In order to identify just how much of the ellipse the parametric curve will cover let’s go back to the parametric equations and see what they tell us about any limits on $$x$$ and $$y$$. A table of values of the parametric equations in Example 10.2.7 along with a sketch of their graph.. Eliminate the Parameter, Set up the parametric equation for to solve the equation for . Before we proceed with the rest of the example be careful to not always just assume we will get the full graph of the algebraic equation. All we need to be able to do is solve a (usually) fairly basic equation which by this point in time shouldn’t be too difficult. Exercise. We’ll start by eliminating the parameter as we did in the previous section. 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. 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. Parametric Equations and Polar Coordinates. Each formula gives a portion of the circle. Example. We just had a lot to discuss in this one so we could get a couple of important ideas out of the way. This is why the table gives the wrong impression. Now, from this work we can see that if we use $$t = - \frac{1}{2}$$ we will get the vertex and so we included that value of $$t$$ in the table in Example 1. In the above formula, f(t) and g(t) refer to x and y, respectively. It can only be used in this example because the “starting” point and “ending” point of the curves are in different places. Observe that the curve is at the point $(3,0)$ when $t_1=-\sqrt{3}$ and $t_2=\sqrt{3}$, so the curve crosses itself at the point $(3,0).$, Since $\frac{dx}{dt}=3t^2-3$ and $$\frac{dy}{dt}=\frac{3(t^2-1)}{2t},$$ at the first point of intersection, $$m_1=\left.\frac{dy}{dx}\right|{t=-\sqrt{3}} =\frac{3(3-1)}{2(-\sqrt{3})} =-\sqrt{3}$$ and an equation of the tangent line is $y-0=-\sqrt{3}(x-3)$ or $y=-\sqrt{3}(x-3).$ At the second point, $$m_2 = \left.\frac{dy}{dx}\right|{t=\sqrt{3}}=\frac{3(3-1)}{2(\sqrt{3})}=\sqrt{3}$$ and an equation of the tangent line is $y-0=\sqrt{3}(x-3)$ or $y=\sqrt{3}(x-3).$. Before we move on to other problems let’s briefly acknowledge what happens by changing the $$t$$ to an nt in these kinds of parametric equations. Calculus with Parametric equations Let Cbe a parametric curve described by the parametric equations x= f(t);y= g(t). However, in order for $$x$$ to decrease, as we know it does in this quadrant, the direction must still be moving a counter-clockwise rotation. x, equals, 8, e, start superscript, 3, t, end superscript. We also put in a few values of $$t$$ just to help illustrate the direction of motion. Consider the cardioid $r= 1 + \cos \theta$. 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. Exercise. Do this by sketching the path, determining limits on $$x$$ and $$y$$ and giving a range of $$t$$’s for which the path will be traced out exactly once (provide it traces out more than once of course). The area between a parametric curve and the x -axis can be determined by using the formula It is more than possible to have a set of parametric equations which will continuously trace out just a portion of the curve. To finish the problem then all we need to do is determine a range of $$t$$’s for one trace. 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. For, plugging in some values of $$n$$ we get that the curve will be at the top point at. 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. Example. If the function f and g are dierentiable and y is also a … This, in turn means that both $$x$$ and $$y$$ will oscillate as well. Here is that work. First we find the derivatives of $x$ and $y$ with respect to $t$: $\frac{dx}{dt}=3t^2-3$ and $\frac{dy}{dt}=2t.$ To find the point(s) where the tangent line is horizontal, set $\frac{dy}{dt}=0$ obtaining $t=0.$ Since $\frac{dx}{dt} \neq 0$ at this $t$ value, the required point is $(0,0).$ To find the point(s) where the tangent line is vertical, set $\frac{dx}{dt}=0$ obtaining $t=\pm 1.$ Since $\frac{dy}{dt}\neq 0$ at either of these $t$-values, the required points are $(2,1)$ and $(-2,1).$, Example. To see this effect let’s look a slight variation of the previous example. In Example 10.2.5, if we let $$t$$ vary over all real numbers, we'd obtain the entire parabola. We’ll see an example of this later. However, what we can say is that there will be a value(s) of $$t$$ that occurs in both sets of solutions and that is the $$t$$ that we want for that point. Here are a few of them. Any of them would be acceptable answers for this problem. So, we will be at the right end point at $$t = \ldots , - 2\pi , - \pi ,0,\pi ,2\pi , \ldots$$ and we’ll be at the left end point at $$t = \ldots , - \frac{3}{2}\pi , - \frac{1}{2}\pi ,\frac{1}{2}\pi ,\frac{3}{2}\pi , \ldots$$ . 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 $$y$$. For example, we could do the following. Show the orientation of the curve. Given the ellipse. 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. 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. Therefore, in this case, we now know that we get a full ellipse from the parametric equations. Here is the sketch of this parametric curve. So, because the $$x$$ coordinate of five will only occur at this point we can simply use the $$x$$ parametric equation to determine the values of $$t$$ that will put us at this point. Find $d^2y/dx^2$ given $x=\sqrt{t}$, $y=1/t$. Before addressing a much easier way to sketch this graph let’s first address the issue of limits on the parameter. Applications of Parametric Equations. Note as well that any limits on $$t$$ given in the problem statement can also affect how much of the graph of the algebraic equation we get. Calculus with Parametric Curves . Use the equation for arc length of a parametric curve. Assign any one of the variable equal to t . Outside of that the tables are rarely useful and will generally not be dealt with in further examples. Section 9.3 Calculus and Parametric Equations ¶ permalink. The position of a particle at time $t$ is $(x,y)$ where $x=\sin t$ and $y=\sin^2 t.$ Describe the motion of the particle as $t$ varies over the time interval $[a,b].$, Solution. We can stop here as all further values of $$t$$ will be outside the range of $$t$$’s given in this problem. We begin by sketching the graph of a few parametric equations. Explain how to find velocity, speed, and acceleration from parametric equations. CALCULUS BC WORKSHEET ON PARAMETRIC EQUATIONS AND GRAPHING Work these on notebook paper. 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 should give a small warning at this point. From a quick glance at the values in this table it would look like the curve, in this case, is moving in a clockwise direction. Let’s see if our first impression is correct. 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? Before we end this example there is a somewhat important and subtle point that we need to discuss first. Namely. To graph the equations, first we construct a table of values like that in the table below. From this analysis we can get two more ranges of $$t$$ for one trace. Suppose that $$x′(t)$$ and $$y′(t)$$ exist, and assume that $$x′(t)≠0$$. 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. For … A reader pointed out that nearly every parametric equation tutorial uses time as its example parameter. OK, so that's our first parametric equation of a line in this class. Solution. 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. 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. Exercise. To finish the sketch of the parametric curve we also need the direction of motion for the curve. It’s starting to look like changing the $$t$$ into a 3$$t$$ in the trig equations will not change the parametric curve in any way. Tangent lines to parametric curves and motion along a curve is discussed. 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. How do you find the parametric equations for a line segment? So, we are now at the point $$\left( {0,2} \right)$$ and we will increase $$t$$ from $$t = \frac{\pi }{2}$$ to $$t = \pi$$. Exercise. It is fairly simple however as this example has shown. The set of points obtained as t varies over the interval I is called the graph of the parametric equations. are called parametric equations and t is called the parameter. David Smith is the CEO and founder of Dave4Math. Calculus. \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. Well recall that we mentioned earlier that the 3$$t$$ will lead to a small but important change to the curve versus just a $$t$$? In this section we'll employ the techniques of calculus to study these curves. In practice however, this example is often done first. Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t: x = t. y = -3t +1.5 Exercise. The derivative from the $$y$$ parametric equation on the other hand will help us. As noted already however, there are two small problems with this method. Getting a sketch of the parametric curve once we’ve eliminated the parameter seems fairly simple. up the path. 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. 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. Parametric curves have a direction of motion. Nothing actually says unequivocally that the parametric curve is an ellipse just from those five points. In this case, we would guess (and yes that is all it is – a guess) that the curve traces out in a counter-clockwise direction. Let’s increase $$t$$ from $$t = 0$$ to $$t = \frac{\pi }{2}$$. Each value of t t defines a point (x,y) = (f (t),g(t)) ( x, y) = ( f ( t), g ( t)) that we can plot. We are now at $$\left( { - 5,0} \right)$$ and we will increase $$t$$ from $$t = \pi$$ to $$t = \frac{{3\pi }}{2}$$. There are many more parameterizations of an ellipse of course, but you get the idea. It is however probably the most important choice of $$t$$ as it is the one that gives the vertex. 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. 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. 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 $t$. The only way for that to happen on this particular this curve will be for the curve to be traced out in both directions. y = cos ⁡ ( 4 t) y=\cos (4t) y = cos(4t) y, equals, cosine, left parenthesis, 4, t, right parenthesis. Because of the ideas involved in them we concentrated on parametric curves that retraced portions of the curve more than once. 10 Curves in the Plane 10.1 Arc Length and Surface Area 10.3 Calculus and Parametric Equations 10.2 Parametric Equations We are familiar with sketching shapes, such as … In the range $$0 \le t \le \pi$$ we had to travel downwards along the curve to get from the top point at $$t = 0$$ to the bottom point at $$t = \pi$$. For the 4th quadrant we will start at $$\left( {0, - 2} \right)$$ and increase $$t$$ from $$t = \frac{{3\pi }}{2}$$ to $$t = 2\pi$$. Calculus and Vectors – How to get an A+ 8.3 Vector, Parametric, and Symmetric Equations of a Line in R3 ©2010 Iulia & Teodoru Gugoiu - Page 2 of 2 D Symmetric Equations The parametric equations of a line may be written as: t R z z tu y y tu In this section we'll employ the techniques of calculus to study these curves. Now, all we need to do is recall our Calculus I knowledge. The line segments between (x0,y0) and (x1,y1) can be expressed as: x(t) = (1− t)x0 + tx1. Most of these types of problems aren’t as long. 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. Do not use your calculator. This is a fairly important set of parametric equations as it used continually in some subjects with dealing with ellipses and/or circles. We begin by sketching the graph of a few parametric equations. We will NOT get the whole parabola. Also note that they won’t all start at the same place (if we think of $$t = 0$$ as the starting point that is). Now, let’s take a look at another example that will illustrate an important idea about parametric equations. Section 10.3 Calculus and Parametric Equations. In a parametric equation, t is the independent variable, and x and y are both dependent variables. Use the equations in the preceding problem to find a set of parametric equations for a circle whose radius is 5 and whose center is (−2, 3). Exercise. Apply the formula for surface area to a volume generated by a parametric curve. To correctly determine the direction of motion we’ll use the same method of determining the direction that we discussed after Example 3. Before we leave this example let’s address one quick issue. So, by starting with sine/cosine and “building up” the equation for $$x$$ and $$y$$ using basic algebraic manipulations we get that the parametric equations enforce the above limits on $$x$$ and $$y$$. 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