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Difference between revisions of "KGS math club/solution 10 1"

(Created page with 'The coefficient of the tangent can be found from implicit derivative formula: <math>dy / dx = - (dz / dx) / (dz / dy) = - (2x + y) / (2y + x) </math> where <math> z = x^2 + y^2 +…')
 
 
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The coefficient of the tangent can be found from implicit derivative formula:
 
The coefficient of the tangent can be found from implicit derivative formula:
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<math>dy / dx = - (dz / dx) / (dz / dy) = - (2x + y) / (2y + x) </math> where <math> z = x^2 + y^2 + x y </math>
 
  
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So we want to find a pair (x, y) such that <math> x^2 + y^2 + x y - 1 = 0 </math>  and  <math> (y - 2) / x = - (2x + y) / (2y + x) </math>, that is, <math> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0 </math>, that is, <math> y^2 + (x - 2) y - x + x^2 = 0 </math>
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<math>dy / dx = - (dz / dx) / (dz / dy) = - (2x + y) / (2y + x) </math>
  
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We notice by magic that (x, y) = (1, 0) and (x, y) = (-1, 1) are the two solutions to the equation.
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where <math> z = x^2 + y^2 + x y </math>
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So we want to find a pair (x, y) such that
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<math> x^2 + y^2 + x y = 1 </math> 
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and 
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<math> (y - 2) / x = - (2x + y) / (2y + x) </math>
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<math> <=> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0 </math>
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<math> <=> y^2 + x y - 2 y - x + x^2 = 0 </math>
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Substituting the first to the second, we get
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<math>  (y^2 + x y + x^2) - 2 y - x = 1 - 2 y - x = 0 </math>
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<math> <=> y = (1 - x) / 2 </math>
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Substituting this back to the first equation, we get
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<math> x^2 + (1 - x)^2 / 4 + x (1 - x) / 2 - 1 = (1 + 1/4 - 1/2) x^2 + (-1/2 + 1/2) x + (1/4 - 1) = 3/4 x^2 - 3/4 = 0 </math>
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<math> <=> x = +-1 </math>
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We have thus found two solutions: <math>(x, y) = (1, 0)</math> and <math>(x, y) = (-1, 1)</math>.
  
 
Verification:  
 
Verification:  
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at (-1, 1), the <math> dy / dx = - (-2 + 1) / (2 - 1) = 1 </math>, so the tangent goes from (-1, 1) to (0, 2)
 
at (-1, 1), the <math> dy / dx = - (-2 + 1) / (2 - 1) = 1 </math>, so the tangent goes from (-1, 1) to (0, 2)
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at (1, 0), the  <math> dy / dx = -(2 + 0) / (0 + 1) = -2 </math>, so the tangent goes from (1, 0) to (0, 2)
 
at (1, 0), the  <math> dy / dx = -(2 + 0) / (0 + 1) = -2 </math>, so the tangent goes from (1, 0) to (0, 2)
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'''This solution is incomplete, please fill in!'''
 

Latest revision as of 15:11, 11 August 2010

The coefficient of the tangent can be found from implicit derivative formula:

$dy / dx = - (dz / dx) / (dz / dy) = - (2x + y) / (2y + x)$

where $z = x^2 + y^2 + x y$

So we want to find a pair (x, y) such that

$x^2 + y^2 + x y = 1$

and

$(y - 2) / x = - (2x + y) / (2y + x)$

$<=> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0$

$<=> y^2 + x y - 2 y - x + x^2 = 0$

Substituting the first to the second, we get

$(y^2 + x y + x^2) - 2 y - x = 1 - 2 y - x = 0$

$<=> y = (1 - x) / 2$

Substituting this back to the first equation, we get

$x^2 + (1 - x)^2 / 4 + x (1 - x) / 2 - 1 = (1 + 1/4 - 1/2) x^2 + (-1/2 + 1/2) x + (1/4 - 1) = 3/4 x^2 - 3/4 = 0$

$<=> x = +-1$

We have thus found two solutions: $(x, y) = (1, 0)$ and $(x, y) = (-1, 1)$.

Verification:

at (-1, 1), the $dy / dx = - (-2 + 1) / (2 - 1) = 1$, so the tangent goes from (-1, 1) to (0, 2)

at (1, 0), the $dy / dx = -(2 + 0) / (0 + 1) = -2$, so the tangent goes from (1, 0) to (0, 2)

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