(adsbygoogle = window.adsbygoogle || []).push({}); 1If [tex] x = t^{3} - 12t [/tex], [tex] y = t^{2} - 1 [/tex]

find [tex] \frac{dy}{dx} [/tex] and [tex] \frac{d^{2}y}{dx^{2}} [/tex]. For what values of [itex] t [/itex] is the curve concave upward.

So [tex] \frac{dy}{dx} = \frac{2t}{3t^{2}-12} [/tex] and

[tex] \frac{d^{2}y}{dx^{2}} = \frac{2}{3t^{2}-12} [/tex]

So [tex] 3t^{2}-12 > 0 [/tex] and [tex] t > 2 [/tex] for the curve to be concave upward?

Is this correct?

2If [tex] x = 2\cos \theta [/tex] and [tex] y = \sin 2\theta [/tex] find the points on the curve where the tangent is horizontal or vertical.

So [tex] \frac{dy}{dx} = -\frac{\cos 2\theta}{\sin \theta} [/tex]. The tangent is horizontal when [tex] -\cos 2\theta = 0 [/tex] and vertical when [tex] \sin \theta = 0 [/tex].

So [tex] \theta = \frac{\pi}{4}+ \pi n [/tex] when the tangent is horizontal and [tex] \theta = \pi n [/tex] when the tangent is vertical? Is this correct?

3At what point does the curve [tex] x = 1-2\cos^{2} t [/tex], [tex] y = (\tan t )(1-2\cos^{2}t) [/tex] cross itself? Find the equations of both tangent at that point. So I set [tex] \tan t = 0 [/tex] and [tex] 1-2\cos^{2}t = 0 [/tex]

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# Homework Help: Parametric Equations homework

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