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

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

So $$t^{2}-4 < 0$$ which gives $$t<-2\mbox{ or }t > 2$$ 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?

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 $$\tan (t_1)=\tan (t_2),\, t_1<>t_2$$ 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 $$t_1<>t_2$$ 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 $$t_1<>t_2$$ 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

$t_1<>t_2$ 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.