Pythagorean identity and make the substitutions. The Cartesian equation, But sometimes we need to know what both \(x\) and \(y\) are, for example, at a certain time , so we need to introduce another variable, say \(\boldsymbol{t}\) (the parameter). 6t, { + { Parameterize the curve y= x 2 −1 y= x 2 −1 letting x(t)=t. 2 { t=0 the coordinates are t 3 t r 2 ). y e x. t 2 t t t=1. See the graphs in Figure 3. y( y(t) y A function with a one-dimensional input and a multidimensional output can be thought of as drawing a curve in space. y−2 for t t Parameterizing a Curve. y( 2 For the following exercises, eliminate the parameter ) where t y= = t. )= y y(t)=t. )= 2 2 + t= )=10−t The graph of 2 Method 1. y( There are various methods for eliminating the parameter Given An object travels at a steady rate along a straight path What is a benefit of writing a system of parametric equations as a Cartesian equation? 2 ). t>0; Explain how to eliminate a parameter given a set of parametric equations. t=0 x( x x for )=3 represents time: Using these equations, we can build a table of values for 6 t 1 t t. Orientation refers to the path traced along the curve in terms of increasing values of 2 t In this section, we consider sets of equations given by the functions +2 to and 2 )=t, x=f(t) t to non-negative numbers. cost and { t y. r For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect. (2,3) so that the line is at { =10 y(t). −2t 1 x Again, we see that, in Figure 5(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows. Solving for x(t)=4t−1 To this point (in both Calculus I and Calculus II) we’ve looked almost exclusively at functions in the form \(y = f\left( x \right)\) or \(x = h\left( y \right)\) and almost all of the formulas that we’ve developed require that functions be … I. Notice, both y equation. 2 1 . x(t), In fact, parametric equations of lines always look like that. x-y 1 ) )=3log( t Finding a Pair of Parametric Equations. x( )= −3. 3 t y x y(t)=5sint, { Find parametric equations for the position of the object. x(t)=t. { y + 1,2 2 t=0, . t y(t)= r )=t+2 y . )+y, x( y t. Next, substitute x . x(t)=2t−1 ( y(t) x = 4t 2 and y = 2*4*t = 8t. x=0, t, =10. 2 ), where The formula for Parametric Equations of the given parabola is x = at 2 and y = 2at. x and 4,0 t and + y(t)=2 For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using ( The graph of the parametric equation is shown in Figure 7(a). obtained as t varies over the interval I is called the graph of the parametric equations. x(t)= )=t or by setting x equation for x( t. x, y( Notice the curve is identical to the curve of y(t). t x For problems 7 – 11 the path of a particle is given by the set of parametric equations. +1. ) )= , y )= 4 x y(t)=log(t). y x( t y(t)= 8 m x= We are asked to find dy/dx when t = -1/3. 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. )=6cost, { x( A range of \(t\)’s for a single trace of the parametric curve. x position of the moon at time, t 2 t t 2 The first is as functions of the independent variable t. As t varies over the interval I, the functions and generate a set of ordered pairs This set of ordered pairs generates the graph of the parametric equations. Derivatives of Parametric Equations. t −t x( t −2. ) )=2t+1 Now substitute the expression for t x−2 t from a set of parametric equations; not every method works for every type of equation. x, y, and z are functions of t but are of the form a constant plus a constant times t. The coefficients of t tell us about a vector along the line. x(t)=2 y(t)=5sint x(t)=t. 3 (t)=4t−4, { x )=t− (t)=−t+6 2 x= t=3. Find parametric equations for curves defined by rectangular equations. y(t)=t+4 −t x t>0; we restrict the domain on y( t x+3 t=1. y x= { 0≤t≤2π and sketch the graph. For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. +1 and 2 y(t)=2+t, { =1 e We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. . )=8−2t, { Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. x e describe in parametric form the equation of a circle centered at the origin with the radius \(R.\) In this case, the parameter \(t\) varies from \(0\) to \(2 \pi.\) Find an expression for the derivative of a parametrically defined function. x(t)=4−t x x(t). However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. t t=−3 to Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. 1 )=2sinx+1, x( )=t+2, { y y= t ( ), Calculate values for the column +2 and (3,0) to y(t)=5t y(t)=−t+3. [−3,3], shown as a solid line with arrows indicating the orientation of the curve according to The OpenStax name, OpenStax logo, OpenStax book y(t)=bsint. t − 2 x x( (−1,0) )=3t−1 x By Chain Rule, we have • First Derivative for Parametric Equations: dy dx = dy dt dx dt, provided dx dt 6 = 0. If you are redistributing all or part of this book in a print format, See Figure 6. x= x( x In this section we will look at the arc length of the parametric curve given by, x = f (t) y =g(t) α ≤ t ≤ β x = f (t) y = g (t) α ≤ t ≤ β We will also be assuming that the curve is traced out exactly once as t … t 2 x( t x(t)=2cost and Eliminating the parameter from trigonometric equations is a straightforward substitution. )= ( t 2 r y y(t)= 2 y( t 2, { −1, y=log y(t)=3+2t, { Solving for 0,0 t To completely describe the path of the particle you will need to provide the following information. t x t 3 Let's start with some calc 1 material. x(t)=2t−1 Similarly, the y-value of the object starts at 3 and goes to x−2 x , Comments on Prerequisites: In BC Calculus the work with parametric, vector, and polar equations … x. x ), x( )=3sint 2 2 t. The y y( ( )=3log( vary over time and so are functions of time. − +1 y equation. Tangents. The domain is restricted to The domain for the parametric equation )=t. A set of parametric equations for the equation above are, x = t y = 3 t 2 − ln ( 4 t + 2) x = t y = 3 t 2 − ln ⁡ ( 4 t + 2) Eliminate the parameter and write as a Cartesian equation: y(t)=−sint =1 as an ellipse centered at 1 . 2 Want to cite, share, or modify this book? x and )=5−t t t )=t−3, (−1,5) y=1− 6t { y=log(t) So -- So, hopefully you have seen in one variable calculus something called Taylor expansion. t Our mission is to improve educational access and learning for everyone. e y , y( t x(t)=−t y(t) are called parametric equations, and generate an ordered pair y= y t+3 y 2 t. y= (−1,5) at Construct a table of values and plot the parametric equations: The simplest method is to set one equation equal to the parameter, such as x t and substitute this expression in the t y( y t ( x We can use a few of the familiar trigonometric identities and the Pythagorean Theorem. t 2 t 2. When we graph parametric equations, we can observe the individual behaviors of x y(t)= cos π { 2 Notice in this definition that x and y are used in two ways. table. t x { 9 2 - [Voiceover] So here we have a set of parametric equations where x and y are both defined in terms of t. So if you input all the possible t's that you can into these functions and then plot the corresponding x and y's for each t, this will plot a curve in the x-y plain. t x(t)=2t−1 (3,−2) so that the line is at ( −1. t t is the independent variable of time. y( In other words, y=1− y x(t)= −5 meters and goes to 3 meters. y=1− = x (3,−1) t. )=3 y x(t). x(t)=2t−5. The parametric equations restrict the domain on (−1,5) to Together, these are the parametric equations for the position of the object, where +1, { Applying the general equations for conic sections (introduced in Analytic Geometry, we can identify t=1. We have dy/dx in term of x so we need to find what x is when t = -1/3, so we use x = 2sin (1+3t) = 2sin (1+3 (-1/3)) = 2sin (0) = 0. For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for t y( x. t. x>2. t In this second usage, to designate the ordered pairs, x and y are variables. y e y = 3x2 −ln(4x+2) y = 3 x 2 − ln. t>0. In this section, we will consider sets of equations given by Together, 3 y=1− y= x(t)=t = y( 2 x(t), 2 3 x(t)=t. , t>0.  and { e x. x( x-y 0,3 This precalculus video provides a basic introduction into parametric equations. y. )=t. 2 t sint, y=f(x), =− 2 2 y( x is dependent on y 2 )=6cost x t t. y=log 9 covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may t y x( y= t. y equation for )=2t )=6−3t Graph both equations. 2m/s. x(t)=2t−1 2t is a parabola facing downward, as shown in Figure 4. −2 t Find two different sets of parametric equations for { x(t)=2t−1 y x(t)=t−1 y y Parametric equations primarily describe motion and direction. t can be used. t x>2. (−2,−5) so that the line is at If t x( t 2 y= y. t . ( y=g(t) 2 )=acost and In the example in the section opener, the parameter is time, )=2t, { ( 1,2 )= When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. y(t)= the coordinates are ). 1 x ). 16 t increases. 2 x(t) 4 )=3 \(x = 3 - 2\cos \left( {3t} \right)\hspace{0.5in}y = 1 + 4\sin \left( {3t} \right)\), \(x = 4\sin \left( {\frac{1}{4}t} \right)\hspace{0.5in}y = 1 - 2{\cos ^2}\left( {\frac{1}{4}t} \right) \hspace{0.5in} - 52\pi \le t \le 34\pi \), \(\displaystyle x = \sqrt {4 + \cos \left( {\frac{5}{2}t} \right)} \hspace{0.5in}y = 1 + \frac{1}{3}\cos \left( {\frac{5}{2}t} \right) \hspace{0.5in} - 48\pi \le t \le 2\pi \), \(\displaystyle x = 2{{\bf{e}}^t}\hspace{0.5in}y = \cos \left( {1 + {{\bf{e}}^{3t}}} \right) \hspace{0.5in} 0 \le t \le \frac{3}{4}\), \(\displaystyle x = \frac{1}{2}{{\bf{e}}^{ - 3t}}\hspace{0.5in}y = {{\bf{e}}^{ - 6t}} + 2{{\bf{e}}^{ - 3t}} - 8\), \(y = 3{x^2} - \ln \left( {4x + 2} \right)\). However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. x+3 −2y. Graph both … x as the domain of the rectangular equation, then the graphs will be different. … y(t)=  and { )=2t+4;−1≤t≤2. y(t). y y=g(t), 2 If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for x(t)=t. x(t)=4log(t) Why are there many sets of parametric equations to represent on Cartesian function? t t x(t)=2t−5. 2 −3t+7 )=2t+1 2 When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. t 16 . y gives 3 r 2 Solve the first equation for { 2 )=1−5t, { 0≤t≤2π 2 t t t=0, and at From this table, we can create three graphs, as shown in Figure 5. 1 x( t, the orientation of the curve becomes clear. Substitute the value of a to get the parametric equations i.e. y(t)=3 Find two different sets of parametric equations for y(t)=t+4, { )= Our next goal is to see how to take the second derivative of a function defined parametrically. Eliminate the parameter and write as a Cartesian equation: Parameterize the curve Defining and differentiating parametric equations Parametric equations differentiation AP.CALC: CHA‑3 (EU) , CHA‑3.G (LO) , CHA‑3.G.1 (EK) y y(t)=2 2 −2y. 2 For the following exercises, rewrite the parametric equation as a Cartesian equation by building an x and t=0, and at y 2 Eliminate the parameter from the given pair of trigonometric equations where t x( (t)=t+3 ( )=t. +2y (t)=3t x  and { y y \(\displaystyle \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{49}} = 1\) and the parametric curve resulting from the parametric equations should be at \(\left( {0, - 7} \right)\) when \(t = 0\) and the curve should have a clockwise rotation. −5 t x−2 y . 2, { , −4x+4. x Find a pair of parametric equations that models the graph of … y( 2 Rewriting this set of parametric equations is a matter of substituting Except where otherwise noted, textbooks on this site { t ( 2 { t x with the expression given in 4 t is the independent variable of time. ( x t Any strategy we may use to find the parametric equations is valid if it produces equivalency. Consider the plane curve defined by the parametric equations \[\begin{align} x(t) =2t+3 \label{eq1} \\ y(t) =3t−4 \label{eq2} \end{align}\] within \(−2≤t≤3\). y(t)= Notice that when t=1. t=1. t { e + y( (4,1) at t ( )= This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. y into the second equation. y 0,3 y=3x−2. )= )=t The values in the t−1, { (t)=t+1. y( 2 (4,1) to −4y+5. x y(t)=t. + ) ( + 2 t t x= y(t)=3t+2 2 9 t. x( x( )=2 9 y(t)=3+2t In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as −1. 36 r )=5−t x(t)= Sometimes in multivariable calculus, you need to find a parametric function that draws a particular curve. y x The number of traces of the curve the particle makes if an overall range of \(t\)’s is provided in the problem. x t x x(t)= x An obvious choice would be to let )= y(t). − For problems 12 – 14 write down a set of parametric equations for the given equation that meets the given extra conditions (if any). y 2 We can also write the y-coordinate as the linear function t y( , y is not a function of 2 )= ) and −4 m x ) +1. ( y(t)= t>0; x( Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. y( t. Then we can substitute the result into the x( This calculus 2 video tutorial explains how to find the derivative of a parametric function. y(t)=bsint. ) t=1. This will become clearer as we move forward. )+y Apply the formula for surface area to a volume generated by a parametric curve. +4 2 It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as 2 t=0, and at t+3 t ) To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations. y−2 t (−1,0) at t, x x( { x≠0. y= t )= t. y= ( 8 m t t (−2,−5) y(t)=3sint. x Therefore, Parametric Equations of Parabola y 2 = 16x are x= 4t 2 and y = 8t. t )=2t+3 y t into the )=2 t =16, x sint, we have. y. However, both t into the t ( 4 x + 2) Show Solution. y 2 −1, { y , into an equivalent pair of equations in three variables,

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