Calculus Vectorial

Quboid

macrumors 6502
Original poster
Oct 16, 2006
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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???
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
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.
 

Genghis Khan

macrumors 65816
Jun 3, 2007
1,202
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Melbourne, Australia
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
 

mkrishnan

Moderator emeritus
Jan 9, 2004
29,641
12
Grand Rapids, MI, USA
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. ;)
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
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
 

Quboid

macrumors 6502
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Oct 16, 2006
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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.
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
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.
It should be an entire ellipse, you'll probably need to check what range you are plotting over
 

siurpeeman

macrumors 603
Dec 2, 2006
6,313
18
the OC
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.
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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Quboid

macrumors 6502
Original poster
Oct 16, 2006
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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.
 

Quboid

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Oct 16, 2006
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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?
 

Rodimus Prime

macrumors G4
Oct 9, 2006
10,132
4
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.
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
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?
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
 

siurpeeman

macrumors 603
Dec 2, 2006
6,313
18
the OC
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?
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?
 

swiftaw

macrumors 603
Jan 31, 2005
6,309
20
Omaha, NE, USA
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.
 

Quboid

macrumors 6502
Original poster
Oct 16, 2006
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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).
Question, what program did you use to draw that graph?
 

Rodimus Prime

macrumors G4
Oct 9, 2006
10,132
4
I just started this course, and these are just porblems similar to the ones i did on the first day of class.
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
 

Quboid

macrumors 6502
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Oct 16, 2006
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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!