MHB Finding the parametrization of the curve

[shamieh]

A particle moves along the curve $$9x^2 + 16y^2 = 144$$

a)Find a parametrization of the curve which corresponds to the particle making one trip around the curve in a clockwise direction starting at (4,0)

so I know that $$cos^2t + sin^2t = 1$$ which is a circle. I also know that $$x^2 + y^2 = 1$$ is a circle.

But I'm not sure where to start... I know I need to let x = t correct? What exactly do I need to solve for here?b) Find a parametrization of the curve which corresponds to the particle making one trip around the curve in a counter-clockwise direction starting at (0,3)

 

[MarkFL]

I would begin by dividing through by 144 to obtain the equation of an ellipse in standard form:

$$\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$$

And to make things a bit easier to see for your parametrization, you may choose to write the equation as:

$$\left(\frac{x}{4} \right)^2+\left(\frac{y}{3} \right)^2=1$$

Now, can you use that Pythagorean identity to form your parametrizations?

 

[shamieh]

so are you saying

$$\frac{y}{3} = \sqrt{1 - (\frac{x}{4})^2}$$

$$y = 3 * \sqrt{1 - (\frac{x}{4})^2}$$

$$y = 0, 3$$ ?

 

[MarkFL]

No, suppose we wish to begin at the point (-4,0) and move in a clockwise direction. I would then let:

$$\frac{x}{4}=\cos(\pi-t)\implies x(t)=-4\cos(t)$$

$$\frac{y}{3}=\sin(\pi-t)\implies y(t)=3\sin(t)$$

Do you see where $\pi-t$ comes from and why I used it, and then rewrote the trig. function with an appropriate identity?

Do you see now what you need to do for parts a) and b)?

 

[shamieh]

I'm not sure where the $$\pi - t$$ comes from, I don;t get it? But here is
what I did, I'm not sure if this is correct or if I just got lucky.

$$9x^2 + 16y^2 = 144$$

By dividing through by 144 I obtained:

$$(\frac{x}{4})^2 + (\frac{y}{3})^2 = 1$$

Since $$\cos^2 t + \sin^2 t = 1$$

I can say

$$(\frac{x}{4})^2 = \cos^2 t$$

square rooting both sides I get:

$$x/4 = \cos t$$

And then finally by clearing out fractions I get $$x = 4\cos t$$

Similarly solving for $$y$$ I got $$y = 3\sin t$$

So if we are "starting" at (4,0) and the x position is positive would that mean that we are automatically going counter-clockwise? So since I need to start at $$4,0$$ and go clockwise, would I just change $$y = 3\sin t$$ to be $$y = -3\sin t$$ so that It would shoot the particle back thus going clockwise?

so : $$x = 4\cos t $$ and $$y = -3\sin t$$
$$0\le t \le 2\pi$$ ?

 

[shamieh]

In addition for part b,

wouldn't I just have x = $$4cost$$ and $$y = 3sint$$

becuase we are starting at $$(0,3)$$ which is essentially $$\pi/2$$

So I can just add $$2\pi$$ because each trip is $$2\pi$$ right?

so I would have

$$x=4cost$$ and $$y = 3sint$$

$$\pi/2 \le t \le 5\pi/2$$ ?

 

[MarkFL]

In the example I gave, the point (-4,0) corresponds to an initial angle of $\pi$ and since we are moving in a clockwise direction we want to move in a negative direction, so the angle is $$\pi-t$$.

So, for part a) what is the initial angle associated with the starting position of (4,0)?

And for part b), what is the initial angle associated with the point (0,3)?

Once you have the initial angles, then decide whether to add or subtract the parameter based on the given direction. After that, use the angle-sum/difference formulas for sine and cosine to simplify the parametrizations.

 

[shamieh]

So, for part a) what is the initial angle associated with the starting position of (4,0)?

And for part b), what is the initial angle associated with the point (0,3)?

Oh I see now why it is going counter clockwise initially because the unit circle goes that way lol wow.

so for Part a the initial angle associated with the startnig position of 4,0 was 45 degrees

and for part b it was 90 degrees right?

 

[MarkFL]

Oh I see now why it is going counter clockwise initially because the unit circle goes that way lol wow.

so for Part a the initial angle associated with the startnig position of 4,0 was 45 degrees

and for part b it was 90 degrees right?

You are correct for part b), although I would use radians. For part a), that starting point corresponds to an initial angle of 0. Plot the point and you will see this. So, can you now fill in the rest? What do you get?

 

[shamieh]

oh wait nevermind I was looking at it wrong from my notes. It would just be $$\pi$$ wouldn't it?

- - - Updated - - -

So for part a, (4,0) my initial angle was $$\pi$$ and for part b, (0,3) my initial angle was $$\frac{\pi}{2} $$

 

[MarkFL]

oh wait nevermind I was looking at it wrong from my notes. It would just be $$\pi$$ wouldn't it?

- - - Updated - - -

So for part a, (4,0) my initial angle was $$\pi$$ and for part b, (0,3) my initial angle was $$\frac{\pi}{2} $$

You are still wrong on part a) regarding the initial angle...plot the point. :D

 

[shamieh]

You are still wrong on part a) regarding the initial angle...plot the point. :D

I rush through things way too fast and don't listen lol. I apologize. My initial angle for part a is definitely not what i stated $$(\pi)$$ my initial angle for part a is definitely 0 degrees. Because I start at (4,0). which is 0 degrees or 0 radians right? Because it is essentially just the x - axis but on the right side of the unit cricle.

 

[MarkFL]

I rush through things way too fast and don't listen lol. I apologize. My initial angle for part a is definitely not what i stated $$(\pi)$$ my initial angle for part a is definitely 0 degrees. Because I start at (4,0). which is 0 degrees or 0 radians right? Because it is essentially just the x - axis but on the right side of the unit cricle.

Yes, that's correct...and you want to move in a clockwise direction, so what do you want the argument of the trig functions to be?

 

[shamieh]

Yes, that's correct...and you want to move in a clockwise direction, so what do you want the argument of the trig functions to be?

So for part a I am moving counter clock-wise and I want to move clockwise so after solving I would get:

$$y = 3\sin t$$ & $$x = 4\cos t $$

and this won't work, so I need to change the $$y$$ value to be negative so it will go clockwise

$$\therefore$$ $$x = 4\cos t$$ & $$y = -3\sin t$$ and then going around once I would just add $$0 + 2\pi$$ thus $$0 \le t \le 2\pi$$Then in part b,

I would be starting at $$\pi/2$$ and I need to go counterclockwise, but I am already going counter clockwise so I could just say x = 4cost & y = 3sint and add 2pi from where I am (pi/2) so $$\frac{\pi}{2} + 2\pi =$$ $$5pi/2$$ so $$\pi/2 \le t \le 5\pi/2$$

 

[MarkFL]

You have the correct end result, but here's how I would look at it:

The angle will be:

$$0-t=-t$$

Hence:

$$x(t)=4\cos(-t)=4\cos(t)$$

$$y(t)=3\sin(-t)=-3\sin(t)$$

You see how I used the identities (stemming from the respective symmetries of cosine and sine):

$$\cos(-\theta)=\cos(\theta)$$ and $$\sin(-\theta)=-\sin(\theta)$$ ?

Now, can you similarly do part b)?

 

[MarkFL]

...
Then in part b,

I would be starting at $$\pi/2$$ and I need to go counterclockwise, but I am already going counter clockwise so I could just say x = 4cost & y = 3sint and add 2pi from where I am (pi/2) so $$\frac{\pi}{2} + 2\pi =$$ $$5pi/2$$ so $$\pi/2 \le t \le 5\pi/2$$

You added this part while I was composing my previous reply...

Yes you begin at an angle of $$\frac{\pi}{2}$$ and because you are moving in a counter-clockwise direction you add $t$ so the argument for the trig. functions is:

$$\frac{\pi}{2}+t$$

Hence:

$$x(t)=4\cos\left(\frac{\pi}{2}+t \right)$$

$$y(t)=3\sin\left(\frac{\pi}{2}+t \right)$$

Now, simplify using the angle-sum identities for cosine and sine...:D

 

[shamieh]

isn't it just

$$x(t) = 4\cos(\frac{\pi}{2} + t) = 4cost$$

and

$$y(t) = 3\sin(\frac{\pi}{2} + t) = 3sint$$

so I just need to like at the sign on t basically?

 

[MarkFL]

isn't it just

$$x(t) = 4\cos(\frac{\pi}{2} + t) = 4cost$$

and

$$y(t) = 3\sin(\frac{\pi}{2} + t) = 3sint$$

so I just need to like at the sign on t basically?

No, apply the angle-sum formulas:

$$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)- \sin(\alpha)\sin(\beta)$$

$$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+ \cos(\alpha)\sin(\beta)$$

What do you get?

 

[shamieh]

No, apply the angle-sum formulas:

$$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)- \sin(\alpha)\sin(\beta)$$

$$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+ \cos(\alpha)\sin(\beta)$$

What do you get?

Ha. I can't be doing this right.. Thought I'd at least give it a try tho.(Dance)

$$x(t) = 4cos(\frac{\pi}{2} + t)$$
$$x(t) = 4[cos(\frac{\pi}{2})*cos(t) - sin(\frac{\pi}{2} * sin(t))]$$
$$x(t) = 4[0 - 1sin(t)]$$
$$x(t) = -4sin(t)$$

$$y(t) = 3sin(\frac{\pi}{2} + t)$$
$$y(t) = 3[sin(\frac{\pi}{2}*cos(t) + cos(\frac{\pi}{2} * sin(t)]$$
$$y(t) = 3\cos t$$

 

[MarkFL]

Yes! That's correct! (Star)

 

[shamieh]

Yes! That's correct! (Star)

Oh I see, so there are multiple ways to do this problem

So $$x = 4\cos t y = 3\sin t 0 \le t \le 2\pi$$ is one solution

and

$$x = -4\sin t ...y = 3\cos t$$ is another solution for part b as well?

Seems peculiar since one of them is negative and one is positive , you would think both would need to be positive for it to go counter clockwise.. Would that mean that when it gets to $$3cost$$ it turns around?

 

[MarkFL]

...
$$x = -4\sin t ...y = 3\cos t$$ is another solution for part b as well?...

No...you found the correct solution.

 

[shamieh]

No...you found the correct solution.

my teacher has the answer for part a as:
$$x = 4cos(t)$$
$$ y = -3sin(t)$$,$$ 0 \le t \le 2\pi$$

and for part b he has:
$$x = 4cos(t)$$
$$ y = 3sin(t)$$, $$\frac{\pi}{2} \le t \le \frac{5\pi}{2}$$

.. Has he made a mistake?

 

[MarkFL]

You teacher has just defined $t$ on a different interval than we did. The two parametrizations are equivalent. We simply incorporated the phase shift of $$\frac{\pi}{2}$$ into our answer so that for us, we have:

$$0\le t<2\pi$$

 

[shamieh]

Yea so I know I've probably drove you crazy with this problem, but I have another question. Today after I solved for x and y, my teacher simply plugged in $$\frac{\pi}{2}$$ into the equation which ended up working out good because:

for Part A: $$x = 4cos(pi/2)$$ & $$y = 3sin(pi/2)$$

x = 0 and y = 3 which we know is NOT the way we need to go, since we want to go clockwise, so then he simply put x = 4cost and y = -3sint to make it go the other way.

Similarly I'm wondering if this would also be correct to say for part B:

Starting at 0,3 and wanting to go counter clock-wise -

plugging in $$\pi$$ for each equation I obtained: x = -4 and y = 0 so could I just say that $$x = 4cost$$ and $$y = 3sint$$ for part B correct? since we know that we will still be going counter clockwise?

 

[MarkFL]

...
Similarly I'm wondering if this would also be correct to say for part B:

Starting at 0,3 and wanting to go counter clock-wise -

plugging in $$\pi$$ for each equation I obtained: x = -4 and y = 0 so could I just say that $$x = 4cost$$ and $$y = 3sint$$ for part B correct? since we know that we will still be going counter clockwise?

If you define the interval for $t$ appropriately as your teacher did, then you could state that.

I just prefer for the parameter $t$ to be defined as $0\le t<2\pi$ and adjust the parametrizations of $x$ and $y$ accordingly. It just makes more sense to me like that. There are an infinite number of ways to correctly define the parameters though.