heckald said:
Homework Statement
Prove that the curve [tex]\vec{r}[/tex](t) = <cost,sint/sqrt(2), sint/sqrt(2)> is at the intersection of a sphere and two elliptic cylinders. Reparametrize the curve with respect to arc length measured from
(0, 1/sqrt(2), 1/sqrt(2)) in the direction of increasing t.
Homework Equations
The Attempt at a Solution
x=cost
y=sint/sqrt(2)
z=sint/sqrt(2)
x^2+y^2+z^2=1
cost^2+sint^2/2+sint^2/2=1
does that show it is in the sphere?
I would order the statements differently. Instead of immediately writing [itex]x^2+ y^2+ z^2= 1[/itex], which is what you
want to prove, start with
[tex]x^2+ y^2+ z^2= (cos(t))^2+ (sin(t)/\sqrt(2))^2+ (sin(t)/\sqrt(2))^2[/tex]
[tex]= cos^2(t)+ sin^2(t)/2+ sin^2(t)/2= cos^2(t)+ sin^2(t)= 1[/tex]
thus proving that this parameterization gives [itex]x^2+ y^2+ z^2= 1[/itex], the equation of a sphere. The curve satisfies the equation of the sphere and so lies on the sphere.
heckald said:
okay so that proves it is within a sphere.
since z is equal to y would saying
cos2t+sin2t/√2=1 show that it is in a cylinder?
with a plane its a bit easier because z can be expressed with x and y. but spheres and cylinders are confusing.
Since y= z, you can write it as [itex]x^2+ y^2+ y^2= x^2+ 2y^2= 1[/itex] which is a relation in only x and y. Since z can be anything for fixed x and y, this is a cylinder (strictly, it is an elliptical cylinder, not a circular cylinder). The curve satisfies the equation of a cylinder and so lies on the cylinder.
But you still haven't got the parameterization in arclength. Remember that the arc length is given by [itex]\int |\vec{T}|dt[/itex] where [itex]\vec{T}[/itex] is the tagent vector at each point
[tex]= \int \sqrt{(dx/dt)^2+ (dy/dt)^2+ (dz/dt)^2}dt[/tex]
[tex]= \int_{\pi/2}^t \sqrt{sin^2(t)+ cos^2(t)/2+ cos^2(t)/2}dt= \int_{\pi/2}^tdt= \int_{\pi/2}^t dt[/tex]
(I chose the lower limit to be [itex]\pi/2[/itex] because [itex]cos(\pi/2)= 0[/itex] and [itex]sin(\pi/2)/2= 1/2[/itex].)
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