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stich3 [128]
2 years ago
13

Please help me with this

Mathematics
1 answer:
creativ13 [48]2 years ago
7 0

Answer:

Step-by-step explanation:

(-3,-2)

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Match each description when z = 9 + 3i. 1. Real part of z 2. Imaginary part of z 3. Complex conjugate of z 4. 3i - z 5. z - 9 6.
sergey [27]
1)\quad \Re(9+3i)=9\qquad\text{F}\\\\
2)\quad \Im(9+3i)=3\qquad\text{E}\\\\
3)\quad \overline{(9+3i)}=9-3i\qquad\text{D}\\\\
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3 years ago
Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line seg
n200080 [17]

\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k

is conservative if there is a scalar function f(x,y,z) such that \nabla f=\vec F. This would require

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

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

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

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

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

\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)

f(x,y,z)=xy+h(z)

\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C

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

so \vec F is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r

=f(5,7,-2)-f(1,2,3)=\boxed{18}

8 0
3 years ago
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