What is the slope of the line tangent to the curve y3−xy2+x3=5 at the point (1,2) ?
Correct. The point (1,2)(1,2) is on the curve since x=1x=1, y=2y=2 satisfies the implicit equation of the curve. The chain rule is the basis for implicit differentiation. 3y2dydx−y2−2xydydx+3×2=03y2dydx−y2−2xydydx+3×2=0 At the point (1,2)(1,2), 12dydx−4−4dydx+3=0⇒8dydx−1=0⇒dydx=1812dydx−4−4dydx+3=0⇒8dydx−1=0⇒dydx=18.
To find the slope of the line tangent to the curve at the given point (1,2), we will need to find the derivative of the curve at that point using implicit differentiation.
Taking the derivative of both sides of the equation y^3 – xy^2 + x^3 = 5 with respect to x, we get:
3y^2 dy/dx – y^2 – 2xy dy/dx + 3x^2 = 0
Simplifying this equation, we get:
dy/dx = (y^2 – 3x^2) / (3y^2 – 2xy)
Plugging in the values of x = 1 and y = 2, we get:
dy/dx = (2^2 – 3(1)^2) / (3(2)^2 – 2(1)(2))
Simplifying further, we get:
dy/dx = -1/10
Therefore, the slope of the line tangent to the curve at the point (1,2) is -1/10.
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