Example. Now, at \(t = 0\) we are at the point \(\left( {5,0} \right)\) and let’s see what happens if we start increasing \(t\). Unless we know what the graph will be ahead of time we are really just making a guess. Find an equation of the tangent line to the curve defined by the parametric equations $x=e^t$ and $y=e^{-t}$ at the point $(1,1).$ Then sketch the curve and the tangent line(s). 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. More than one parameter can be employed when necessary. That parametric curve will never repeat any portion of itself. So, we saw in the last two examples two sets of parametric equations that in some way gave the same graph. And, I hope you see it's not extremely hard. Then the derivative \(\dfrac{dy}{dx}\)is given by \[\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{y′(t)}{x′(t)}. The curve does change in a small but important way which we will be discussing shortly. \end{align} as desired. We did include a few more values of \(t\) at various points just to illustrate where the curve is at for various values of \(t\) but in general these really aren’t needed. 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 \(\left( { - \frac{1}{4}, - 2} \right)\). Therefore, it is best to not use a table of values to determine the direction of motion. Very often we can think of the trajectory as that of a particle moving through space and the parameter as time. 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. Here is the sketch of this parametric curve. You may find that you need a parameterization of an ellipse that starts at a particular place and has a particular direction of motion and so you now know that with some work you can write down a set of parametric equations that will give you the behavior that you’re after. The equation involving only \(x\) and \(y\) will NOT give the direction of motion of the 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. Parametric equations provide us with a way of specifying the location \((x,y,z)\) ... We're now ready to discuss calculus on parametric curves. Since $\frac{dx}{d\theta}=2(1-\cos \theta)$ and $\frac{dy}{d\theta}=2\sin \theta.$ The slope of the tangent line at $\theta=\pi/6$ is\begin{align} \left.\frac{dy}{dx}\right|{\theta=\pi/6} & =\left.\frac{dy/d\theta}{dx/d\theta}\right|{\theta=\pi/6} \\ & =\left.\frac{2\sin \theta}{2(1-\cos \theta)}\right|_{\theta=\pi/6} =\frac{1}{2\left.(1-\frac{\sqrt{3}}{2}\right)}=2+\sqrt{3}. All we need to do is graph the equation that we found by eliminating the parameter. If we had put restrictions on which \(t\)’s to use we might really have ended up only moving in one direction. We also have the following limits on \(x\) and \(y\). We will need to be very, very careful however in sketching this parametric curve. And, I hope you see it's not extremely hard. Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. Doing this gives the following equation and solution. Section 9.3 Calculus and Parametric Equations ¶ permalink. We’ll eventually see an example where this happens in a later section. This is not the only range that will trace out the curve however. 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. In some of the later sections we are going to need a curve that is traced out exactly once. To get the direction of motion it is tempting to just use the table of values we computed above to get the direction of motion. This is why the table gives the wrong impression. 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 ˙ : Show the orientation of the curve. In that case we had sine/cosine in the parametric equations as well. 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. We just didn’t compute any of those points. 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 \(t = \pi \). Note as well that the last two will trace out ellipses with a clockwise direction of motion (you might want to verify this). It is now time to take a look at an easier method of sketching this parametric curve. Well recall that we mentioned earlier that the 3\(t\) will lead to a small but important change to the curve versus just a \(t\)? However, that is all that would be at this point. 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. Let’s work with just the \(y\) parametric equation as the \(x\) will have the same issue that it had in the previous example. Every curve can be parameterized in more than one way. Therefore, we will continue to move in a counter‑clockwise motion. 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. 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. 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}. 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 \(0 \le t \le 2\pi \). x, equals, 8, e, start superscript, 3, t, end superscript. Calculus; Parametric Differentiation; Parametric Differentiation . 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. 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). 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 \(\left( {5,0} \right)\). 1. Given the ellipse. Then, the given equation can be rewritten as y = t 2 + 5 . Use the equation for arc length of a parametric curve. 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. Doing this gives. 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 \(y\). To finish the problem then all we need to do is determine a range of \(t\)’s for one trace. Take, for example, a circle. Then eliminate the parameter. Before we end this example there is a somewhat important and subtle point that we need to discuss first. Each parameterization may rotate with different directions of motion and may start at different points. Contrast this with the sketch in the previous example where we had a portion of the sketch to the right of the “start” and “end” points that we computed. The previous section defined curves based on parametric equations. Parametric equations 1. Some authors choose to use x(t) and y(t), but this can cause confusion. Let’s see if our first impression is correct. In the above formula, f(t) and g(t) refer to x and y, respectively. We have the \(x\) and \(y\) coordinates of the vertex and we also have \(x\) and \(y\) parametric equations for those coordinates. For example, two functions His work helps others learn about subjects that can help them in their personal and professional lives. That however, can only happen if we are moving in a counter‑clockwise direction. Example. Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. Each formula gives a portion of the circle. The previous section defined curves based on parametric equations. Here’s a final sketch of the curve and note that it really isn’t all that different from the previous sketch. 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. 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. The only way for this to happen is if the curve is in fact tracing out in a counter-clockwise direction initially. If we were moving in a clockwise direction from the point \(\left( {5,0} \right)\) we can see that \(y\) would have to decrease! So, let’s plug in some \(t\)’s. Let’s take a look at an example to see one way of sketching a parametric curve. x = 8 e 3 t. x=8e^ {3t} x = 8e3t. So, it looks like we have a parabola that opens to the right. Since $\frac{dx}{dt}=e^t$ and $\frac{dy}{dt}=-e^{-t}$, \begin{equation} \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-e^{-t}}{e^t}=-\frac{1}{e^{2t}}. Exercise. As you can probably see there are an infinite number of ranges of \(t\) we could use for one trace of the curve. 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). It is important to note however that we won’t always be able to do this. Let’s take a quick look at the derivatives of the parametric equations from the last example. Many, if not most parametric curves will only trace out once. 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. Section 10.3 Calculus and Parametric Equations. 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. So, in this case there are an infinite number of ranges of \(t\)’s for one trace. Sketch the graph of the parametric equations $x=3 t$ and $y=9t^2$, then indicate the direction of increasing $-\infty

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