Parametric Equations:
They describe how the \(y\)-values are changing with respect to the \(x\)-values, they are useful in making approximations, and they indicate instantaneous direction of travel.The slope of the tangent line is still \(\frac{dy}{dx}\), and the \[\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}.\]\[\frac{dy}{dx} = \frac{dy}{dt}\Bigg/\frac{dx}{dt} = \frac{g^\prime (t)}{f^\prime (t)},\]provided that \(f^\prime (t)\neq 0\). Following the work established in Section 3.4, we look at values of \(t\) greater/less than \(3/5\) on a number line:Reviewing Example 9.3.1, we see that when \(t=3/5=0.6\), the graph of the parametric equations has a vertical tangent line. This is not trivial, as equations that mix polynomials and trigonometric functions generally do not have "nice'' solutions.

The graph of the parametric functions is concave up when \(\frac{d^2y}{dx^2} > 0\) and concave down when \(\frac{d^2y}{dx^2} <0\). It seems reasonable that these lines be defined (one can draw a line tangent to the "right side'' of a circle, for instance), so we add the following to the above definition.Let \(x=5t^2-6t+4\) and \(y=t^2+6t-1\), and let \(C\) be the curve defined by these equations.Find the equation of the tangent line to the astroid \(x=\cos^3 t\), \(y=\sin^3t\) at \(t=0\), shown in Figure 9.31.We can, however, examine the slopes of tangent lines near \(t=0\), and take the limit as \(t\to 0\).We found the slope of the tangent line at \(t=0\) to be 0; therefore the tangent line is \(y=0\), the \(x\)- axis. 3

Viewed 441 times 0 $\begingroup$ The question is: A curve is defined by the parametric equations $$ x = t^2 + a $$ $$ y = t(t-a)^2 $$ Find the range of values for t … x = 2t , y = 3t - 1 Determine the \(t\)- intervals on which the graph is concave up/down.In Example 9.3.1, we found \(\frac{dy}{dx} = \frac{2t+6}{10t-6}\) and \(f^\prime (t) = 10t-6\). 0 0
These approximations are not very good, made only by looking at the graph. Likewise, when the normal line is horizontal, the tangent line is undefined. x

Parametric Equations: 0

t

x = 2t

-3

We continue to analyze curves in the plane by considering their concavity; that is, we are interested in \(\frac{d^2y}{dx^2}\), "the second derivative of \(y\) with respect to \(x\).''

Get the free "Second Parametric Derivative (d^2)y/dx^2" widget for your website, blog, Wordpress, Blogger, or iGoogle.

1 When the tangent line is horizontal, the normal line is undefined by the above definition as \(g^\prime (t_0)=0\).

Using Key Idea 39, we see we need for \(g(t)\geq 0\) on \([-1,1]\), and this is not the case. To do so, we create a new coordinate system, called Gregory Hartman (Virginia Military Institute). Letting \(x=f(t)\) and \(y=g(t)\), we know that \(\frac{dy}{dx} = g^\prime (t)/f^\prime (t)\). where - the derivative of the parametric equation y (t) by the parameter t and - the derivative of the parametric equation x (t), by the parameter t. Our online calculator finds the derivative of the parametrically derined function with step by step solution.

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Inflection Points and Concavity Calculator. Note:In words, to find \(\frac{d^2y}{dx^2}\), we first take the derivative of \(\frac{dy}{dx}\) Let \(x=f(t)\) and \(y=g(t)\) be twice differentiable functions on an open interval \(I\), where \(f^\prime (t)\neq 0\) on \(I\). y = 3t - 1

We want \(\frac{d}{dx}[h(t)]\); that is, we want \(\frac{dh}{dx}\).

It will also be useful to calculate the differential of \(x\):\[dx = f^\prime (t)dt \qquad \Rightarrow \qquad dt = \frac{1}{f^\prime (t)}\cdot dx.\]Starting with the arc length formula above, consider:Note the new bounds (no longer "\(x\)" bounds, but "\(t\)" bounds).

We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined.As the numerator of \( -\frac{9}{(5t-3)^3}\) is never 0, \(\frac{d^2y}{dx^2} \neq 0\) for all \(t\). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University.

In this section we'll employ the techniques of calculus to study these curves. It shows that it has zeros at approximately \(t=0.5,\ 3.5,\ 6.5,\ 9.5,\ 12.5\) and \(16\).

Find the surface area if this shape is rotated about the \(x\)- axis, as shown in Figure 9.3.8.The teardrop shape is formed between \(t=-1\) and \(t=1\).

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Recall in Section 7.4 we found the arc length of the graph of a function, from \(x=a\) to \(x=b\), to be\[L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx.\]We can use this equation and convert it to the parametric equation context. The surface area is: Once again we arrive at an integral that we cannot compute in terms of elementary functions.

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