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 +…') |
|||
| Line 1: | Line 1: | ||
The coefficient of the tangent can be found from implicit derivative formula: | The coefficient of the tangent can be found from implicit derivative formula: | ||
| − | |||
| − | 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> | + | <math>dy / dx = - (dz / dx) / (dz / dy) = - (2x + y) / (2y + x) </math> |
| + | |||
| + | where <math> z = x^2 + y^2 + x y </math> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | <math> <=> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0 </math> | ||
| + | |||
| + | <math> <=> y^2 + (x - 2) y - x + x^2 = 0 </math> | ||
We notice by magic that (x, y) = (1, 0) and (x, y) = (-1, 1) are the two solutions to the equation. | We notice by magic that (x, y) = (1, 0) and (x, y) = (-1, 1) are the two solutions to the equation. | ||
Verification: | Verification: | ||
| + | |||
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) | ||
| + | |||
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) | ||
'''This solution is incomplete, please fill in!''' | '''This solution is incomplete, please fill in!''' | ||
Revision as of 14:02, 11 August 2010
The coefficient of the tangent can be found from implicit derivative formula:
where 
So we want to find a pair (x, y) such that
and
We notice by magic that (x, y) = (1, 0) and (x, y) = (-1, 1) are the two solutions to the equation.
Verification:
at (-1, 1), the 
, so the tangent goes from (-1, 1) to (0, 2)
at (1, 0), the 
, so the tangent goes from (1, 0) to (0, 2)
This solution is incomplete, please fill in!
