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solong [7]
3 years ago
13

Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,

Mathematics
1 answer:
gogolik [260]3 years ago
5 0

\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k

We're looking for a scalar function f such that \nabla f=\vec f. This means

\dfrac{\partial f}{\partial x}=yze^{xz}

\dfrac{\partial f}{\partial y}=e^{xz}

\dfrac{\partial f}{\partial z}=xye^{xz}

In the second PDE, integrate both side with respect to y:

\dfrac{\partial f}{\partial y}=e^{xz}\implies f(x,y,z)=ye^{xz}+g(x,z)

Differentiate both sides of this with respect to x:

yze^{xz}=yze^{xz}+\dfrac{\partial g}{\partial x}

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

\implies f(x,y,z)=ye^{xz}+h(z)

Differentiate this with respect to z:

xye^{xz}=xye^{xz}+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=K

for some constant K, so that

f(x,y,z)=ye^{xz}+K

By the fundamental theorem of calculus, this line integral has a value of

\displaystyle\int_C\nabla f\cdot\mathrm d\vec r=f(\vec r(5))-f(\vec r(0))=f(30,22,20)-f(5,-3,0)=\boxed{22e^{600}+3}

since

\vec r(5)=30\,\vec\imath+22\,\vec\jmath+20\,\vec k

\vec r(0)=5\,\vec\imath-3\,\vec\jmath

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