Parametric Curve from the intersection of 2 surfaces

[heckald]

Homework Statement

Prove that the curve $$\vec{r}$$(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?

 

[tiny-tim]

welcome to pf!

hi heckald! welcome to pf!

(have a square-root: √ and try using the X² icon just above the Reply box )

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?

yup!

 

[heckald]

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.

 

[heckald]

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.

By intuition i got
x²+2y²=1 and x²+2z²=1
since this seems to work out. however, i have no idea how to jump from the x,y,z to these equations.

 

[HallsofIvy]

Homework Statement

Prove that the curve $$\vec{r}$$(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 $x^2+ y^2+ z^2= 1$, which is what you want to prove, start with

$$x^2+ y^2+ z^2= (cos(t))^2+ (sin(t)/\sqrt(2))^2+ (sin(t)/\sqrt(2))^2$$
$$= cos^2(t)+ sin^2(t)/2+ sin^2(t)/2= cos^2(t)+ sin^2(t)= 1$$

thus proving that this parameterization gives $x^2+ y^2+ z^2= 1$, the equation of a sphere. The curve satisfies the equation of the sphere and so lies on the sphere.

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 $x^2+ y^2+ y^2= x^2+ 2y^2= 1$ 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 $\int |\vec{T}|dt$ where $\vec{T}$ is the tagent vector at each point
$$= \int \sqrt{(dx/dt)^2+ (dy/dt)^2+ (dz/dt)^2}dt$$
$$= \int_{\pi/2}^t \sqrt{sin^2(t)+ cos^2(t)/2+ cos^2(t)/2}dt= \int_{\pi/2}^tdt= \int_{\pi/2}^t dt$$

(I chose the lower limit to be $\pi/2$ because $cos(\pi/2)= 0$ and $sin(\pi/2)/2= 1/2$.)

 

[tiny-tim]

since z is equal to y would saying
cos²t+sin²t/√2=1 show that it is in a cylinder?

By intuition i got
x²+2y²=1 and x²+2z²=1
since this seems to work out. however, i have no idea how to jump from the x,y,z to these equations.

i don't understand what's bothering you about this

in 2D, x²+2y²=1 is an ellipse,

so in 3D x²+2y²=1 is an elliptical cylinder (the same 2D cross-section for all z)

similarly in 3D x²+2z²=1 is an elliptical cylinder …

that's it! ​

 

[heckald]

Thanks for the help i think i understand it now. Now i just need practice. Too bad my final is today lol.

thanks again.