Implicit Differentiation of 2x^2 + xy - y^2 + 18 = 0

[milkyways]

A curve has the equation;

2x[sup:3i0b4lvd]2[/sup:3i0b4lvd] + xy - y[sup:3i0b4lvd]2[/sup:3i0b4lvd] + 18 = 0

Find the co-ordinates of the points where the tangent to the curve is parallel to the x-axis.

I differentiated to get:

dy/dx = y+4x/2y -x = 0

which is hopefully correct, but unfortunately I am not sure where to go next...

 

[pka]

Re: Implicit Differentiation Question

You wants the points where \(\displaystyle \frac {dy} {dx} =0\).
Or where \(\displaystyle y=-4x\).
Use that in the first equation and solve for x.

 

[galactus]

Re: Implicit Differentiation Question

Differentiate implicitly.

Can you do that OK?. You should get \(\displaystyle y'=\frac{-(4x+y)}{x-2y}\)

Set this to 0 and solve for y.

Then sub that back into the original so it'll be entirely in terms of x.

Then, set to 0 and solve for x and you should have two solutions. Let me know what you get.

Another thing you can do is solve the equation for y and differentiate the old-fashioned way. Then set to 0 and solve for x.

 

[Deleted member 4993]

Re: Implicit Differentiation Question

A curve has the equation;

2x[sup:1b63yo0x]2[/sup:1b63yo0x] + xy - y[sup:1b63yo0x]2[/sup:1b63yo0x] + 18 = 0

Find the co-ordinates of the points where the tangent to the curve is parallel to the x-axis.

I differentiated to get:

dy/dx = (y+4x)/(2y -x) = 0

You need to put parenthesis as indicated - otherwise it will mean completely different expression.

Now as indicated by others - just put the numerator equal to zero. That makes the whole expression equal to zero.




which is hopefully correct, but unfortunately I am not sure where to go next...