The previous section defined curves based on parametric equations. We also put in a few values of $$t$$ just to help illustrate the direction of motion. y = cos ⁡ ( 4 t) y=\cos (4t) y = cos(4t) y, equals, cosine, left parenthesis, 4, t, right parenthesis. The direction vector from (x0,y0) to (x1,y1) is. Comparing (a) and (b) we see that the graph of a curve may have more than one parametrization. \end{eqnarray*} Here, the parameter $\theta$ represents the polar angle of the position on a circle of radius $3$ centered at the origin and oriented counterclockwise. 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. A curve in the plane is defined parametrically by the equations. We can’t just jump back up to the top point or take a different path to get there. This is a fairly important set of parametric equations as it used continually in some subjects with dealing with ellipses and/or circles. Finally, even though there may not seem to be any reason to, we can also parameterize functions in the form $$y = f\left( x \right)$$ or $$x = h\left( y \right)$$. We have one more idea to discuss before we actually sketch the curve. However, we’ll need to note that the $$x$$ already contains a $${\sin ^2}t$$ and so we won’t need to square the $$x$$. We shall apply the methods for Cartesian coordinates to ﬁnd their generalized statements when using parametric equations instead. So, how can we eliminate the parameter here? Since $\frac{dx}{dt}=\frac{t}{\sqrt{t^2+1}}$ and $\frac{dy}{dt}=1+\ln t$ $$\frac{dy}{dx}=\frac{ dy/dt}{ dx/dt}=\frac{1+\ln t}{t/\sqrt{t^2+1}}=\frac{\sqrt{t^2+1}(1+\ln t)}{t}. CALCULUS BC WORKSHEET ON PARAMETRIC EQUATIONS AND GRAPHING Work these on notebook paper. . Exercise. However, the curve only traced out in one direction, not in both directions. Starting at $$\left( {5,0} \right)$$ no matter if we move in a clockwise or counter-clockwise direction $$x$$ will have to decrease so we haven’t really learned anything from the $$x$$ derivative. When we parameterize a curve, we are translating a single equation in two variables, such as $x$ and $y$, into an equivalent pair of equations in three variables, $x,y$, and $t$. So, plug in the coordinates for the vertex into the parametric equations and solve for $$t$$. Be careful with the above reasoning that the oscillatory nature of sine/cosine forces the curve to be traced out in both directions. Now, let’s write down a couple of other important parameterizations and all the comments about direction of motion, starting point, and range of $$t$$’s for one trace (if applicable) are still true. Parametric Equations and Polar Coordinates. Find the slope of the tangent line to the curve x=2\sin \theta, y=3\cos \theta at the point corresponding to the value of the parameter \theta=\pi/4. That doesn’t help with direction much as following the curve in either direction will exhibit both increasing and decreasing $$x$$. His work helps others learn about subjects that can help them in their personal and professional lives.$$ Next, \begin{align} & \frac{d}{dt}\left[ \frac{(t^2+1)^{1/2}(1+\ln t)}{t}\right] \\ & \qquad = \frac{t\left[ \frac{1}{2} (t^2+1)^{-1/2} (2t) (1+\ln t)+\frac{ \left( t^2+1\right)^{1/2}}{t}\right]-(t^2+1)^{1/2}(1+\ln t)}{t^2} \\ & \qquad =\frac{ \frac{t^2(1+\ln t)}{(t^2+1)^{1/2}}+(t^2+1)^{1/2}-(t^2+1)^{1/2}(1+\ln t)}{t^2} \\ & \qquad = \frac{ \frac{t^2(1+\ln t)}{(t^2+1)^{1/2}}+\frac{t^2+1}{(t^2+1)^{1/2}}-\frac{(t^2+1)(1+\ln t)}{(t^2+1)^{1/2}}}{t^2} \\ & \qquad = \frac{t^2+t^2\ln t+t^2+1-t^2-t^2\ln t-1-\ln t}{t^2(t^2+1)^{1/2}} \\ & \qquad = \frac{t^2-\ln t}{t^2\sqrt{t^2+1}} \end{align} and so \frac{d^2y}{dx^2}=\frac{ \frac{d}{dt}\left( \frac{dy}{dx}\right)}{ \frac{dx}{dt}} = \frac{t^2-\ln t}{t^2\sqrt{t^2+1}}\cdot \frac{\sqrt{t^2+1}}{t}=\frac{t^2-\ln t}{t^3}. In this case, these also happen to be the full limits on $$x$$ and $$y$$ we get by graphing the full ellipse. Often we would have gotten two distinct roots from that equation. The first is direction of motion. However, there are times in which we want to go the other way. Note that the $$x$$ derivative isn’t as useful for this analysis as it will be both positive and negative and hence $$x$$ will be both increasing and decreasing depending on the value of $$t$$. Given the nature of sine/cosine you might be able to eliminate the diamond and the square but there is no denying that they are graphs that go through the given points. Notice that with this sketch we started and stopped the sketch right on the points originating from the end points of the range of $$t$$’s. (a) Sketch the graph of the curve $$C_1: x=t, y=1-t$$ on $[0,1]$ by plotting values for $t;$ and then check your graph by finding an equation in $x$ and $y$ only and then graphing. The table seems to suggest that between each pair of values of $$t$$ a quarter of the ellipse is traced out in the clockwise direction when in reality it is tracing out three quarters of the ellipse in the counter-clockwise direction. Given a function or equation we might want to write down a set of parametric equations for it. Some authors choose to use x(t) and y(t), but this can cause confusion. up the path. Here are a few of them. EXAMPLE 10.1.1 Graph the … Can you think of another set of parametric equations that gives the same graph? Sketch the graph of the curve $$C_3: x=\cos 2t, y=\sin 2t$$ on $\left[0,\frac{\pi }{2}\right]$ by plotting values for $t;$ and then check your graph by finding an equation in $x$ and $y$ only and then graphing. Each value of t t defines a point (x,y) = (f (t),g(t)) ( x, y) = ( f ( t), g ( t)) that we can plot. Parametric equations are commonly used to express the coordinates of the points that make up a geometric object such as a curve or surface, in which case the equations are collectively called a parametric representation or parameterization of the object. As $t$ increases from $t=a$ to $t =b$, the particle traverses the curve in a specific direction called the orientation of a curve, eventually ending up at the terminal point $(f(b), g(b))$ of the curve. So we now know that we will have an ellipse. Section 10.3 Calculus and Parametric Equations. This is directly counter to our guess from the tables of values above and so we can see that, in this case, the table would probably have led us to the wrong direction. We begin by sketching the graph of a few parametric equations. So, we see that we will be at the bottom point at. Solution. and. Therefore, $2x^2=3$ and so $x=\pm \sqrt{3/2}$ and $y=\pm \sqrt{1/2}.$. 1. We’d be correct. Increasing $$t$$ again until we reach $$t = 3\pi$$ will take us back down the curve until we reach the bottom point again, etc. The best method, provided it can be done, is to eliminate the parameter. Exercise. In this case, the parametric curve is written ( x ( t ); y ( t ); z ( t )), which gives the position of the particle at time t . If we were moving in a clockwise direction from the point $$\left( {5,0} \right)$$ we can see that $$y$$ would have to decrease! We are still interested in lines tangent to points on a curve. . We only have cosines this time and we’ll use that to our advantage. In this quadrant the $$y$$ derivative tells us nothing as $$y$$ simply must decrease to move from $$\left( {0,2} \right)$$. In this section we'll employ the techniques of calculus to study these curves. Our pair of parametric equations is. We will often use parametric equations to describe the path of an object or particle. We’ll start by eliminating the parameter as we did in the previous section. Example. Find the rectangular equations for the curve represented by $(1) \quad x=4\cos \theta$ and $y=3\sin\theta$, $0\leq \theta \leq 2\pi$.$(2) \quad x=\sin t$ and $y=\sin2t$, $0\leq t \leq 2\pi$.$(3) \quad C: x=t^2$, $y=t-1$; $0\leq t \leq 3$$(4) \quad C: x=t^2+1, y=2t^2-1; -2\leq t\leq 2, Exercise. If $$n > 1$$ we will increase the speed and if $$n < 1$$ we will decrease the speed. It will also be useful to calculate the differential of $$x$$: In the above formula, f(t) and g(t) refer to x and y, respectively. Definition 4.1.2. Parametric equations are a set of functions of one or more independent variables called parameters and are used to express the coordinates of the points that make up a geometric object such as a curve or surface. It is fairly simple however as this example has shown. As you can probably see there are an infinite number of ranges of $$t$$ we could use for one trace of the curve. At this point our only option for sketching a parametric curve is to pick values of $$t$$, plug them into the parametric equations and then plot the points. We should always find limits on $$x$$ and $$y$$ enforced upon us by the parametric curve to determine just how much of the algebraic curve is actually sketched out by the parametric equations. We’ll discuss an alternate graphing method in later examples that will help to explain how these values of $$t$$ were chosen. So, once again, tables are generally not very reliable for getting pretty much any real information about a parametric curve other than a few points that must be on the curve. As noted just prior to starting this example there is still a potential problem with eliminating the parameter that we’ll need to deal with. That is not correct however. All “fully traced out” means, in general, is that whatever portion of the ellipse that is described by the set of parametric curves will be completely traced out.$$ (a) once clockwise (b) once counterclockwise (c) twice clockwise (d) twice counterclockwise. We have the $$x$$ and $$y$$ coordinates of the vertex and we also have $$x$$ and $$y$$ parametric equations for those coordinates. x2+y2 = 36 x 2 + y 2 = 36 and the parametric curve resulting from the parametric equations should be at (6,0) (6, 0) when t =0 t = 0 and the curve should have a counter clockwise rotation. For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation. (b) Find the points on the cardioid where the tangent lines are horizontal and where the tangent lines are vertical. The reality is that when writing this material up we actually did this problem first then went back and did the first problem. Each parameterization may rotate with different directions of motion and may start at different points. To find the slope of the tangent line to the graph of$r=f(\theta)$at the point$P(r, \theta)$, let$P(x, y)$be the rectangular representation of$P$. (a) Find a rectangular equation whose graph contains the curve$C$with the parametric equations $$x=\frac{2t}{1+t^2} \qquad \text{and}\qquad y=\frac{1-t^2}{1+t^2}$$ and (b) sketch the curve$C$and indicate its orientation. d=Va*t, where d is the distance,and Va means the average velocity. Parametric equations are incredibly intuitive, especially when the parameter represents time. Let’s increase $$t$$ from $$t = 0$$ to $$t = \frac{\pi }{2}$$. There are also a great many curves out there that we can’t even write down as a single equation in terms of only $$x$$ and $$y$$. Don’t Think About Time. Let’s take a look at just what that change is as it will also answer what “went wrong” with our table of values. Using implicit differentiation we have, $$2 \left(x^2+y^2\right) \left(2x+2y\frac{dy}{dx}\right)=8x-8y\frac{dy}{dx}$$ and so $$\frac{dy}{dx}=-\frac{x \left(-2+x^2+y^2\right)}{y \left(2+x^2+y^2\right)}\$$ we need to find all$(x,y)$where$dy/dx=0.$Clearly, the point$(0,0)$is ruled out and so$-2+x^2+y^2=0$; that is$x^2+y^2=2.$Using$x^2+y^2=2$with the original we see$x^2-y^2=1$also. Therefore, it is best to not use a table of values to determine the direction of motion. Tangent lines to parametric curves and motion along a curve is discussed. How do you find the parametric equations for a line segment? The problem is that tables of values can be misleading when determining a direction of motion as we’ll see in the next example. Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Parametric Equations Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as "parameters." Finding Parametric Equations for Curves Defined by Rectangular Equations. Exercise. This, in turn means that both $$x$$ and $$y$$ will oscillate as well. Calculus of Parametric Equations July Thomas , Samir Khan , and Jimin Khim contributed The speed of a particle whose motion is described by a parametric equation is given in terms of the time derivatives of the x x x -coordinate, x ˙ , \dot{x}, x ˙ , and y y y -coordinate, y ˙ : \dot{y}: y ˙ : In this section we'll employ the techniques of calculus to study these curves. So, it is clear from this that we will only get a portion of the parabola that is defined by the algebraic equation. Calculus; Parametric Differentiation; Parametric Differentiation . We just didn’t compute any of those points. If we set the $$y$$ coordinate equal to zero we’ll find all the $$t$$’s that are at both of these points when we only want the values of $$t$$ that are at $$\left( {5,0} \right)$$. Find equations of the tangent lines to the curve at that point. Consider the cardioid$r= 1 + \cos \theta$. So, why did our table give an incorrect impression about the direction? Also note that they won’t all start at the same place (if we think of $$t = 0$$ as the starting point that is). They are. Note that while this may be the easiest to eliminate the parameter, it’s usually not the best way as we’ll see soon enough. Therefore, in the first quadrant we must be moving in a counter-clockwise direction. In Example 4 we were graphing the full ellipse and so no matter where we start sketching the graph we will eventually get back to the “starting” point without ever retracing any portion of the graph. A reader pointed out that nearly every parametric equation tutorial uses time as its example parameter. Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Show the orientation of the curve. Let’s move on to the second quadrant. The curve starts at$(1,0)$and follows the upper part of the unit circle until it reaches the other endpoint of$(-1,0).$Can you think of another set of parametric equations that give the same graph? Problem Statement Solution So, we are now at the point $$\left( {0,2} \right)$$ and we will increase $$t$$ from $$t = \frac{\pi }{2}$$ to $$t = \pi$$. Take, for example, a circle. Unfortunately, we usually are working on the whole circle, or simply can’t say that we’re going to be working only on one portion of it. If x and y are continuous functions of t on an interval I, then the equations. Tangent lines to parametric curves and motion along a curve is discussed. Find an equation for the line tangent to the curve$x=t$and$y=\sqrt{t}$at$t=1/4.$Also, find the value of$\frac{d^2y}{dx^2}$at this point. Outside of that the tables are rarely useful and will generally not be dealt with in further examples. Therefore, we must be moving up the curve from bottom to top as $$t$$ increases as that is the only direction that will always give an increasing $$y$$ as $$t$$ increases. Because of the ideas involved in them we concentrated on parametric curves that retraced portions of the curve more than once. Sure enough from our Algebra knowledge we can see that this is a parabola that opens to the right and will have a vertex at $$\left( { - \frac{1}{4}, - 2} \right)$$. What, if anything, can be said about the values of$g'(-5)$and$f'(g(-5))?$. The collection of points that we get by letting $$t$$ be all possible values is the graph of the parametric equations and is called the parametric curve. Doing this gives. In the previous example we didn’t have any limits on the parameter. Most of these types of problems aren’t as long. Section 9.3 Calculus and Parametric Equations ¶ permalink. We will need to be very, very careful however in sketching this parametric curve. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities. Calculus with Parametric equations Let Cbe a parametric curve described by the parametric equations x = f(t);y = g(t). Parametric Equations - examples, solutions, practice problems and more. To correctly determine the direction of motion we’ll use the same method of determining the direction that we discussed after Example 3. Because the “end” points on the curve have the same $$y$$ value and different $$x$$ values we can use the $$x$$ parametric equation to determine these values. Note that in the process of determining a range of $$t$$’s for one trace we also managed to determine the direction of motion for this curve. Note that this is not always a correct analogy but it is useful initially to help visualize just what a parametric curve is. This precalculus video provides a basic introduction into parametric equations. Rewrite the equation as . In other words, changing the argument from $$t$$ to 3$$t$$ increase the speed of the trace and the curve will now trace out three times in the range $$0 \le t \le 2\pi$$! So, let’s plug in some $$t$$’s. We will often give the value of $$t$$ that gave specific points on the graph as well to make it clear the value of $$t$$ that gave that particular point. How do you find the parametric equations for a line segment? Calculus of Parametric Equations July Thomas , Samir Khan , and Jimin Khim contributed The speed of a particle whose motion is described by a parametric equation is given in terms of the time derivatives of the x x x -coordinate, x ˙ , \dot{x}, x ˙ , and y y y -coordinate, y ˙ : \dot{y}: y ˙ : When we are dealing with parametric equations involving only sines and cosines and they both have the same argument if we change the argument from $$t$$ to nt we simply change the speed with which the curve is traced out. Exercise. In mathematics, a parametric equation defines a group of quantities as functions of one or more independent variables called parameters. In some cases, only one of the equations, such as this example, will give the direction while in other cases either one could be used. The point $$\left( {x,y} \right) = \left( {f\left( t \right),g\left( t \right)} \right)$$ will then represent the location of the ping pong ball in the tank at time $$t$$ and the parametric curve will be a trace of all the locations of the ping pong ball. The only way for that to happen on this particular this curve will be for the curve to be traced out in both directions. The speed of the tracing has increased leading to an incorrect impression from the points in the table. Since x^2+y^2 =\frac{4t^2+(1-2t^2+t^4)}{(1+t^2)^2} =\frac{1+2t^2+t^4}{(1+t^2)^2}=1 and also$x(0)=0$,$y(0)=0$and$x(1/2)=4/5$,$y(1/2)=3/5$we see the graph of the given parametric equations represents the unit circle with orientation counterclockwise. 9.3 Parametric Equations Contemporary Calculus 1 9.3 PARAMETRIC EQUATIONS Some motions and paths are inconvenient, difficult or impossible for us to describe by a single function or formula of the form y = f(x). It is also possible that, in some cases, both derivatives would be needed to determine direction. Consider the plane curve defined by the parametric equations $$x=x(t)$$ and $$y=y(t)$$. Here is a quick sketch of the portion of the parabola that the parametric curve will cover. We’ll solve one of the of the equations for $$t$$ and plug this into the other equation. We can usually determine if this will happen by looking for limits on $$x$$ and $$y$$ that are imposed up us by the parametric equation. Exercise. Further increasing $$t$$ takes us back down the path, then up the path again etc. x = 8 e 3 t. x=8e^ {3t} x = 8e3t. Calculus. Recall we said that these tables of values can be misleading when used to determine direction and that’s why we don’t use them. The derivatives of the parametric equations are. If the starting/ending point is the same then we generally need to go through the full derivative argument to determine the actual direction of motion. So, if we start at say, $$t = 0$$, we are at the top point and we increase $$t$$ we have to move along the curve downwards until we reach $$t = \pi$$ at which point we are now at the bottom point. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. See videos from Calculus 2 / BC on Numerade In addition,we know that the difference of velocity Vdelta=Vf-Vi=g*t. So,Vf=g*t+Vi,since Vi=0, so Vf=g*t+Vi=g*t+0=g*t. … ( −2 , 3 ) . Do not use your calculator. Finding Parametric Equations for Curves Defined by Rectangular Equations. So, in general, we should avoid plotting points to sketch parametric curves. In Example 10.2.5, if we let $$t$$ vary over all real numbers, we'd obtain the entire parabola. Parametric equations define trajectories in space or in the plane. The second problem with eliminating the parameter is best illustrated in an example as we’ll be running into this problem in the remaining examples. For example, two functions We will sometimes call this the algebraic equation to differentiate it from the original parametric equations. In that case we had sine/cosine in the parametric equations as well. As we will see in later examples in this section determining values of $$t$$ that will give specific points is something that we’ll need to do on a fairly regular basis. If we take Examples 4 and 5 as examples we can do this for ellipses (and hence circles). Now, let’s take a look at another example that will illustrate an important idea about parametric equations. Second Order Linear Equations, take two 18 Useful formulas We have already seen how to compute slopes of curves given by parametric equations—it is how we computed slopes in polar coordinates. This, however, doesn’t really help us determine a direction for the parametric curve. So, we saw in the last two examples two sets of parametric equations that in some way gave the same graph. while Va= (Vf+Vi)/2, where Vf is the final velocity and Vi is the initial velocity (in this case Vi=0). Note that if we further increase $$t$$ from $$t = \pi$$ we will now have to travel back up the curve until we reach $$t = 2\pi$$ and we are now back at the top point. Again, given the nature of sine/cosine you can probably guess that the correct graph is the ellipse. To a volume generated by a parametric curve a parametric equation defines a group quantities! 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Curve and note that it would be acceptable answers for this problem first went... Arrows in both directions as shown in the plane is defined using the same ellipse that we will the! Define function by the pair of parametric equations of lines always look like that danger of sketching a curve! Simple however as this example will also parameterize the function is defined parametrically by the equations, first let s. One way of sketching a parametric equation context are continuous functions of t on an interval,... The parabola that opens upward will not always be able to do this for (. Dt and dy dt we say that we have the following algebraic equation easy to address problems. Of important ideas out of the equation for \ ( 0 \le t \le \pi ). In fact, it won ’ t always be able to... Model motion in the is! ( x0, y0 ) to look at an easier method of determining the direction motion! Plug in some subjects with dealing with ellipses and/or circles it can determined! 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