What is the slope of the tangent line of # (xy-y^2)(1+x) =C #, where C is an arbitrary constant, at #(-3,1)#?

Question

Christopher Agresta · Accepted Answer

#3x-5y+14=0#. Tangent-inclusive graph is inserted.

As# P(-3, 1) is on the graph, C = ((-3)(1)-1^2)(1-3)=0#. Differentiating, #(xy'+y-2yy')(1+x)+(xy-y^2)(1)=0#. At P, #(-3y'+1-2y')(1-3)+(-3-1)=0#, giving the slope of the tangent at P #y'=3/5#. So, the equation of the tangent is #y-1=3/5(x+3)#, giving #3x-5y+14=0#. I have used a parallel line, in proximity of the tangent, to keep off the gap, at the point of contact. In this graphics method, P appears as a gap, for the exact equation of the tangent. For the interested reader, this graph is also included. In this graph, the pixels at P do not glow. graph{((xy-y^2)(1+x)-8)(3x-5y+13.7)=0 [-9.79, 9.785, -4.895, 4.895]} graph{((xy-y^2)(1+x)-8)(3x-5y+14)=0 [-9.79, 9.785, -4.895, 4.895]}

Luke Alvarez · Answer

undefined To find the slope of the tangent line at the point (-3, 1) for the curve $ (xy - y^2)(1+x) = C $, we need to differentiate the equation implicitly with respect to x and then evaluate it at the given point (-3, 1).

Differentiating implicitly, we get:

$$ (xy - y^2)(1+x) = C $$

$$ (y + xy' - 2yy')(1 + x) + (xy - y^2)(1) = 0 $$

$$ (y + xy' - 2yy' + xy - y^2) + x(y + xy' - 2yy') = 0 $$

$$ xy' - 2yy' + xy - y^2 + xy + x^2y' - 2xyy' = 0 $$

$$ y'(x + x^2 - 2y - 2xy) = y^2 - xy $$

$$ y' = \frac{y^2 - xy}{x^2 - 2xy + x} $$

Now, substitute (-3, 1) into the expression for $ y' $ to find the slope of the tangent line at that point.

$$ y' = \frac{(1)^2 - (-3)(1)}{(-3)^2 - 2(-3)(1) - 3} $$

$$ y' = \frac{1 + 3}{9 + 6 - 3} $$

$$ y' = \frac{4}{12} $$

$$ y' = \frac{1}{3} $$

So, the slope of the tangent line at the point (-3, 1) is $ \frac{1}{3} $.