Find a vector function, $r (t)$ $r(t)$ , that represents the curve of intersection of the two surfaces. The cylinder $x^{2} + y^{2} = 81$ $x^2 + y^2 = 81$ and the surface $z = x y$ $z = xy$ ?

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Find a vector function, $r (t)$ $r(t)$ , that represents the curve of intersection of the two surfaces. The cylinder $x^{2} + y^{2} = 81$ $x^2 + y^2 = 81$ and the surface $z = x y$ $z = xy$ ?

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Answer 1 · 2018-02-22T16:40:55

The curve of intersection may be parametrized as $(z, r) = ((\frac{81}{2}) sin 2 θ, 9)$ $(z,r) = ((81/2) sin2\theta, 9)$ .

Explanation:

I am not sure what you mean by vector function. But I understand it that you seek to represent the curve of intersection between the two surfaces in the question statement.

Since the cylinder is symmetric around the $z$ $z$ axis, it may be easier to express the curve in cylindrical coordinates.

Change to cylindrical coordinates:
$x = r cos θ$ $x = r cos\theta$
$y = r sin θ$ $y = r sin\theta$
$z = z$ $z = z$ .
$r$ $r$ is the distance from the $z$ $z$ axis and $θ$ $\theta$ is the counter-clockwise angle from the $x$ $x$ axis in the $x, y$ $x,y$ plane.

Then the first surface becomes
$x^{2} + y^{2} = 81$ $x^2 + y^2 = 81$
$r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ = 81$ $r^2cos^2\theta + r^2sin^2\theta = 81$
$r^{2} = 81$ $r^2=81$
$r = 9$ $r=9$ ,
because of the Pythagorean trigonometric identity.

The second surface becomes
$z = x y$ $z = xy$
$z = r cos θ r sin θ$ $z = rcos\theta rsin\theta$
$z = r^{2} sin θ cos θ$ $z= r^2sin\theta\cos\theta$ .
We learned from the equation of the first surface that the intersecting curve must be at a squared distance $r^{2} = 81$ $r^2=81$ from the first surface, giving that
$z = 81 sin θ cos θ$ $z = 81 sin\theta cos \theta$ ,
$z = (\frac{81}{2}) sin 2 θ$ $z = (81/2) sin2\theta$ ,
a curve parametrized by $θ$ $\theta$ . The last step is a trigonometric identity and is done just from personal preference.

From this expression we see that the curve is indeed a curve, as it has one degree of freedom.

All, in all, we can write the curve as
$(z, r) = ((\frac{81}{2}) sin 2 θ, 9)$ $(z,r) = ((81/2) sin2\theta, 9)$ ,
which is a vector valued function of a single variable $θ$ $\theta$ .

Answer 2 · 2018-02-22T17:17:10

See below.

Explanation:

Considering the intersection of

$C_1->{(x^2+y^2=r^2),(z in RR):}$

with

$C_2-> z = x y$

or $C_1 nn C_2$

we have

${(x^2+y^2=r^2),(x^2y^2= z^2):}$

now solving for $x^2,y^2$ we obtain the parametric curves

${(x^2=1/2(r^2-sqrt(r^2-4 z^2))),(y^2=1/2(r^2+sqrt(r^2-4 z^2))):}$ or

${(x = pm sqrt(1/2(r^2-sqrt(r^2-4 z^2)))),(y= pm sqrt(1/2(r^2+sqrt(r^2-4 z^2)))):}$

which are real for

$r^2-4 z^2 ge 0 rArr z lepm( r/2)^2$

Attached a plot showing the intersection curve in red (one leaf).

enter image source here

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Find a vector function, $r (t)$ $r(t)$ , that represents the curve of intersection of the two surfaces. The cylinder $x^{2} + y^{2} = 81$ $x^2 + y^2 = 81$ and the surface $z = x y$ $z = xy$ ?

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Explanation:

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Find a vector function, r(t)r(t)r(t), that represents the curve of intersection of the two surfaces. The cylinder x^2 + y^2 = 81x2+y2=81x^2 + y^2 = 81 and the surface z = xyz=xyz = xy?

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Explanation:

Explanation:

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Find a vector function, $r (t)$ $r(t)$ , that represents the curve of intersection of the two surfaces. The cylinder $x^{2} + y^{2} = 81$ $x^2 + y^2 = 81$ and the surface $z = x y$ $z = xy$ ?