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

Question

Ezra Adams · Accepted Answer

I got #m = -3/23#

Given: # 3y^2+y/x+x^2/y =C #, and #(2,2)# lies on the graph.

Find: the slope of the tangent line to the graph at the point #(2,2)#.

Solution:

#C = 3(2)^2+2/2+(2)^2/2 = 15#

So # 3y^2+y/x+x^2/y =15 #,

and #3xy^3+y^2+x^3 = 15xy#.

Differentiating with respect to #x# yields,

#3y^3+9xy^2 dy/dx + 2y dy/dx + 3x^2 = 15y+15x dy/dx#.

Therefore,

#(9xy^2 + 2y - 15x) dy/dx = 15y-3y^3- 3x^2#.

And, so,

#(9(2)(2)^2 + 2(2) - 15(2)) dy/dx = 15(2)-3(2)^3- 3(2)^2#.

#dy/dx = (30-24-12)/(72+4-30) = -6/46 = -3/23#

Isaiah Allen · Answer

#dy/dx=-3/23#

This can also be done without evaluating the constant and using quotient rule:

#6ydy/dx+(dy/dx(x)-y)/x^2+(2xy-dy/dx(x^2))/y^2=0#

Plug in the point #(2,2)# and solve for #dy/dx#.

#12dy/dx+(2dy/dx-2)/4+(8-4dy/dx)/4=0#
#48dy/dx+2dy/dx-2+8-4dy/dx=0#
#46dy/dx=-6#
#dy/dx=-3/23#

Emilia Aguilar · Answer

undefined The slope of the tangent line of the curve $3y^2 + \frac{y}{x} + \frac{x^2}{y} = C$ at the point (2,2) can be found by first finding the derivative of the curve with respect to $x$, then evaluating it at the given point.

Given curve: $3y^2 + \frac{y}{x} + \frac{x^2}{y} = C$

Taking the derivative of the curve with respect to $x$, we get:

$$6yy' + \frac{-y}{x^2} + \frac{2x}{y}y' - \frac{x^2}{y^2}y' = 0$$

$$6yy' + \frac{-y^2 + 2x^2}{xy}y' - \frac{x^2}{y^2}y' = 0$$

$$6yy' + \frac{-y^3 + 2x^2y}{xy^2} - \frac{x^2}{y^2}y' = 0$$

$$6yy' + \frac{-y^3 + 2x^2y - x^2y'}{xy^2} = 0$$

$$y'(6y - \frac{x^2}{y^2}) = \frac{y^3 - 2x^2y}{xy^2}$$

$$y' = \frac{xy^2}{6y - \frac{x^2}{y^2}} $$

Now, substituting $x = 2$ and $y = 2$ into the equation, we get:

$$y' = \frac{2 \cdot 2^2}{6 \cdot 2 - \frac{2^2}{2^2}} = \frac{8}{12 - 1} = \frac{8}{11}$$

Therefore, the slope of the tangent line of the curve at the point (2,2) is $\frac{8}{11}$.