Find the point of intersection of 2 curves

[sigma]

Find the point at which the curves
\(\displaystyle \
\L\
r_1 (t) = (e^t ,2\sin \left( {t + \frac{\pi }{2}} \right),t^2 - 2)
\\)

and
\(\displaystyle \
\L\
r_2 (t) = t\vec i + 2\vec j + (t^3 - 3)\vec k
\\)

intersect and find the angle of intersection.

I've tried a few things like equating the corrisponding x, y and z components and didn't really get anywhere. Not sure what to do now.

 

[arthur ohlsten]

I do not believe the two curves intersect for positive t.
For the two curves to intersect ther I,J,K values must be the same at time t.
this does not occur.
I do not believe the curves cross for positive t.
Arthur

 

[petrol.veem]

i believe this question is impossible unless the two t values are different (ie t1=t and t2=s). then it becomes easy. as for the angle of intersection, i am not sure.

do you know how to find the the point(s) of intersection of r(t)=(t,t^2,t^3) and the plane 4x+2y+z=24? how about the angle of intersection of the curve with the normal to the plane?

i can't figure out how to find the angle of intersection. using a dot/cross product perhaps? but with which values?

 

[sigma]

lol I can see your in 2720 as well? Well I actually figured out the question (I think) so for r2(t) I just let t = s and solved to find values for both t and s and plugged them back into both curves and got a point (1, 2, -2) which if you plug in the value for t that you found into r1(t) you get (1,2,-2) and you get the same values for plugging s into r2(t) so I found an intersection point but I still can't figure out how to find the angle does anybody know?

For the question you were wondering about, let x, y and z equal their corrisponding values for r(t) then put that into 4x + 2y + z = 24 so 4x would be 4t, 2y would be 2t^2 and so on. Factor from there and you will get a polynominal which can't be factored but thats ok. Because this equaiton equals 24 you just have to find what value of t would satisfy this equation to equal 24. Thats not too hard. Now that you have found t just put it back into r(t) then you have your intersection points! From there you use that formula (cosTheta=vectorA dot vectorB all over the norm of vectorA times the norm of vectorB) to find the angle and your done! Hope that helps.