Question

Consider the curve defined by 2x2+3y2−4xy=36 2 x 2 + 3 y 2 − 4 x y = 36 .
(a) Show that dy/dx=2y−2x / 3y−2x

Answer 1

Answer:

Step-by-step explanation:

Given a curve defined by the function 2x²+3y²−4xy=36

The total differential of this function with respect to a variable x makes the function an implicit function because it contains two variables.

Differentiating both sides of the equation with respect to x we have:

4x+6ydy/dx-(4xd(y)/dx+{d(4x)/dx(y))} = 0

4x + 6ydy/dx -(4xdy/dx +4y) = 0

4x + 6ydy/dx - 4xdy/dx -4y = 0

Collecting like terms

4x-4y+6ydy/dx - 4xdy/dx = 0

4x-4y+(6y-4x)dy/dx = 0

4x-4y = -(6y-4x)dy/dx

4y-4x = (6y-4x)dy/dx

dy/dx = (4y-4x)/6y-4x

dy/dx = 2(2y-2x)/2(3y-2x)

dy/dx = 2y-2x/3y-2x proved!

Answer 2

Following are the calculation to the given curve:

Given:

$2x^2+3y^2-4xy=36$

To find:

curve =?

Solution:

Since this function involves two variables, its cumulative difference in respect to a variable x qualifies this as an implicit function.

When dividing all sides of the issue by x, we get:

$\to 4x+6y \frac{dy}{dx}-(4x\frac{d(y)}{dx}+{\frac{d(4x)}{dx(y)} } = 0\\\\\to 4x + 6y\frac{dy}{dx} -(4x \frac{dy}{dx} +4y) = 0\\\\\to 4x + 6y\frac{dy}{dx} - 4x\frac{dy}{dx} -4y = 0$

Collecting similar terms:

$\to 4x-4y+6y\frac{dy}{dx} - 4x\frac{dy}{dx} = 0\\\\\to 4x-4y+(6y-4x)\frac{dy}{dx} = 0\\\\ \to 4x-4y = -(6y-4x)\frac{dy}{dx}\\\to 4y-4x = (6y-4x)\frac{dy}{dx}\\\\\to \frac{dy}{dx} = \frac{(4y-4x)}{6y-4x}\\\\\to \frac{dy}{dx}= \frac{2(2y-2x)}{2(3y-2x)}\\\\\to \frac{dy}{dx} = \frac{2y-2x}{3y-2x}$

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