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3 years ago

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Convert y-11=-2(x-2)^2 from vertex to standard form

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

$y-11=-2(x-2)^2\\
y-11=-2(x^2-4x+4)\\
y-11=-2x^2+8x-8\\
y=-2x^2+8x+3\\$

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Evaluate the integral Integral from (5 comma 2 comma 4 )to (7 comma 9 comma negative 3 )y dx plus x dy plus 3 dz by finding para

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$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_{(5,2,4)}^{(7,9,-3)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz$

Parameterize the line segment (call it $C$ ) by

$\vec r(t)=(1-t)(5\,\vec\imath+2\,\vec\jmath+4\,\vec k)+t(7\,\vec\imath+9\,\vec\jmath-3\,\vec k)$

$\vec r(t)=(2t+5)\,\vec\imath+(7t+2)\,\vec\jmath+(4-7t)\,\vec k$

with $0\le t\le1$ . Then

$\vec r'(t)=2\,\vec\imath+7\,\vec\jmath-7\,\vec k$

and the line integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1((7t+2)\,\vec\imath+(2t+5)\,\vec\jmath+3\,\vec k)\cdot\vec r'(t)\,\mathrm dt$

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Alternatively, if we can show that $\vec F$ is conservative, then we can apply the fundamental theorem of calculus. We need to find $f$ such that $\nabla f=\vec F$ , which requires

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

$\dfrac{\partial f}{\partial z}=3$

Integrating both sides of the first equation with respect to $x$ gives

$f(x,y,z)=xy+g(y,z)$

Differentiating both sides wrt $y$ gives

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}$

$\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

Differentiating wrt $z$ gives

$\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}$

$\implies h(z)=3z+C$

So we have

$f(x,y,z)=xy+3z+C$

and $\vec F$ is conservative. By the FTC, we find

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=f(7,9,-3)-f(5,2,4)=\boxed{32}$

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