Question

Find the extremums of the function? `y^2+2x2y+4x-3=0` Thanks much..

Answer 1

Please see below.

Explanation:

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`y^2+4xy+4x-3=0`

We solve for `y` using the quadratic formula:

`y=(-4x+-sqrt(16x^2-4(4x-3)))/2=(-4x+-2sqrt(4x^2-4x+3))/2`

`y=-2x+-sqrt(4x^2-4x+3)`

To find the extrema of the function, we take its derivative, set it equal to `0`, and solve for `x`:

`dy/dx=-2+-(8x-4)/(2sqrt(4x^2-4x+3))=-2+-(2(2x-1))/sqrt(4x^2-4x+3)=0`

`+-(2(2x-1))/sqrt(4x^2-4x+3)=2`

`(4(2x-1)^2)/(4x^2-4x+3)=4`

`((2x-1)^2)/(4x^2-4x+3)=1`

`4x^2-4x+1=4x^2-4x+3`

`1=3`

This indicates that there are no solutions for `x`. As such, the function does not have any extrema. It can be argued that its extrema occur at `x=+-oo` as we will see later.

We know that the slope of a tangent line to a curve at a specific point can be found by evaluating its derivative at that point. Since the function does not have any extrema and it is of higher order than one, it must have aymptotes.

This means that the asymptotes are tangent to the curve at `+-oo`.

To evaluate the derivative at `+-oo`, we calculate its limit when `x->+-oo`.

`Lim_(x->oo)-2+-(2(2x-1))/sqrt(4x^2-4x+3)=Lim_(x->oo)-2+-(4-2/x)/sqrt(4-4/x+3/x^2)=-2+-2, :. m=0 and -4`

This means there are two asymptotes with slopes of `0` and `-4` as follows:

`slope=0` is a horizontal asymptote. To find its equation, we evaluate the function when `x->+-oo`

`Lim_(x->oo)-2x+sqrt(4x^2-4x+3)=Lim_(x->oo)(-2x+sqrt(4x^2-4x+3))*(2x+sqrt(4x^2-4x+3))/(2x+sqrt(4x^2-4x+3))=Lim_(x->oo)(4x^2-4x+3-4x^2)/(2x+sqrt(4x^2-4x+3))=Lim_(x->oo)((-4x+3)/(2x+sqrt(4x^2-4x+3)))=Lim_(x->oo)((-4+3/x)/(2+sqrt(4-4/x+3/x^2)))=-4/(2+-2)=-1 and oo`

`y=-1` is the horizontal asymptotte.

`y=oo` indicates that the function has an oblique (slant) asymptote.

If we repeat the above process for:

`Lim_(x->oo)-2x-sqrt(4x^2-4x+3)` as well as repeating it for both of them with `x->-oo`

we will obtain the results of `-1 and -oo` which indicates that the function is in two pieces and asymptotic in all four directions.

In order to find the equation of the oblique asymptote, we will make it much easier if we could avoid dealing with radicals. As such, we will find `x` as a function of `y` and perform a long division:

`y^2+4xy+4x-3=0`

`y^2+4x(y+1)-3=0`

`4x(y+1)=3-y^2`

`x=(3-y^2)/(4y+4)=-1/4y+1/4+2/(4(1+y))`

`4x=-y+1`

`y=-4x+1`

As you can see, this is a straight line with slope `m=-4`

Now, we can graph both asymptotes first, then find the coordinates of a number of points using the equation of the function and connect them to construct the full graph as shown below:

`x=0, :. y=+-sqrt3, :. (0,sqrt3) and (0,-sqrt3)` are the `y` intercepts.

`y=0, :. x=3/4, :. (3/4,0)` is the `x` intercept.

The function is a hyperbola with the center located at the intersection of the two asymptotes.

We can find the center by setting the equations of the asymptotes equal to each other and solving for `x` and `y`:

`-4x+1=-1`

`x=1/2`, `y=-1`

`C(1/2,-1)`

In order to find the equation of this hyperbola in its standard form, we would have to get rid of the `4xy` term in the function so that we can separate the `x's` ans `y's`.

That process requires rotation of the coordinates axes by an angle `theta` and using trigonometric functions to obtain the equation of the function in the new coordinate system; and after simplification, convert it back to our original coordinate system.

I spared us from the pain.

Answer 2

`pi/4(sqrt(R^2-a^2)-sqrt(R^2-b^2))`

Explanation:

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Hi Lany,

I am finally getting the chance to look at the problem you sent me. I am not sure what you meant to input. Especially, at the bottom I was not sure what you indicated for the interval of `x`.

So, I am going to make some guesses and try to figure this out based on some assumptions on my part.

`int_Gint dA`, with `G` under both integral signs, yields the area of the region `G`.

`int_Gint f(x,y)dA`, with `G` under both integral signs, yields the volume of a solid over the region `G`.

In the cartesian coordinate system, the increment of the area `dA=dxdy` and we integrate with respect to `x` and `y`.

In the polar coordinate system, `dA=rdrd theta` and we integrate with respect to `r` and `theta`.

In your problem, the integral is in the cartesian coordinate system:

`int_Gint(dxdy)/sqrt(R^2-x^2-y^2)`

with the following boundary information(as far as I could figure out):

The region `G` is bounded by:

`x^2+y^2=a^2`, `x^2+y^2=b^2`

`y=x`, `y=sqrt(3x)`, `a < b < R,`, `y >= 0`, and `x >=0`

The last piece was unclear to me. I will just use `x >=0`

Let's see what this region looks like. We will graph `y=x` in red, and `y=sqrt(3x)` in green:

So, the region lies in between the two curves. Now let's graph `x^2+y^2=a^2` in blue and `x^2+y^2=b^2` in purple. Let's say `a=1, and b=2`:

These two circles further restrict `G` to the area shaded in yellow.

The given integral calculates the volume of a solid that has the yellow area `G` as its base and height of `z` that has a lower and upper `z`-plane defined by lower and upper values of `z`.

Let's look at the graph of the function inside the integral with `R=3`:

`1/sqrt(9-x^2-y^2)`:

The real part is:

The imaginary part is:

Imagine the yellow shaded region `G` shown on the two-dimensional graph above being inside the base of this three-dimensional object, with a vertical side coming up parallel to the `z`-axis.

Additionally, imagine two `z`-planes cutting through the object. I have let the two planes be:

`z=2.5` and `z=4`

The volume of the piece bounded by the cylinder I described above and these two planes is the result of the integral given in our problem. It looks like this:

Let's convert the integral to a polar one:

We set the two given boundary functions equal to each other and solve for their intersection points:

`y=x`

`y=sqrt(3x)`

`x=sqrt(3x)`

`x^2=3x`

`x^2-3x=0`

`x(x-3)=0`

`x=0, x=3`

In a polar system:

`x=rcostheta` and `y=rsintheta`

`rsintheta=rcostheta`

`sintheta=costheta`

`tantheta=1`

`theta=arctan(1), :. theta=pi/4`

`x=rcostheta, if x=0, :. r=0, theta=pi/4`

`rsintheta=sqrt(3rcostheta)`

`r^2sin^2theta=3rcostheta`

`rsin^2theta=3costheta`

`r=(3costheta)/sin^2theta`

`theta=pi/4, :. r=(3(sqrt2/2))/((sqrt2/2))^2=3sqrt2`

`x^2+y^2=r^2`, and `dA=rdrd theta`

`int_Gint(dxdy)/sqrt(R^2-x^2-y^2)=int_Gint(rdrd theta)/sqrt(R^2-r^2)=`

`int_0^(pi/4)int_a^b1/sqrt(R^2-r^2)rdrd theta=int_0^(pi/4)I`

`I=int_a^br/sqrt(R^2-r^2)dr=int_a^br(R^2-r^2)^(-1/2)dr`

Let `u=R^2-r^2, :. du=-2rdr, :. rdr=-(du)/2`

`I=int-1/2u^(-1/2)du=-1/2intu^(-1/2)du=-1/2*2u^(1/2)=-1u^(1/2)`

`I=-1(R^2-r^2)^(1/2)=(-sqrt(R^2-r^2))_a^b=`

`-1(sqrt(R^2-b^2)-sqrt(R^2-a^2))`

`int_0^(pi/4)I=int_0^(pi/4)(sqrt(R^2-a^2)-sqrt(R^2-b^2))d theta=`

`(theta(sqrt(R^2-a^2)-sqrt(R^2-b^2)))_0^(pi/4)=`

`pi/4(sqrt(R^2-a^2)-sqrt(R^2-b^2))`

Answer 3

There are no critical points but the function has critical point behaviour at its asymptotes of `y=+-1`

Explanation:

We have:

`y^2+2x2y + 4x-3=0`

Differentiating implicitly wrt `x`:

`2y dy/dx + 4(xdy/dx+y)+4 = 0`

At an extremum, or critical point, the first derivative vanishes, so putting `dy/dx=0`, we have:

`0 + 4(0+y)+4 = 0 => 4y+4 = 0`
`" " => y=-1`

Substituting `y=-1` into the original equation we have:

`1-4x + 4x-3=0 => 0 =0 =>` no `y` solution

If we examine the graph of the function, this becomes clear as `y=-1` is an asymptote:

graph{y^2+2x2y + 4x-3=0 [-10, 10, -5, 5]}