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richyw
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Find the points on \(x^2+xy+y^2=3\) which are closest and farthest from \((0,0)\)
 one year ago
 one year ago
richyw Group Title
Find the points on \(x^2+xy+y^2=3\) which are closest and farthest from \((0,0)\)
 one year ago
 one year ago

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AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Hmm... What sort of class are you doing this problem for?
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
refresher course in calculus.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
this is a multivariable optimization problem.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Ah. I had a feeling it was calc. ;P I was thinking of maybe using rotation of axes and trying to use conic section properties. I'm not as familiar with the multivar calc methods I saw before.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
Would you be able to do it without calc? Any method that get's the answer is fine by me haha.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
It's been a while since I did rotation of axes, but what it basically does is removes the obnoxious 'xy' term by using more obnoxious trig. With any luck though, we could possibly get a nice conic section like an ellipse, where we can identify the major and minor axes for those maximum/minimum distances
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
This PDF has some pretty sweet pictures and explanation, I thought it was really helpful: http://www.stewartcalculus.com/data/CALCULUS%20Early%20Vectors/upfiles/RotationofAxes.pdf
 one year ago

cruffo Group TitleBest ResponseYou've already chosen the best response.0
I have a suggestion that may not require much more than calc 1...
 one year ago

cruffo Group TitleBest ResponseYou've already chosen the best response.0
We want to maximize and minimize the distance between the curve and the point. \[\large \text{Distance}^2 = (x0)^2 + (y0)^2 = x^2 + y^2\] problem is that the curve is not easily solved for y. But If you use quadratic formula and solve for y thus, \[\large y^2 +xy + (x^2 3) = 0\] you may have some luck..
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
hmm. that could work. but will keep bumping for a calc answer! Thanks for all your help though.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Alright, I got my method to work out, although it is quite long and tedious. lol  w = r cos t z = r sin t x = r cos(t + p) y = r sin(t + p) x = r cos t cos p  r sin t sin p y = r sin t cos p + r cos t sin p x = w cos p  z sin p y = z cos p + w sin p (w cos p  z sin p)^2 + (w cos p  z sin p)(z cos p + w sin p) + (z cos p + w sin p)^2 = 3 w^2 cos^2 p  2 wz sin p cos p + z^2 sin^2 p + wz cos^2 p + w^2 sin p cos p  z^2 sin p cos p  wz sin^2 p + z^2 cos^2 p + 2 wz cos p sin p + w^2 sin^2 p = 3 w^2 (cos^2 p + sin p cos p + sin^2 p) + wz(2sin p cos p + cos^2 p  sin^2 p + 2sin p cos p) + z^2 (sin^2 p  sin p cos p + cos^2 p) = 3 w^2 (1 + sin p cos p) + wz(cos^2 p  sin^2 p) + z^2(1  sin p cos p) = 3 w^2 (1 + sin p cos p) + wz cos(2p) + z^2 (1  sin p cos p) = 3 w^2 (1 + 1/2 sin(2p)) + wz cos(2p) + z^2 (1  1/2 sin(2p)) = 3 cos(2p) = 0 2p = pi/2 p = pi/4 w^2 (1 + 1/2 sin 2(pi/4)) + wz cos(2 pi/4) + z^2 (1  1/2 sin(2 pi/4)) = 3 w^2 (1 + 1/2) + 0wz + z^2(1  1/2) = 3 3/2 w^2 + 1/2 z^2 = 3 1/2 w^2 + 1/6 z^2 = 1 Our rotation of axes doesn't affect (0, 0)'s position, so we can ignore messing with it. sqrt(2) is 1/2 the length of our minor axis, and it is along the xaxis. (+ sqrt(2), 0) are our minor vertices. sqrt(6) is 1/2 the length of our major axis, and it is along the yaxis. (0, + sqrt(6) are our major vertices. So, the last thing to do now is convert back to the original system. (w, z) = (sqrt(2), 0) x = sqrt(2) cos(pi/4)  0 sin(pi/4) y = 0 cos(pi/4) + sqrt(2) sin (pi/4) x = sqrt(2)* sqrt(2)/2 = 1 y = sqrt(2) sqrt(2)/2 = 1 (1, 1) is the minimum point, along with (1, 1). (w, z) = (0, sqrt(6)) x = 0 cos(pi/4)  sqrt(6) sin(pi/4) y = sqrt(6) cos(pi/4) + 0 sin(pi/4) x = sqrt(6) * sqrt(2)/2 = sqrt(3) y = sqrt(6) * sqrt(2)/2 = sqrt(3) (sqrt(3), sqrt(3)), and (sqrt(3), sqrt(3)) Conclusion: Points closest to (0, 0): (1, 1) and (1, 1). Points farthest from (0, 0): (sqrt(3), sqrt(3)) and (sqrt(3), sqrt(3)). 
 one year ago

cruffo Group TitleBest ResponseYou've already chosen the best response.0
Got the same thing using the same method :)
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
holy! thanks for putting in that effort. I really appreciate it.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
I looked into a method of solving this using Lagrange Multipliers. I'm no expert with this multivariable calc, but here's my attempt: Let \(D = d^2 = x^2 + y^2\) be the equation we are trying to optimize, and our constraint is \(x^2 + xy + y^2 = 3\). In general, this Lagrange Multiplier stuff says that: \(\nabla f(x,y) = \lambda \nabla g(x,y) \) We can then break this up into individual partial derivatives in each variable. \( f_x(x,y) = \lambda g_x(x,y) \), and \( f_y(x,y) = \lambda g_y(x,y) \). Pairing this up with our constraint, we have a threevariable system of equations: \( f_x(x,y) = \lambda g_x(x,y) \), \( f_y(x,y) = \lambda g_y(x,y) \), \( g(x,y) = k \) Now, we'll go back to our problem and we'll apply this information. \(f(x,y) = D\), \(g(x,y) = x^2 + xy + y^2\), and \(k = 3\). \(D_x = \lambda g_x\), \(D_y = \lambda g_y\) \(g = k\) \(2x = \lambda \left(2x + y\right)\) \(2y = \lambda \left(2y + x\right) \) \( x^2 + xy + y^2 = 3 \) From what I can see, solving this system gives you the answers. I suppose there are some checks that should be carried out to verify that your answers are correct from there. http://www.wolframalpha.com/input/?i=solve+system+%7B2x%3Dz%282x%2By%29%2C+2y%3Dz%282y%2Bx%29%2C+x%5E2%2Bxy%2By%5E2%3D3%7D
 one year ago
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