… Eliminating the parameter this time will be a little different. Section 10.3 Calculus and Parametric Equations. Here is an approach which only needs information about dx dt and dy dt. It will always be dependent on the individual set of parametric equations. Doing this gives. Let’s increase \(t\) from \(t = 0\) to \(t = \frac{\pi }{2}\). It is important to remember that each parameterization will trace out the curve once with a potentially different range of \(t\)’s. Here are a few of them. 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. While they may at first seem foreign and confusing, parametric functions are just a more Calculus; Parametric Differentiation; Parametric Differentiation . \end{equation} The point $(1,1)$ corresponds to $t=0.$ The slope of the tangent line is $$ \left.\frac{dy}{dx}\right|_{t=0}=-1, $$ so an equation is $y-1=-1(x-1)$ or just $-x+2.$. The only way for this to happen is if the curve is in fact tracing out in a counter-clockwise direction initially. We’ll see an example of this later. y(t) = (1 −t)y0 +ty1, where 0 ≤ t ≤ 1. \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. All these limits do is tell us that we can’t take any value of \(t\) outside of this range. We can check our first impression by doing the derivative work to get the correct direction. Step-by-Step Examples. The speed of the tracing has increased leading to an incorrect impression from the points in the table. We end with parametric equations expressed in polar form. 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 $P$. It is also possible that, in some cases, both derivatives would be needed to determine direction. Note that the only difference here is the presence of the limits on \(t\). Parametric equations are incredibly intuitive, especially when the parameter represents time. Unless we know what the graph will be ahead of time we are really just making a guess. Solution. Calculus. So, in this case there are an infinite number of ranges of \(t\)’s for one trace. ( −2 , 3 ) . The first few values of \(t\) are then. The problem is that not all curves or equations that we’d like to look at fall easily into this form. 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 in one of these two forms. We’ll solve one of the of the equations for \(t\) and plug this into the other equation. It is always possible that the parametric curve is only a portion of the ellipse. Before discussing that small change the 3\(t\) brings to the curve let’s discuss the direction of motion for this curve. Calculus; Parametric Differentiation; Parametric Differentiation . Exercise. Our pair of parametric equations is. This time the algebraic equation is a parabola that opens upward. We’ll discuss an alternate graphing method in later examples that will help to explain how these values of \(t\) were chosen. EXAMPLE 10.1.1 Graph the … Convert $\left(x^2+y^2\right)^2=4\left(x^2-y^2\right)$ to polar form to find all points on the lemniscate of Bernoulli where the tangent line is horizontal. In this case however, based on the table of values we computed at the start of the problem we can see that we do indeed get the full ellipse in the range \(0 \le t \le 2\pi \). Given a function or equation we might want to write down a set of parametric equations for it. Find an equation for the line tangent to the curve $x=t-\sin t$ and $y=1-\cos t$ at $t=\pi /3.$ Also, find the value of $\frac{d^2y}{dx^2}$ at this point. 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. Both the \(x\) and \(y\) parametric equations involve sine or cosine and we know both of those functions oscillate. The derivative from the \(y\) parametric equation on the other hand will help us. x = 8 e 3 t. x=8e^ {3t} x = 8e3t. So, it is clear from this that we will only get a portion of the parabola that is defined by the algebraic 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. 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. Again, as we increase \(t\) from \(t = 0\) to \(t = \frac{\pi }{2}\) we know that cosine will be positive and so \(y\) must be increasing in this range. We can think of $t$ on the closed interval $[a, b]$ as representing time, and in doing so we can interpret the parametric equations in terms of the motion of a particle as follows: at time $t=a$ the particle is at the initial point $(f(a), g(a))$ of the curve or trajectory $C$. 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. In Example 4 as we trace out the full ellipse both \(x\) and \(y\) do in fact oscillate between their two “endpoints” but the curve itself does not trace out in both directions for this to happen. Okay, that was a really long example. We should give a small warning at this point. In this section we'll employ the techniques of calculus to study these curves. (a) The graph starts at the point $(0,1)$ and follows the line ${y=1-x}$ until it reaches the other endpoint at $(1,0).$ (b) The graph starts at the point $(1,0)$ and follows the line $x=1-y$ until it reaches the other endpoint at $(0,1).$. Parametric equations provide us with a way of specifying the location \((x,y,z)\) ... We're now ready to discuss calculus on parametric curves. Then, the given equation can be rewritten as y = t 2 + 5 . To graph the equations, first we construct a table of values like that in the table below. More than one parameter can be employed when necessary. A sketch of the algebraic form parabola will exist for all possible values of \(y\). 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. So, in general, we should avoid plotting points to sketch parametric curves. 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. This is the second potential issue alluded to above. Assign any one of the variable equal to t . Exercise. Getting a sketch of the parametric curve once we’ve eliminated the parameter seems fairly simple. This set of parametric equations will trace out the ellipse starting at the point \(\left( {a,0} \right)\) and will trace in a counter-clockwise direction and will trace out exactly once in the range \(0 \le t \le 2\pi \). The rest of the examples in this section shouldn’t take as long to go through. Can you think of another set of parametric equations that gives the same graph? Sketching a parametric curve is not always an easy thing to do. The best method, provided it can be done, is to eliminate the parameter. Since $\frac{dx}{dt}=\frac{t}{\sqrt{t^2+1}}$ and $\frac{dy}{dt}=1+\ln t$ \begin{equation} \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}. For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation. Now, we could continue to look at what happens as we further increase \(t\), but when dealing with a parametric curve that is a full ellipse (as this one is) and the argument of the trig functions is of the form nt for any constant \(n\) the direction will not change so once we know the initial direction we know that it will always move in that direction. Example. 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. This, however, doesn’t really help us determine a direction for the parametric curve. So, what is this telling us? Dave4Math » Calculus 1 » Parametric Equations and Calculus (Finding Tangent Lines). y = cos ( 4 t) y=\cos (4t) y = cos(4t) y, equals, cosine, left parenthesis, 4, t, right parenthesis. Notice that we made sure to include a portion of the sketch to the right of the points corresponding to \(t = - 2\) and \(t = 1\) to indicate that there are portions of the sketch there. 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. (say x = t ). In Example 10.2.5, if we let \(t\) vary over all real numbers, we'd obtain the entire parabola. At \(t = 0\) the derivative is clearly positive and so increasing \(t\) (at least initially) will force \(y\) to also be increasing. In fact, this curve is tracing out three separate times. a set of parametric equations for it would be. Applications of Parametric Equations. up the path. The previous section defined curves based on parametric equations. In this case the curve starts at \(t = - 1\) and ends at \(t = 1\), whereas in the previous example the curve didn’t really start at the right most points that we computed. The first question that should be asked at this point is, how did we know to use the values of \(t\) that we did, especially the third choice? From this analysis we can get two more ranges of \(t\) for one trace. Often we would have gotten two distinct roots from that equation. Yet, because they traced out the graph a different number of times we really do need to think of them as different parametric curves at least in some manner. 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. So, first let’s get limits on \(x\) and \(y\) as we did in previous examples. They are. Finding Parametric Equations for Curves Defined by Rectangular Equations. 1. x t y t 2 1 and 1 … Take, for example, a circle. 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. While it is often easy to do we will, in most cases, end up with an equation that is almost impossible to deal with. Here is a quick sketch of the portion of the parabola that the parametric curve will cover. Example. Can you see the problem with doing this? Well back in Example 4 when the argument was just \(t\) the ellipse was traced out exactly once in the range \(0 \le t \le 2\pi \). In the first example we just, seemingly randomly, picked values of \(t\) to use in our table, especially the third value. So, it looks like we have a parabola that opens to the right. Then, using the trig identity from above and these equations we get. 240 Chapter 10 Polar Coordinates, Parametric Equations Just as we describe curves in the plane using equations involving x and y, so can we describe curves using equations involving r and θ. Make a table of values and sketch the curve, indicating the direction of your graph. Do not use your calculator. It is fairly simple however as this example has shown. (b) Find the points on the cardioid where the tangent lines are horizontal and where the tangent lines are vertical. This gives. 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. Had we simply stopped the sketch at those points we are indicating that there was no portion of the curve to the right of those points and there clearly will be. However, in the previous example we’ve now seen that this will not always be the case. The curve does change in a small but important way which we will be discussing shortly. In this case the algebraic equation is a parabola that opens to the left. 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. 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. Now, if we start at \(t = 0\) as we did in the previous example and start increasing \(t\). That parametric curve will never repeat any portion of itself. In this range of \(t\)’s we know that sine is always positive and so from the derivative of the \(x\) equation we can see that \(x\) must be decreasing in this range of \(t\)’s. In this range of \(t\) we know that cosine is negative (and hence \(y\) will be decreasing) and sine is also negative (and hence \(x\) will be increasing). Here is the sketch of this parametric curve. This means that we had to go back
The previous section defined curves based on parametric equations. Exercise. The first one we looked at is a good example of this. Despite the fact that we said in the last example that picking values of \(t\) and plugging in to the equations to find points to plot is a bad idea let’s do it any way. However, at \(t = 2\pi \) we are back at the top point on the curve and to get there we must travel along the path. There really was no apparent reason for choosing \(t = - \frac{1}{2}\). The area between a parametric curve and the x -axis can be determined by using the formula 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. Make a table of values and sketch the curve, indicating the direction of your graph. Definition. Tangent lines to parametric curves and motion along a curve is discussed. In this range of \(t\) we know that cosine is positive (and hence \(y\) will be increasing) and sine is negative (and hence \(x\) will be increasing). We will need to be very, very careful however in sketching this parametric curve. Solution. It is important to note however that we won’t always be able to do this. We won’t bother with a sketch for this one as we’ve already sketched this once and the point here was more to eliminate the parameter anyway. while Va= (Vf+Vi)/2, where Vf is the final velocity and Vi is the initial velocity (in this case Vi=0). Since \begin{equation} 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 \end{equation} 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. … The line segments between (x0,y0) and (x1,y1) can be expressed as: x(t) = (1− t)x0 + tx1. This is generally an easy problem to fix however. A curve in the plane is defined parametrically by the equations. However, we will never be able to write the equation of a circle down as a single equation in either of the forms above. We’ll see in later examples that for different kinds of parametric equations this may no longer be true. The derivative of the \(y\) parametric equation is. Find a parametrization for the curve whose graph is the lower half of the parabola $x-1=y^2.$, Exercise. All travel must be done on the path sketched out. Show the orientation of the curve. Solution. The only differences are the values of \(t\) and the various points we included. Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. Parametric equation, a type of equation that employs an independent variable called a parameter (often denoted by t) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable. Solution. Suppose that $u=g(x)$ is differentiable at $x=-5,$ $y=f(u)$ is differentiable at ${u=g(-5)}$ and $(f\circ g)'(-5)$ is negative. Although rectangular equations in x and y give an overall picture of an object's path, they do not reveal the position of an object at a specific time. Very often we can think of the trajectory as that of a particle moving through space and the parameter as time. Now that we can describe curves using parametric equations, we can analyze many more curves than we could when we were restricted to simple functions. 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. Exercise. However, the parametric equations have defined both \(x\) and \(y\) in terms of sine and cosine and we know that the ranges of these are limited and so we won’t get all possible values of \(x\) and \(y\) here. The only way for that to happen on this particular this curve will be for the curve to be traced out in both directions. Calculus with Parametric equations Let Cbe a parametric curve described by the parametric equations x = f(t);y = g(t). Parametric equations define trajectories in space or in the plane. 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. Note that we put direction arrows in both directions to clearly indicate that it would be traced out in both directions. The derivatives of the parametric equations are. The only way to get from one of the “end” points on the curve to the other is to travel back along the curve in the opposite direction. 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 \(y\). Therefore, $2x^2=3$ and so $x=\pm \sqrt{3/2}$ and $y=\pm \sqrt{1/2}.$. The derivative of a vector valued function is defined using the same definition as first semester calculus. Example. The collection of points that we get by letting t t be all possible values is the graph of the parametric equations and is called the parametric curve. We will sometimes call this the algebraic equation to differentiate it from the original parametric equations. Most common are equations of the form r = f(θ). Parametric Equations and Calculus July 7, 2020 December 22, 2018 Categories Formal Sciences , Mathematics , Sciences Tags Calculus 1 , Latex By David A. Smith , Founder & CEO, Direct Knowledge First, because a circle is nothing more than a special case of an ellipse we can use the parameterization of an ellipse to get the parametric equations for a circle centered at the origin of radius \(r\) as well. (b) Sketch the graph of the curve $$ C_2: x=1-t^2, y=t^2 $$ 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. 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