Describe the locus in space. points 4 mm from a line

1 answer:

3 0

-- On a flat, 2-dimensional (x-y) graph, the locus of all points that are
4mm from a given line consists of two more lines, one on each side
of the given line, and all three lines are parallel.

-- If we want to talk about 3-dimensional space, then the locus of all
points that are 4mm from a given line is a cylinder, with radius of 8mm,
and the given line is the axis of the cylinder.

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Katyanochek1 [597]

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Evaluate the line integral ∫CF⋅dr, where F(x,y,z)=5sinxi+4cosyj−2xzk and C is given by the vector function r(t)=t3i−t2j+t1k , 0≤

pickupchik [31]

$\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j+z(t)\,\mathbf k$
$\implies\begin{cases}x(t)=t^3\\y(t)=-t^2\\z(t)=t\end{cases}$
$\implies\dfrac{\mathrm d\mathbf r}{\mathrm dt}=3t^2\,\mathbf i-2t\,\mathbf j+\mathbf k$

with $0\le t\le1$ . With this parameterization, the vector field can be written as

$\mathbf F(x,y,z)=\mathbf F(x(t),y(t),z(t))=5\sin(t^3)\,\mathbf i+4\cos(t^2)\,\mathbf j-2t^4\,\mathbf k$

Now the line integral is given by

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\mathbf F(x(t),y(t),z(t))\cdot\dfrac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$
$=\displaystyle\int_0^1(15t^2\sin(t^3)-8t\cos(t^2)-2t^4)\,\mathrm dt$
$=-5\cos1-4\sin1+\dfrac{23}5$