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3 years ago

Use lagrange multipliers to find the points on the given surface that are closest to the origin. y2 = 64 + xz

1 answer:

8 0

The distance between an arbitrary point on the surface and the origin is

$d(x,y,z)=\sqrt{x^2+y^2+z^2}$

Recall that for differentiable functions $g(x)$ and $h(x)$ , the composition $g(h(x))$ attains extrema at the same points that $h(x)$ does, so we can consider an augmented distance function

$D(x,y,z)=x^2+y^2+z^2$

The Lagrangian would then be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)$

We have partial derivatives

$\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}$

Set each partial derivative to 0 and solve the system to find the critical points.

From the second equation it follows that either $y=0$ or $\lambda=-1$ . In the first case we arrive at a contradiction (I'll leave establishing that to you). If $\lambda=-1$ , then we have

$\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0$

This means $y^2=64\implies y=\pm8$

so that the points on the surface closest to the origin are $(0,\pm8,0)$ .

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Yuliya22 [10]

Solving a 5th grade polynomial

We want to find the answer of the following polynomial:

$x^5+3x^4+3x^3+19x^2-54x-72=0$

We can see that the last term is -72

We want to find all the possible numbers that can divide it. Those are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

We want to factor this polynomial in order to find all the possible x-values. In order to factor it we will have to find some binomials that can divide it using the set of divisors of -72.

We know that if

(x - z) is a divisor of this polynomial then z might be a divisor of the last term -72.

We will verify which is a divisor using synthetic division. If it is a divisor then we can factor using it:

Let's begin with

(x-z) = (x - 1)

We want to divide

$\frac{(x^5+3x^4+3x^3+19x^2-54x-72)}{x-1}$

Using synthetic division we have that if the remainder is 0 it will be a factor

We can find the remainder by replacing x = z in the polynomial, when it is divided by (x - z). It is to say, that if we want to know if (x -1) is a factor of the polynomial we just need to replace x by 1, and see the result:

If the result is 0 it is a factor

If it is different to 0 it is not a factor

Replacing x = 1

If we replace x = 1, we will have that:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ 1^5+3\cdot1^4+3\cdot1^3+19\cdot1^2-54\cdot1-72 \\ =1+3+3+19-54-72 \\ =-100 \end{gathered}$

Then the remainder is not 0, then (x - 1) is not a factor.

Similarly we are going to apply this until we find factors:

(x - z) = (x + 1)

We replace x by -1:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ (-1)^5+3\cdot(-1)^4+3\cdot(-1)^3+19\cdot(-1)^2-54\cdot(-1)-72 \\ =-1+3-3+19+54-72 \\ =0 \end{gathered}$

Then, (x + 1) is a factor.

Using synthetic division we have that:

Then:

$x^5+3x^4+3x^3+19x^2-54x-72=(x+1)(x^4+2x^3+x^2+18x-72)$

Now, we want to factor the 4th grade polynomial.

Let's remember our possibilities:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

Since we verified ±1, let's try with ±2 as we did before.

(x - z) = (x - 2)

We want to divide:

$\frac{x^4+2x^3+x^2+18x-72}{x-2}$

We replace x by z = 2:

$\begin{gathered} x^4+2x^3+x^2+18x-72 \\ \downarrow \\ 2^4+2\cdot2^3+2^2+18\cdot2-72 \\ =16+16+4+36-72 \\ =0 \end{gathered}$

Then (x - 2) is a factor. Let's do the synthetic division:

Then,

$x^4+2x^3+x^2+18x-72=(x-2)(x^3+4x^2+9x+36)$

Then, our original polynomial is:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =\mleft(x+1\mright)\mleft(x^4+2x^3+x^2+18x-72\mright) \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \end{gathered}$

Now, let's prove if (x +2) is a factor, using the new 3th grade polynomial.

(x - z) = (x + 2)

We replace x by z = -2:

$\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-2)^3+4(-2)^2+9(-2)+36 \\ =-8+16-18+36 \\ =26 \end{gathered}$

Since the remainder is not 0, (x +2) is not a factor.

All the possible cases are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

let's prove with +4

(x - z) = (x + 4)

We want to divide:

$\frac{x^3+4x^2+9x+36}{x+4}$

Let's replace x by z = -4 in order to find the remainder:

$\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-4)^3+4(-4)^2+9(-4)+36 \\ =-64+64-36+36 \\ =0 \end{gathered}$

Then (x + 4) is a factor. Let's do the synthetic division:

Then,

$x^3+4x^2+9x+36=(x+4)(x^2+9)$

Since

x² + 9 cannot be factor, we have completed our factoring:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \\ =(x-1)(x-2)(x+4)(x^2+9) \end{gathered}$

Now, we have the following expression:

$(x-1)(x-2)(x+4)(x^2+9)=0$

Then, we have five posibilities:

(x - 1) = 0

or (x - 2) = 0

or (x + 4) = 0

or (x² + 9) = 0

Then, we have five solutions;

x - 1 = 0 → x₁ = 1

x - 2 = 0 → x₂ = 2

x + 4 = 0 → x₃ = -4

x² + 9 = 0 → x² = -9 → x = ±√-9 = ±√9√-1 = ±3i

→ x₄ = 3i

→ x₅ = -3i

<h2><em>Answer- the solutions of the polynomial are: x₁ = 1, x₂ = 2, x₃ = -4, x₄ = 3i and x₅ = -3i</em></h2>

7 0

11 months ago

I don't need you to work it out, just theorem(s) i need to reach the answer :)

VikaD [51]

Not sure why such an old question is showing up on my feed...

Anyway, let $x=\tan^{-1}\dfrac43$ and $y=\sin^{-1}\dfrac35$ . Then we want to find the exact value of $\cos(x-y)$ .

Use the angle difference identity:

$\cos(x-y)=\cos x\cos y+\sin x\sin y$

and right away we find $\sin y=\dfrac35$ . By the Pythagorean theorem, we also find $\cos y=\dfrac45$ . (Actually, this could potentially be negative, but let's assume all angles are in the first quadrant for convenience.)

Meanwhile, if $\tan x=\dfrac43$ , then (by Pythagorean theorem) $\sec x=\dfrac53$ , so $\cos x=\dfrac35$ . And from this, $\sin x=\dfrac45$ .

So,

$\cos\left(\tan^{-1}\dfrac43-\sin^{-1}\dfrac35\right)=\dfrac35\cdot\dfrac45+\dfrac45\cdot\dfrac35=\dfrac{24}{25}$

7 0

3 years ago

Please van you work out 5/6 of 54

Talja [164]

Answer:

Step-by-step explanation:

1/6 of 54= 9

9*5=45

5 0

3 years ago

Analyzing a Graph In Exercise, analyze and sketch the graph of the function. Lable any relative extrema, points of inflection, a

Anit [1.1K]