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kari74 [83]
2 years ago
10

Let F(x, y, z) = (5ex sin(y))i + (5ex cos(y))j + 7z2k. Evaluate the integral C F · ds, where c(t) = 8 t , t3, exp( t ) , 0 ≤ t ≤

1. (Note that exp(u) = eu.)
Mathematics
1 answer:
maria [59]2 years ago
4 0

I'm going to assume this reads

\vec F(x,y,z)=5e^x\sin y\,\vec\imath+5e^x\cos y\,\vec\jmath+7z^2\,\vec k

and the path C is parameterized by

\vec c(t)=8t\,\vec\imath+t^3\,\vec\jmath+e^t\,\vec k

with 0\le t\le1. Under this parameterization,

\vec F(x,y,z)=\vec F(x(t),y(t),z(t))=5e^{8t}\sin(t^3)\,\vec\imath+5e^{8t}\cos(t^3)\,\vec\jmath+7e^{2t}\,\vec k

and

\mathrm d\vec c=\dfrac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt=(8\,\vec\imath+3t^2\,\vec\jmath+e^t\,\vec k)\,\mathrm dt

Then in the integral,

\displaystyle\int_C\vec F\cdot\mathrm d\vec s=\int_0^1\vec F(x(t),y(t),z(t))\cdot\frac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt

=\displaystyle\int_0^1(7e^{3t}+15e^{8t}t^2\cos(t^3)+40e^{8t}\sin(t^3))\,\mathrm dt\approx\boxed{12586.5}

(It's unlikely that an exact answer can be found in terms of elementary functions)

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Answer:

The option which is:

On a coordinate plane, a parabola opens up. It goes through (2, 10), has a vertex at (5, 1), and goes through (8, 10).

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See the attached figure which represent the graph of f(x) and g(x)

As shown g(x) goes through (2, 10), has a vertex at (5, 1), and goes through (8, 10)

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