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2 years ago

1. Write a linear equation with the given information

Mathematics

1 answer:

Readme [11.4K]2 years ago

3 0

1) The equation of the parallel line in Slope Intercept Form is; y = -x - 1

2) The equation of the perpendicular line in Slope Intercept Form is;

y = ³/₂x - 4

<h3>How to write equation in Slope Intercept Form?</h3>

1) When two lines are parallel, it means their slopes are the same.

Formula for equation in slope intercept form is;

y = mx + c

where;

m is slope

c is y-intercept

Thus, slope of y = -x + 1 is m = -1.

Equation of parallel line is;

(y - (-5))/(x - 4) = -1

y + 5 = -x + 4

y = -x - 1

2) When two lines are perpendicular, it means their slopes are the negative inverse of each other. Thus;

Slope of line = -2/3 and so slope of perpendicular line = 3/2. Thus, equation of new line is;

(y + 1)/(x - 2) = 3/2

2y + 2 = 3x - 6

2y = 3x - 8

y = ³/₂x - 4

Read more about slope intercept form at; brainly.com/question/1884491

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$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

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The curl is

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where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

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