x²+y²-4x-6y-25=0,Find dy/dx.
Answers
we have to find dy/dx of given equation x² + y² - 4x - 6y - 25 = 0
differentiating equation with respect to x,
d(x² + y² - 4x - 6y - 25)/dx = 0
⇒d(x²)/dx + d(y²)/dx - 4d(x)/dx - 6d(y)/dx - d(25)/dx = 0
⇒2x × dx/dx + 2y × dy/dx - 4 × dx/dx - 6 × dy/dx = 0
⇒2x + 2y (dy/dx) - 4 - 6(dy/dx) = 0
⇒(2x - 4 ) + (2y - 6)(dy/dx) = 0
⇒(2x - 4) = -(2y - 6)(dy/dx)
⇒(2x - 4) = (6 - 2y)(dy/dx)
⇒dy/dx = (2x - 4)/(6 - 2y)
hence, dy/dx = (2x - 4)/(6 - 2y)
[interesting point : the above mentioned equation is equation of circle, and centre of circle is (2, 3).
here you found dy/dx = (2x - 4)/(6 - 2y)
2x - 4 = 0 ⇒x = 2
and 6 - 3y = 0 ⇒y = 3 i.e., centre of circle
interesting isn't ? :))) ]
We have to find dy/dx of given equation x² + y² - 4x - 6y - 25 = 0
differentiating equation with respect to x,
d(x² + y² - 4x - 6y - 25)/dx = 0
⇒d(x²)/dx + d(y²)/dx - 4d(x)/dx - 6d(y)/dx - d(25)/dx = 0
⇒2x × dx/dx + 2y × dy/dx - 4 × dx/dx - 6 × dy/dx = 0
⇒2x + 2y (dy/dx) - 4 - 6(dy/dx) = 0
⇒(2x - 4 ) + (2y - 6)(dy/dx) = 0
⇒(2x - 4) = -(2y - 6)(dy/dx)
⇒(2x - 4) = (6 - 2y)(dy/dx)
⇒dy/dx = (2x - 4)/(6 - 2y)
hence, dy/dx = (2x - 4)/(6 - 2y)
[interesting point : the above mentioned equation is equation of circle, and centre of circle is (2, 3).
here you found dy/dx = (2x - 4)/(6 - 2y)
2x - 4 = 0 ⇒x = 2
and 6 - 3y = 0 ⇒y = 3 i.e., centre of circle
interesting isn't ? :))) ]