Calculus III; standard form of & angle to eliminate the cross produc

[Molderish]

Hi; I've been trying to solve the problem myself but i really don't what could be wrong;

The problem says :

Make the change of variables
x=ucos−vsin
y=usin+vcos

where the angle 0<(phi)<2 is chosen in order to eliminate the cross product term in
x^2+xy+y^2=6

Then find the standard form of equation in the (uv) variables. (Enter a function of (uv).)

---------------=1

well what I've found the angle is (pi/4) which would eliminates the cross product terms "uv" when i make the substituion..

then I've tried to reorder the equation: x^2+xy+y^2=6 which is an elipse with center at (0,0) ; semimajor axis=2sqrt3 & semiminor axis=2

then i got the equation (x^2/12)+(y^2/4)=1 then change variables again and i got ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4.
its incorrect.

If you could make a step by step solution , would be great , thanks in advanced.

 

[SammyS]

Hi; I've been trying to solve the problem myself but i really don't what could be wrong;

The problem says :

Make the change of variables
x=ucos−vsin
y=usin+vcos

where the angle 0<(phi)<2 is chosen in order to eliminate the cross product term in
x^2+xy+y^2=6

Then find the standard form of equation in the (uv) variables. (Enter a function of (uv).)

---------------=1

well what I've found the angle is (pi/4) which would eliminates the cross product terms "uv" when i make the substituion..

then I've tried to reorder the equation: x^2+xy+y^2=6 which is an elipse with center at (0,0) ; semimajor axis=2sqrt3 & semiminor axis=2

then i got the equation (x^2/12)+(y^2/4)=1 then change variables again and i got ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4.
its incorrect.

If you could make a step by step solution , would be great , thanks in advanced.

We don't provide step by step solutions at PF.

Why does the equation you get,

(x²/12)+(y²/4)=1,

have x & y in it, rather than u & v ?



To do the integration with u,v, don't forget your Jacobian -- although in this case it's a convenient value.

 

[Molderish]

Because what i tried to do its to re-write the equation " x^2+xy+y^2=6" which i found easier. then i plugged into it the "x" and "y" the problems gives.

but what i am not really sure if it means "elipse standard form"; i think the angle i got it's right since " ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4." & "x^2+xy+y^2=6" are the same elipse.. but i don't why its incorrect.

 

[SammyS]

Because what i tried to do its to re-write the equation " x^2+xy+y^2=6" which i found easier. then i plugged into it the "x" and "y" the problems gives.

but what i am not really sure if it means "elipse standard form"; i think the angle i got it's right since " ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4." & "x^2+xy+y^2=6" are the same elipse.. but i don't why its incorrect.

Sure, you have the correct angle.

plugging in u & v, I get

u²/4 + v²/12 = 1 ,

which is the ellipse you describe, with major axis along the v axis.

 

[Molderish]

but wouldn't it be x=ucos−vsin--> so ((ucos(pi/4)-vsin(pi/4))^2)/12 & y=usin+vcos----> ((usin(pi/4))+(vcos(pi/4)))^2/4?? did you reduce factors?

 

[vela]

The presence of the xy term is what causes the ellipse to be rotated with respect to the x and y axes. You're trying to find a rotated set of coordinates in which the ellipse isn't rotated. You don't want the ellipse to look exactly the same in both sets of coordinates.

 

[Molderish]

oh ok! i see.. well i think is a minor detail i couldn't see , thanks so much!

 

[SammyS]

but wouldn't it be x=ucos−vsin--> so ((ucos(pi/4)-vsin(pi/4))^2)/12 & y=usin+vcos----> ((usin(pi/4))+(vcos(pi/4)))^2/4?? did you reduce factors?

Take the original equation

x²+xy+y² = 6

Substitute as directed to express it in terms of u & v.

As a result of that, I got u²/4 + v²/12 = 1


It looks like you may have done that, then switched back to x & y , because using u & v was unfamiliar to you. -- then you tried the substitution again into that result.

Is that what you did ?

 

[Molderish]

yeah actually i make the change of variables & i end up.. with "uv" terms.

then i found the angle to make them 0. because it was the cross product terms then i just got a u^2 and v^2 with sin & cos =6 at this point i didn't know how to reorder the equation.

that's why i go back to the original equation. i though you needed the "angle" to make the direct substituion and have the values of cosines and sines once you have the "standard form of the given elipse".

 

[SammyS]

yeah actually i make the change of variables & i end up.. with "uv" terms.

then i found the angle to make them 0. because it was the cross product terms then i just got a u^2 and v^2 with sin & cos =6 at this point i didn't know how to reorder the equation.

Your result at this point is what is asked for in the original problem: a function of (u,v) .

that's why i go back to the original equation. i though you needed the "angle" to make the direct substituion and have the values of cosines and sines once you have the "standard form of the given elipse".

 

[Molderish]

Your result at this point is what is asked for in the original problem: a function of (u,v) .

Actually at first a tried with this result. but it said ,it was "incorrect" i guess because i didn't reorder the equation.