- anonymous

x^2/4 + (y+2)^2 = 1
The task is; "Parametrize the curve"
What does it mean? What is expected? I have tried looking for several examples, but none are similar. Both got two functions (x and y) and such. Is the task to find the two functions (x and y)? How is this solved (or rather, how is this expected to be solved)?

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- inkyvoyd

hmm

- inkyvoyd

a parametrization is accomplished with the introduction of a third variable. Say, t

- inkyvoyd

the equation you've provided is one of a ellipse centered at (0,-2).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- inkyvoyd

so, we can actually define t such that we get the same equation of the ellipse as we sweep t across values. to give you an idea, a parametrization of a circle x^2+y^2=25 would be x=5 sin t and y = 5 cos t. now, because x^2+y^2=5^2* sin^2t+5^2*cos^2t=25(1)=25, you know you ahve the equation of a circle in terms of a parameter t.

- inkyvoyd

now, if you are familiar with vector valued functions you can actually express this parametrization as a single vector valued function. In this example that would be r(t)=<5sin(t),5cos(t)> note that r would normally in bold to represent that it is a vector. Another thing to notice is that parametrizations are not unique. q(s)=<5cos(s),5sin(s)> comes out to the same thing... it also draws a circle, though it starts at a different point when s=0 versus when t=0

- inkyvoyd

Using my example with the circle, do you have any ideas as to how you would parametrize the equation of your ellipse?

- anonymous

Not yet. Reviewing old material

- anonymous

Slowly getting there.... !

- inkyvoyd

May I ask, what is the class youa re in and is there a chapter name for this topic? there are many ways to approach this topic and I am not sure how your material expects you to.

- anonymous

"Parametric Curves"

- inkyvoyd

is this under calculus 2?

- anonymous

I dont know what number; but it is Calculus

- inkyvoyd

Ok. let's work through this. With the equation x^2/4 + (y+2)^2 = 1 you can tweak my example slightly to get the same thing why don't we rewrite the equation as the sum of squares to apply the identity sin^2 x+cos^2 x in our parametrization.

- anonymous

Wondering if there is a formula for conversion from x^2/9 to 3cos t etc

- inkyvoyd

there probably is, but I try to do so algebraically - generally speaking there are so many formulas in calculus that it does not make sense to memorize every one unless you will use it constantly... derivations are more important.

- inkyvoyd

anyways, the first term is not in perfect square form... can you rewrite it in the form (blah)^2?

- anonymous

first step for me is to find out "how" to convert it. Second step is to understand how (in order for me to learn). Everyone learns differently - I learn by examples and step by step solutions

- anonymous

calculus 2 hard 5 me

- inkyvoyd

Yes. I am working with you right now on this. there are a few more steps after rewriting the first term as a perfect square. As I have explained before, you'll want to write your equation in a more familiar form if you want any luck with it.
I would agree that everyone learns differently, but I would also like to add that there are two ways to get through calculus, and math in general. One can struggle repeatedly with a problem and figure it out with just the principles involved, or, one can look in the book and follow the same exact problem with different numbers and go through it step by step. The former is mathematics. The second is following rules. Most people can get by with the second, but a word of warning: typically tests go after the former.

- anonymous

I am somehow able to get A's and B's with this method - dont ask me how... xP

- inkyvoyd

If that is all you want, I can give you the answer and let you be done with it. I hope for your sake you learn before upper level university classes.

- anonymous

Actually, I dont need an answer - I just need to understand what is being asked by the question "Parametrize the curve" ... ! I do not understand the question. Is it asking me to rewrite it in order to get a single function of t?

- inkyvoyd

the other problem with doing textbook examples step by step is that you forget them. if you have to remember EVERY single problem type, it would be anightmare. But if you understood the fundamentals deeply, it might take you a bit more time, but you would reach the conclusions.
Now I am not saying it is practical to derive everything on spot. But I think this problem is doable with a bit of insight and a bit of thinking.

- anonymous

Appreciate the help btw

- inkyvoyd

Well, I tried to clarify that by making another example and solving it step by step

- inkyvoyd

you can check the wiki article if you want an idea of what you are doing.
https://en.wikipedia.org/wiki/Parametric_equation

- anonymous

... !
x = 2cost
y = sint - 2
tâ‚¬(0, 2pi) ?

- anonymous

I think I am getting closer... !

- IrishBoy123

" Is it asking me to rewrite it in order to get a single function of t?"
yes, sort of
but you will get x = x(t) and y = y(t)

- inkyvoyd

I think you have it.

- anonymous

Ellipse:
\[\frac{ (x-x_0)^2 }{ a^2 } + \frac{ (y-y_0)^2 }{ a^2 } = 1\]
Parametrized form(?):
\[x = a*cost + x_0\]\[y = b*sint+ y_0\]

- inkyvoyd

yes. the second a^2 should be a b^2 but otherwise yes. Can you visualize it?

- anonymous

There is no edit button! x.x

- inkyvoyd

Well, given that you've deduced what I think is the correct answer, I'm going to head off for the night. Feel free to let me know if you have any additional questions.

- anonymous

I am wondering if it would be correct that \[f(t) = 2cost + sint - 2\]

- inkyvoyd

no.

- anonymous

Damn - Wondering if it is possible to create a single function of t

- inkyvoyd

yes. but you need knowledge of vector valued functions.

- inkyvoyd

it would be r(t)=<2cos t,sin t - 2>

- inkyvoyd

note that you are actually returning multiple scalars (a vector) for a single scalar input. This isn't a function in the form taht you are used to seeing, because it maps R to R^2 rather than R to R

- inkyvoyd

https://en.wikipedia.org/wiki/Vector-valued_function
you can extend it to three dimensions. if we took
r(t)=<2cos t,sin t - 2,t> we would have the equation of a elliptical helix.

- inkyvoyd

anyways, I am off for the night. If you are curious, you can check paul's online math notes
http://tutorial.math.lamar.edu/
or khan academy
https://www.khanacademy.org/
for some particularly good tutorials on vector valued functions and the general geometry of space.

- anonymous

Thanks - good night

- anonymous

For reference; this is each and every step I took in order to get the x(t) and y(t) (which, I assume; is parametrizing the curve - PLEASE CORRECT ME IF I AM WRONG, OR MISSING INFORMATION).
1. Looking at trigometric formulas; I can see that \[\cos^2x + \sin^2x = 1\]
2. I set \[\cos^2x = x^2/4\], set root on both sides and get; \[cosx = x/2\] Then I put x on one side and get \[x(t) = 2cost\]
3. I set the other; \[\sin^2x = (y+2)^2\], set root on both sides and get; \[sinx = y+2\] Then I put y on one side and get \[y(t) = sinx - 2\]
y = y(t) and x = x(t)
Crossing fingers that this is all that was asked by the task!

- IrishBoy123

for
\(\cos^2x + \sin^2x = 1\)
write
\(\cos^2t + \sin^2t = 1\)
and the same applies all the way down from these, ie introduced your new parameter, t
so for \(\cos^2x = x^2/4\)
write \(\cos^2t = x^2/4\)

- anonymous

My bad, wrote X because it was easier than alpha, and I forgot x was already in use!

Looking for something else?

Not the answer you are looking for? Search for more explanations.