How Do You Calculate the Concavity and Tangents for Parametric Equations?

[sherlockjones]

1 If x = t^{3} - 12t, y = t^{2} - 1

find \frac{dy}{dx} and \frac{d^{2}y}{dx^{2}}. For what values of t is the curve concave upward.
So \frac{dy}{dx} = \frac{2t}{3t^{2}-12} and

\frac{d^{2}y}{dx^{2}} = \frac{2}{3t^{2}-12}So 3t^{2}-12 > 0 and t > 2 for the curve to be concave upward?

Is this correct?

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

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

So \theta = \frac{\pi}{4}+ \pi n when the tangent is horizontal and \theta = \pi n when the tangent is vertical? Is this correct?
3 At what point does the curve x = 1-2\cos^{2} t, y = (\tan t )(1-2\cos^{2}t) cross itself? Find the equations of both tangent at that point. So I set \tan t = 0 and 1-2\cos^{2}t = 0

 

[HallsofIvy]

1 If x = t^{3} - 12t, y = t^{2} - 1

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



So \frac{dy}{dx} = \frac{2t}{3t^{2}-12}

Yes, \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

and

\frac{d^{2}y}{dx^{2}} = \frac{2}{3t^{2}-12}

How did you get this? \frac{d^2y}{dx^2}= \frac{\frac{d(\frac{dy}{dx})}{dt}}{\frac{dx}{dt}}
I get \frac{d^2y}{dx^2}= -\frac{2}{3}\frac{1}{x^2- 4}

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

Is this correct?

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

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

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

Okay, that looks good.

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

Why? Are you assuming that y= 0 where the curve crosses itself? "Crossing itself" only means that x and y are the same for two or more different values of t. Solve the pair of equations 1- 2cos^2 t= 1- 2cos^2 s and (tan t)(1- 2cos^2 t)= (tan s)(1- 2cos^2 s) for t and s.

 

[benorin]

1 If x = t^{3} - 12t, y = t^{2} - 1

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

So \frac{dy}{dx} = \frac{2t}{3t^{2}-12} and

\frac{d^{2}y}{dx^{2}} = \frac{2}{3t^{2}-12}

For parametric equations, we have \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{3t^{2}-12}

so \frac{dy}{dx} = \frac{2t}{3t^{2}-12} is correct, but

for the second derivative, we consider y as a function of x and apply the chain rule to the first derivative to get

\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)

\Rightarrow\frac{d^2y}{dx^2}\cdot\frac{dx}{dt} = \frac{\frac{d^2y}{dt^2}\frac{dx}{dt}- \frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^2 }

which gives

\frac{d^2y}{dx^2}= \frac{\frac{d^2y}{dt^2}\frac{dx}{dt}- \frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^3} = <br /> \frac{2(3t^2-12)- 2t(6t)}{(3t^2-12)^3}
=-\frac{6}{27}\frac{t^2+4}{(t^2-4)^3}

So t^{2}-4 &lt; 0 which gives t&lt;-2\mbox{ or }t &gt; 2 for the curve to be concave upward.

Is this correct?
2 If x = 2\cos \theta and y = \sin 2\theta find the points on the curve where the tangent is horizontal or vertical.

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

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

Absolutely.

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

Notice that y = x\tan t. Now for two points to be the same for different values of t, say t_1 and t_2, and the y and x values are the same for the values of t so we have

y-y = x\left(\tan (t_1)-\tan (t_2)\right) \Rightarrow \tan (t_1)-\tan (t_2)=0​

so we solve \tan (t_1)=\tan (t_2),\, t_1&lt;&gt;t_2 which has the solution t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots, you can get it from here...

 

[sherlockjones]

So y = x\tan t_{1} and y = x\tan t_{2} \Rightarrow \tan t_{1} = \tan t_{2}.

What does t_1&lt;&gt;t_2 mean? Also how did you get the solution t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots. Do I just substitute this back in for the equations for x and y? How would you find the equations of both tangents? I have only one point.

Thanks

 

[HallsofIvy]

So y = x\tan t_{1} and y = x\tan t_{2} \Rightarrow \tan t_{1} = \tan t_{2}.

What does t_1&lt;&gt;t_2 mean? Also how did you get the solution t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots. Do I just substitute this back in for the equations for x and y? How would you find the equations of both tangents? I have only one point.

Thanks

t_1&lt;&gt;t_2 means "t₁ is NOT equal to t₂. It's a computer programming notation.

He got t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots by remembering that tan(x) is periodic with period \pi!

Well, you can't substitute t_1 =t_2+k\pi,\, \, k=\pm 1, \pm2,\ldots because you don't know what t₂ is! Substitute those back into your other equation to solve for both t₁ and t₂. Use either to determine x and y.

 

[sherlockjones]

Sorry for my stupidity, but do you mean to do this:

\tan t_{1} = \tan (t_{1}-k\pi)?

t_{1} = \arctan (t_{1}-k\pi)

 

[HallsofIvy]

Sorry for my stupidity, but do you mean to do this:

\tan t_{1} = \tan (t_{1}-k\pi)?

Yes, that's true because tangent has period \pi

t_{1} = \arctan (t_{1}-k\pi)

No, that's not true. Even
arctan(tan t_1)= \arctan(tan(t_1- k\pi))
so that t_1= t_1- k\pi isn't correct. Since tangent is periodic it is not one-to-one and arctan is not a true inverse.