It is given that the equation of a curve is y^2+2xy=4
(a) Express dy/dx in terms of x and y
(b)Hence, find the equation of the tangent to hte curve at the point(3/2 , 1)
applications of differentiatio
2011-03-10 6:14 am
回答 (2)
2011-03-10 6:27 am
2011-03-10 6:28 am
(a) Differentiation on both sides w.r.t x
(2y)(dy/dx) + 2x(dydx) + 2y = 0
dy/dx = -y/(x + y)
(b) At point (3/2,1) dy/dx = -2/5
By point - slope form
y - 1 = (-2/5)(x - 3/2)
5y - 5 = -2x + 3
2x + 5y - 8 = 0
(2y)(dy/dx) + 2x(dydx) + 2y = 0
dy/dx = -y/(x + y)
(b) At point (3/2,1) dy/dx = -2/5
By point - slope form
y - 1 = (-2/5)(x - 3/2)
5y - 5 = -2x + 3
2x + 5y - 8 = 0
收錄日期: 2021-04-23 23:28:15
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