# Parametric Equations homework

• sherlockjones
In summary: You want two DIFFERENT points, not the same point twice. Since you have two points, you can find the two tangents. y= x tan t_1 and y= x tan t_2 are both equations for lines passing through the point (x,y).
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 $$1-2\cos^{2}t = 0$$

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sherlockjones said:
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.

Last edited by a moderator:
sherlockjones said:
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.

sherlockjones said:
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.

sherlockjones said:
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...

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

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sherlockjones said:
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.

Last edited by a moderator:
Sorry for my stupidity, but do you mean to do this:

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

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

sherlockjones said:
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.

## 1. What are parametric equations and how are they different from regular equations?

Parametric equations are a set of equations that express a set of variables, usually x and y, in terms of a third variable, usually t. This third variable is known as the parameter, and it allows us to graph and analyze equations that cannot be expressed as a function of x and y. Regular equations, on the other hand, are expressed in terms of x and y only and can be graphed on a Cartesian plane.

## 2. How do I graph parametric equations?

To graph parametric equations, you need to plot the x and y values for different values of the parameter, t. This will create a set of points that can be connected to form a curve. It is also helpful to use a table of values to plot the points and to determine the domain and range of the graph.

## 3. What are the applications of parametric equations in real life?

Parametric equations are used in many fields, including physics, engineering, and computer graphics. They can be used to describe the motion of objects, such as projectiles or planets, and to create visual effects in video games and animations. They are also used in parametric design, where equations are used to create complex shapes and designs.

## 4. How do I solve parametric equations?

To solve parametric equations, you need to eliminate the parameter, t, and solve for x and y. This can be done by either eliminating t algebraically or by using substitution. It is also important to check for any restrictions on the domain of the parametric equations, as some values of t may not produce real solutions.

## 5. How do parametric equations relate to vectors?

Parametric equations can be thought of as vector equations, where the x and y values are the components of a vector. This allows us to use vector operations, such as addition and scalar multiplication, to manipulate and solve parametric equations. Additionally, the derivative of a parametric equation is equivalent to the velocity vector, and the second derivative is equivalent to the acceleration vector.

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