Implicit differentiation

[Rumor]

Here's the problem:

"Consider the curve defined by y^2+xy+x^2=15.

a) Calculate dy/dx.

b) Show that at the two points where y=0, the tangent lines at those points are parallel.

c) Find the coordinates of all points at which there is a horizontal tangent line. Show your work."

For dy/dx, I got dy/dx= -(2x-y)/(2y+x). Is this correct?

Also, could someone show me how to do parts b) and c)?

 

[Deleted member 4993]

Here's the problem:

"Consider the curve defined by y^2+xy+x^2=15.

a) Calculate dy/dx.

b) Show that at the two points where y=0, the tangent lines at those points are parallel.

c) Find the coordinates of all points at which there is a horizontal tangent line. Show your work."

For dy/dx, I got dy/dx= -(2x-y)/(2y+x). Is this correct? <<< Almost - you have problem with "sign"
Also, could someone show me how to do parts b) and c)?

b) Show that at the two points where y=0, the tangent lines at those points are parallel.

at y =0 , ? x^2 = 15 ? x = ±sqrt(15)

Now find slopes of tangents at (sqrt(15), 0) and (-sqrt(15), 0).

c) What is the value of dy/dx when the tangent line is horizontal?

 

[Rumor]

Hm, okay.

When there's a horizontal tangent line, that means the numerator = 0, right?

 

[Deleted member 4993]

Hm, okay.

When there's a horizontal tangent line, that means the numerator = 0, right? <<< Numerator of what? Why?

 

[Rumor]

Ah, never mind that. I'm getting concepts mixed up.

It's asking for the points where the derivative equals 0?

 

[BigGlenntheHeavy]

$\displaystyle y^{2}+xy+x^{2}=15, \ Implicit \ differentation \ yields: \ 2yy'+y+xy'+2x=0$

$\displaystyle 2yy'=xy' \ = \ -2x-y, \ y'=-\frac{2x+y}{x+2y}$

$\displaystyle Horizontal \ Tangents: \ 2x+y \ = \ 0 \ \implies \ y \ = \ -2x \ (black \ lines)$

$\displaystyle (-2x)^{2}+x(-2x)+x^{2} \ = \ 15, \ 4x^{2}-2x^{2}+x^{2} \ = \ 15, \ 3x^{2} \ = \ 15, \ \implies \ x \ = \ \pm\sqrt5$

$\displaystyle (\sqrt5,-2\sqrt5), \ m \ =0 \ and \ (-\sqrt5,2\sqrt5), \ m \ = \ 0 \ gives \ y \ = \ 2\sqrt5 \ and \ y \ = \ -2\sqrt5$

$\displaystyle Vertical \ Tangents: \ x+2y \ = \ 0 \ \implies \ x \ = \ -2y \ (green \ lines)$

$\displaystyle y^{2}+(-2y)y+(-2y)^{2} \ = \ 15, \ y^{2}-2y^{2}+4y^{2} \ = \ 15, \ 3y^{2} \ = \ 15 \ \implies \ y \ = \ \pm\sqrt5$

$\displaystyle Hence, \ x \ = \ -2\sqrt5 \ and \ x \ = \ 2\sqrt5, \ see \ graph.$

[attachment=0:esmgwyzi]good.jpg[/attachment:esmgwyzi]

 

[Rumor]

Ah, I see..

Thank you both!