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標題:

A-MATHS

發問:

The point A(1,3) is on the curve y2 -2y√x =3 , where x>0 (a)find the value of dy/dx at A (b)find the equation to the curve a A

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a) y2 -2y√x =3 2y ( dy / dx ) - 2y ( d / dx )( x^0.5 ) - 2x^0.5 ( d / dx )( y ) = 0 2y ( dy / dx ) - 2y ( 0.5 x^-0.5 ) - 2x^0.5 ( dy / dx ) = 0 2 ( y - x^0.5 )( dy/dx ) = x^-0.5y dy / dx = ( x^-0.5 y ) / 2 ( y - x^0.5 ) dy / dx ( 1, 3 ) = ( 1 )( 3 ) / 2 ( 3 - 1 ) = 3 / 4 b) Equation: ( y - 3 ) / ( x - 1 ) = 3 / 4 3x - 4y + 9 = 0 2008-02-05 14:48:11 補充： I think you want to find the equation of tangent to the curve at A, but if it is normal, then the equation will be:4x + 3y - 13 = 0

其他解答:

3x - 4y + 9 = 0