Question

Consider the curve given by the equation y^2-2x^2y=3.

Answer 1

Answer:

Part A)

$\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}$

Part B)

$y=x-2$

Part C)

$(0, \sqrt{3})\text{ and } (0, -\sqrt3)$

Part D)

$\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}$

Step-by-step explanation:

We have the equation:

$y^2-2x^2y=3$

Part A)

We want to find the derivative of the equation. So, dy/dx.

Let’s take the derivative of both sides with respect to x. Therefore:

$\displaystyle \frac{d}{dx}\Big[y^2-2x^2y\Big]=\frac{d}{dx}[3]$

Differentiate. We will need to implicitly different on the left. The second term will also require the product rule. Therefore:

$\displaystyle 2y\frac{dy}{dx}-4xy-2x^2\frac{dy}{dx}=0$

Rearranging gives:

$\displaystyle \frac{dy}{dx}\Big(2y-2x^2\Big)=4xy$

Therefore:

$\displaystyle \frac{dy}{dx}=\frac{4xy}{2y-2x^2}$

And, finally, simplifying:

$\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}$

Part B)

We want to write the equation for the line tangent to the curve at the point (1, -1).

So, we will require the slope of the tangent line at (1, -1). Substitute these values into our derivative. So:

$\displaystyle \frac{dy}{dx}_{(1, -1)}=\frac{2(1)(-1)}{(-1)-(1)^2}=1$

Now, we can use the point-slope form:

$y-y_1=m(x-x_1)$

Substitute:

$y-(-1)=1(x-1)$

Simplify:

$y+1=x-1$

So:

$y=x-2$

Part C)

If the line tangent to the curve is horizontal, this means that dy/dx=0. Hence:

$\displaystyle 0=\frac{2xy}{y-x^2}$

Multiplying both sides by the denominator gives:

$0=2xy$

Assuming y is not 0, we can divide both sides by y. Hence:

$2x=0$

Then it follows that:

$x=0$

Going back to our original equation, we have:

$y^2-2x^2y=3$

Substituting 0 for x yields:

$y^2=3$

So:

$y=\pm\sqrt{3}$

Therefore, two points where the derivative equals 0 is:

$(0, \sqrt{3})\text{ and } (0, -\sqrt3)$

However, we still have to test for y. Let’s go back. We have:

$0=2xy$

Assuming x is not 0, we can divide both sides by x. So:

$0=2y$

Therefore:

$y=0$

And going back to our original equation and substituting 0 for y yields:

$(0)^2-2x^2(0)=0\neq3$

Since this is not true, we can disregard this case.

So, our only points where the derivative equals 0 is at:

$(0, \sqrt{3})\text{ and } (0, -\sqrt3)$

Part D)

Our first derivative is:

$\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}$

Let’s take the derivative of both sides again. Hence:

$\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\Big[\frac{2xy}{y-x^2}\Big]$

Utilize the quotient and product rules and differentiate:

$\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2x\frac{dy}{dx})(y-x^2)-2xy(\frac{dy}{dx}-2x)}{(y-x^2)^2}$

Let dy/dx=y’. Therefore:

$\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2xy^\prime)(y-x^2)-2xy(y^\prime-2x)}{(y-x^2)^2}$

For (1, -1), we already know that y’ is 1 at (1, -1). Therefore:

$\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=\frac{(2(-1)+2(1)(1))((-1)-(1)^2)-2(1)(-1)((1)-2(1))}{((-1)-(1)^2)^2}$

Evaluate:

$\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}$