Mathematics, 14.12.2020 01:50, khikhi1705

Consider the curve given by the equation y^2-2x^2y=3. a) Find dy/dx .
b) Write an equation for the line tangent to the curve at the point (1, –1).
c) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is horizontal.
d) Evaluate d^2y/dx^2 at the point (1, –1)

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answered: Pitts1971

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:

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

answered: Guest

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

$4/2 = 2\\7/3.5=2\\8/4=2\\10/5=2$

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

$y=\frac{x}{2}$

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.

answered: Guest

I’m vary good with science if you need with that