im about to study calculus and have no clue about it.what the heck is this? b^2*x^2 + a^2*y^2 = a^2*b^2 can you solve it for me step by step?
2006-08-02 07:16:43 · 14 answers · asked by Nafertiti 2 in Science & Mathematics ➔ Mathematics
I'm afraid you've been misled; this is not a calculus problem at all. In fact, it doesn't even seem to be a valid question - the question doesn't even specify what exactly are we supposed to solve for, and even so, there are too many unknowns (four - a, b, x, and y) to solve for all variables - we would need four equations to solve for them all.
2006-08-02 07:28:44 · answer #1 · answered by Anonymous · 0⤊ 0⤋
solve for x? y?
http://www.khake.com/page47 is a jump page with mountains of math help. Take a couple of hours, go through the links, and see which ones can help you.
It goes from addition through calculus, and has definitions.
http://www.mathworld.wolfram.com is a mathematician's help site. The guy who runs it is the advisor for the TV show "Numbers" - the cop show where the cop's brother is a mathematician.
Hyperphysics site will get you through physics.
You won't need luck if you use these.
2006-08-02 14:29:53 · answer #2 · answered by helixburger 6 · 0⤊ 0⤋
There is nothing to solve.
This is the equation for an ellipse centred at the origin with the two semiaxises a (in x-direction) and b (in y-direction).
So if you choose values for the semiaxis lengths a and b, the pairs of x and y that solve the equation are the coordinates of points on the ellipse.
2006-08-02 14:50:05 · answer #3 · answered by Anonymous · 0⤊ 0⤋
i'm assuming you're trying to find x and y?
doesn't look like a standard calculus question but if it's calculus i'm assuming you have to take the derivative of these numbers to be able to solve the equation.
that's all i can say.. it's been too long since i took calculus
2006-08-02 14:25:57 · answer #4 · answered by angry_fruit 2 · 0⤊ 0⤋
b^2*x^2+a^2*y^2=a^2*b^2
b^2*x^2=a^2*b^2-a^2*y^2
If b=0 then
0 = a^2*y^2
If a=0 then answer is: any x, any y
if a<>0 then y = 0, answer is: any x, y=0
If b<>0 then
x^2=(a^2*b^2-a^2*y^2)/b^2
x^2=a^2*(b^2-y^2)/b^2
x^2=a^2*(1 - (y/b)^2)
answer: choose any y so |y/b|<=1 then x calculated by
x=+/- |a|*sqrt(1 - (y/b)^2)
If let (y/b)=cos(t) then x=+/- |a|*sqrt(1 - cos^2(t))
x=+/- |a|*sqrt(sin^2(t))
x=+/- |a*sin(t)|
2006-08-02 14:49:59 · answer #5 · answered by Anonymous · 0⤊ 0⤋
b^2x^2 + a^2y^2 = a^2b^2
Divide both sides by a^2b^2
((x^2)/(a^2)) + ((y^2)/(b^2)) = 1
This is an ellipse
2006-08-02 16:30:56 · answer #6 · answered by Sherman81 6 · 0⤊ 0⤋
That's all just a blur to me... but then again I suck at math.
2006-08-02 14:20:55 · answer #7 · answered by Amanda 4 · 0⤊ 0⤋
Why don't you call your math teacher and ask ALL these questions?
2006-08-02 14:20:49 · answer #8 · answered by Anonymous · 0⤊ 0⤋
This is a fake problem
2006-08-02 14:21:13 · answer #9 · answered by Hawtman1092 3 · 0⤊ 0⤋
this a fake problem and i am in geometry
2006-08-02 14:24:18 · answer #10 · answered by briander_101 1 · 0⤊ 0⤋