Curve of intersection with 2 surfaces

[steller]

Show that the curve with vector equation \(\displaystyle r(t)=<2cos^{2} t, sin(2t), 2sint>\)
is the curve of intersection of the surfaces

and

.

Attempt:

(1)
\(\displaystyle (x-1)^2+y^2=1\)
\(\displaystyle y^2 = 1-(x-1)^2\)

(2)

\(\displaystyle x^2+y^2+z^2=4\)

plug (1) into (2) for y


\(\displaystyle x^2+1 -(x-1)^2+z^2=4\)

Expanding and canceling


\(\displaystyle x^2+1-x^2+2x-1+z^2=4\)

(3)
\(\displaystyle 2x+z^2=4\)

vector equation \(\displaystyle r(t)=<2cos^{2} t, sin(2t), 2sint>\)

(4)
\(\displaystyle x= 2cos^{2} t\)
\(\displaystyle y= sin(2t)\)
\(\displaystyle z= 2sint\)

plug (4) into (3)

\(\displaystyle 2(2cos^2(t)) +(2sint)^2=4\)
\(\displaystyle 4cos^2(t) +4sin^{2}t=4\)

divide out the 4

\(\displaystyle \frac{4cos^2(t) +4sin^{2}(t)=4}{4}\)

\(\displaystyle cos^2(t) +sin^{2}(t)=1\)

identity
\(\displaystyle cos^2(t) +sin^{2}(t) = 1\)
therefore 1= 1

Did i even do this correct?
If not where did i go wrong?
Thank you

 

[DrPhil]

Show that the curve with vector equation \(\displaystyle r(t)=<2cos^{2} t, sin(2t), 2sint>\)
is the curve of intersection of the surfaces

and

.

Attempt:

(1)
\(\displaystyle (x-1)^2+y^2=1\)
\(\displaystyle y^2 = 1-(x-1)^2\)

(2)

\(\displaystyle x^2+y^2+z^2=4\)

plug (1) into (2) for y


\(\displaystyle x^2+1 -(x-1)^2+z^2=4\)

Expanding and canceling


\(\displaystyle x^2+1-x^2+2x-1+z^2=4\)

(3)
\(\displaystyle 2x+z^2=4\)

vector equation \(\displaystyle r(t)=<2cos^{2} t, sin(2t), 2sint>\)

(4)
\(\displaystyle x= 2cos^{2} t\)
\(\displaystyle y= sin(2t)\)
\(\displaystyle z= 2sint\)

plug (4) into (3)

\(\displaystyle 2(2cos^2(t)) +(2sint)^2=4\)
\(\displaystyle 4cos^2(t) +4sin^{2}t=4\)

divide out the 4

\(\displaystyle \frac{4cos^2(t) +4sin^{2}(t)=4}{4}\)

\(\displaystyle cos^2(t) +sin^{2}(t)=1\)

identity
\(\displaystyle cos^2(t) +sin^{2}(t) = 1\)
therefore 1= 1

Did i even do this correct?
If not where did i go wrong?
Thank you

What you have is correct, but I think there is a flaw. Your equation (3) by itself is not a complete description of the intersection: all points on the intersection do satisfy equation (3), but not all points satisfying (3) are on the intersection - unless you include the constraint on \(\displaystyle y\). It takes two equations to complete the proof. For instance, does \(\displaystyle r(t)\) satisfy \(\displaystyle y^2 = 1 - (x - 1)^2\)? Showing that would complete your proof.

Instead of combining the two surface equations, I would start with the two separate surfaces, and show that \(\displaystyle r(t)\) satisfies each of those two equations.
\(\displaystyle x= 2 \cos^{2} t\)
\(\displaystyle y= \sin{2t}\)
\(\displaystyle z= 2 \sin{t}\)

\(\displaystyle \displaystyle (x - 1)^2 + y^2\ =\ (2\ \cos^{2} t - 1)^2 + (\sin{2t})^2\ =\ \cdot \cdot \cdot \ =?\ 1 \)

\(\displaystyle \displaystyle x^2 + y^2 + z^2 = (2\ \cos^{2} t)^2 + (\sin{2t})^2 + (2\ \sin{t})^2\ =\ \cdot \cdot \cdot \ =?\ 4 \)

 

[steller]

I am having a hard time working out your solution.
what i get for trying to solve the first equation is

and this does not equal 1

 

[steller]

Could i use my original solution 1=1
then use the surface

y^2 = 1-(x-1)^2

and plug in x,y and see if the equal?

 

[steller]

Do you have a graphing program? I would love to see this graph!
Could you graph it on a program and attach it?