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4 years ago

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What is x if y Is 5 x^2 (2y +5)=240

1 answer:

UkoKoshka [18]4 years ago

6 0

X^2(2y+5)=240
x^2(2(5)+5)=240
x^2+(10+5)=240
x^2+15=240
minus 15 both sides
x^2=225
sqrt both sides
x=15

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Let φ(x, y) = arctan (y/x) .

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Answer:

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b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

So

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b)

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

c)

Here we want to find all the points such that

$\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)$

where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

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Step-by-step explanation:

I'm not good at explaining but I'll try.

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Answer:

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Step-by-step explanation:

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