# Calculus Vectorial

Discussion in 'Community Discussion' started by Quboid, Aug 18, 2007.

1. ### Quboid macrumors 6502

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#1
Hey all,

Stuck on another calculus question. It seems prettty simple but I can't seem to figure it out.

Draw the curve for
x=4 cos t
y=5 sen t

I plugged in values and drew hte curve with was pretty easy but i can't figure out the ecuation for the curve.

This is what i did so far:

x/4=cos t
y/5=sen t

t[0,180] (I couldn't find the pi symbol)

(x/4)^2=cos^2 t
(y/5)^2=sen^2 t

(x/4)^2+(y/5)^2=cos^2 t + sent^2 t

(x/4)^2+(y/5)^2 = 1

....from this is see a circle with radius 1 and center in th origin. But the curve i got from plugging in values is clearly a parabola.

any help???

2. ### mkrishnan Moderator emeritus

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#2
Can I start by asking a stupid question? What function is "sen"? Secant?

3. ### swiftaw macrumors 603

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#3
It's an equation of an ellipse

Edit: I'm presuming that sen is actually the sine function seeing as you said cos^2 t + sen^2 t = 1

Edit 2: An ellipse that has equation x^2 / a^2 + y^2 / b^2 = 1 has parametric equation x = a cos t and y = b sin t.

4. ### Genghis Khan macrumors 65816

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#4
god it's been almost a year since i did this stuff...

what you've done so far seems right (i'm taking that the sen = sine)

my advice is check all the settings on the calculator (hoping you're using texas instruments)

EDIT: if sen is indeed secant you have something very not circle

5. ### mkrishnan Moderator emeritus

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#5
If "sen" = sine, then you did the trig correctly, but what you're experiencing is that your graphing program assumes a function (the ellipse equation is not a function -- it has two y values for every x), and so it just draws one quarter of the ellipse, which looks like a parabola.

If "sen" = secant (normally "sec" in the US, at least), then remember the definition of secant... you're on the wrong track.

6. ### swiftaw macrumors 603

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#6
As I said above, if by sen you meant sine then what you have is an ellipse.

However, if by sen you meant secant, then you have

x = 4 cos t
y = 5 sec t

so x*y = 4*5*cos t*sec t = 20

so y = 20/x

Edit: Either way, you definitely do not have a parabola, the parametric equations of a parabola are not trig functions, they are of the form y = 2at, x = 2at^2

7. ### mkrishnan Moderator emeritus

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#7
Careful with your math... y = 5 sec (t) = 5 / cos (t) = 5 / (x/4) = 20/x.

8. ### swiftaw macrumors 603

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#8
Yeah, I had it right in my head, just wrong when i typed it, changed it now

9. ### Quboid thread starter macrumors 6502

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#9
Sen is indeed sine, (its seno in spanish.....long story). So if i had the trig right, how do i explain the curve i get. It's half of hte eclips when plotted in hte given range.

10. ### mkrishnan Moderator emeritus

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#10
Read the first part of my post #5 again.

11. ### swiftaw macrumors 603

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#11
It should be an entire ellipse, you'll probably need to check what range you are plotting over

12. ### siurpeeman macrumors 603

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#12
the thing you entered is half of an ellipse. the reason why you get the top half only is because you set the domain of t to (0,pi), where sin is strictly non-negative. if you want the whole ellipse, your domain would need to be [0,2 pi].

*edit*
it might look an awful lot like a parabola, because the y-axis acts as the major axis with a length of 10 (though only 5 is shown).

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13. ### Quboid thread starter macrumors 6502

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#13
Thanks alot guys. I was thinking the equation was a circle with a radius of one. It didn't occur to me that it was an elipse equation.

14. ### Quboid thread starter macrumors 6502

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#14
One more question:

This is an enirely different problem:

Draw the graph for the parametric ecautions:

x(t)=t
y(t)=sqrt(4-t^2)

t[-4 to 4]

Now for this one i just said that x=t so I contructed an equation:
y=sqrt(4-x^2)

Now as i get ready to plot, I realise that there are indeterminations at x equal to -4,-3,4 and 3. All the other points plot a nice parabola. For those x values that give "unreal" y values I just put vertical lines extending from them.

Is that the write way to solve this problem?

15. ### Rodimus Prime macrumors G4

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Oct 9, 2006
#15
also I might want to add you calculator can only draw 1/2 of the figure.

As for help on it I think everyone else covered what I could do. It been over 3 years since I even touch the vector calculus and even then I never really understood it that and my calculus book is about 600 miles from my current location so I can not even look anything up.

16. ### swiftaw macrumors 603

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#16
y is only real for t in the range [-2,2], outside of that range y is complex because you are square rooting a negative number

17. ### siurpeeman macrumors 603

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#17
like the problem before, it actually isn't a parabola, it's a semi-circle (top half). reason is that y must be non-negative since it's sqr rt (junk). you don't get points less than -2 or greater than +2, because it doesn't work in your second equation. graphically, it's a circle centered at the origin with a radius of 2 (again, top half), so you wouldn't have any points outside [-2,2] anyway.

p.s. i gotta ask, how is this related to vector calculus anyway?

18. ### swiftaw macrumors 603

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#18
as the last poster said, if you square both sides, you get y^2 = 4 - x^2

or in other words, x^2 + y^2 = 4. This is a circle, centered at the origin with radius 2. Now, the way the formula is given, you'll only get the top half.

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#19

20. ### Quboid thread starter macrumors 6502

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#20
Question, what program did you use to draw that graph?

21. ### siurpeeman macrumors 603

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#21
grapher. all macs ship with it. i'm not very familiar with it, though.

22. ### swiftaw macrumors 603

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#22
Looks like Grapher to me, it's in your /Applications/Utilities/ folder

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23. ### siurpeeman macrumors 603

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#23
just to point out, you can make a square root with option-v.

24. ### Rodimus Prime macrumors G4

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#24
it is that way because you are not longer using the y=mx+b format. Now you are solving for both x and y with the variable t and with t you are out putting (x,y)

Later on in the course you will start doing triple integration.

Cal III is mostly cal I in 3d. It is a heck of a lot easier in my book then cal II. Sadly you will need to understand cal II to understand cal III

25. ### Quboid thread starter macrumors 6502

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#25
Re-Kindle

Hey there,

Can i molest you guys with yet another calculus question? ok here goes:'

Find parametric equations of the ellipse: x^2/a^2+y^2/b^2=1

I know it looks trivial, but the abtraction is making it difficult for me.

Thanks!