Respuesta :
Answer:
Part A)
[tex]\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}[/tex]
Part B)
[tex]y=x-2[/tex]
Part C)
[tex](0, \sqrt{3})\text{ and } (0, -\sqrt3)[/tex]
Part D)
[tex]\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}[/tex]
Step-by-step explanation:
We have the equation:
[tex]y^2-2x^2y=3[/tex]
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:
[tex]\displaystyle \frac{d}{dx}\Big[y^2-2x^2y\Big]=\frac{d}{dx}[3][/tex]
Differentiate. We will need to implicitly different on the left. The second term will also require the product rule. Therefore:
[tex]\displaystyle 2y\frac{dy}{dx}-4xy-2x^2\frac{dy}{dx}=0[/tex]
Rearranging gives:
[tex]\displaystyle \frac{dy}{dx}\Big(2y-2x^2\Big)=4xy[/tex]
Therefore:
[tex]\displaystyle \frac{dy}{dx}=\frac{4xy}{2y-2x^2}[/tex]
And, finally, simplifying:
[tex]\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}[/tex]
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:
[tex]\displaystyle \frac{dy}{dx}_{(1, -1)}=\frac{2(1)(-1)}{(-1)-(1)^2}=1[/tex]
Now, we can use the point-slope form:
[tex]y-y_1=m(x-x_1)[/tex]
Substitute:
[tex]y-(-1)=1(x-1)[/tex]
Simplify:
[tex]y+1=x-1[/tex]
So:
[tex]y=x-2[/tex]
Part C)
If the line tangent to the curve is horizontal, this means that dy/dx=0. Hence:
[tex]\displaystyle 0=\frac{2xy}{y-x^2}[/tex]
Multiplying both sides by the denominator gives:
[tex]0=2xy[/tex]
Assuming y is not 0, we can divide both sides by y. Hence:
[tex]2x=0[/tex]
Then it follows that:
[tex]x=0[/tex]
Going back to our original equation, we have:
[tex]y^2-2x^2y=3[/tex]
Substituting 0 for x yields:
[tex]y^2=3[/tex]
So:
[tex]y=\pm\sqrt{3}[/tex]
Therefore, two points where the derivative equals 0 is:
[tex](0, \sqrt{3})\text{ and } (0, -\sqrt3)[/tex]
However, we still have to test for y. Let’s go back. We have:
[tex]0=2xy[/tex]
Assuming x is not 0, we can divide both sides by x. So:
[tex]0=2y[/tex]
Therefore:
[tex]y=0[/tex]
And going back to our original equation and substituting 0 for y yields:
[tex](0)^2-2x^2(0)=0\neq3[/tex]
Since this is not true, we can disregard this case.
So, our only points where the derivative equals 0 is at:
[tex](0, \sqrt{3})\text{ and } (0, -\sqrt3)[/tex]
Part D)
Our first derivative is:
[tex]\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}[/tex]
Let’s take the derivative of both sides again. Hence:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\Big[\frac{2xy}{y-x^2}\Big][/tex]
Utilize the quotient and product rules and differentiate:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2x\frac{dy}{dx})(y-x^2)-2xy(\frac{dy}{dx}-2x)}{(y-x^2)^2}[/tex]
Let dy/dx=y’. Therefore:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2xy^\prime)(y-x^2)-2xy(y^\prime-2x)}{(y-x^2)^2}[/tex]
For (1, -1), we already know that y’ is 1 at (1, -1). Therefore:
[tex]\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}[/tex]
Evaluate:
[tex]\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}[/tex]
