Find an equation for the line tangent to the graph of y=x3+ √xy + y3=3 at the point (1,1).
SOLUTION :
Curve : x^3 + sqrt(xy) + y^3 = 3
=> x^3 + y^3 + x^(1/2) * y^(1/2) = 3
Differentiate w.r.to x :
=> 3x^2 + 3y^2 dy/dx - (1/(2x^(1/2) * y^(1/2) + x^(1/2) / (2y^(1/2) dy/dx) = 0
=> (3y^2 - 1/2 (x/y)^(1/2)) dy/dx = (-3x^2 + 1/2 (y/x)^(1/2))
=> dy/dx = (-3x^2 + 1/2 (y/x)^(1/2) / (3y^2 - 1/2 (x/y)^(1/2))
So, slope of tangent at point (1.1) of the curve
= (-3 + 1/2 * 1)/ (3 - 1/2 * 1)
= (-5/2) / (5/2)
= - 1
Equation of this tangent through point (1, 1) :
=> (y - 1) = - 1 (x - 1) (point-slope form (y- y1) = m(x - x1))
=> y - 1 = - x + 1
=> y = - x + 2 (ANSWER)
Find an equation for the line tangent to the graph of y=x3+ √xy + y3=3 at the point (1,1).
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