Find a piecewise smoother parameterization of C

[whynot314]

I am having a real tough time trying to build a parametrization of the path c. its a really simple one but my professor just briefly went over it, and I can't find the thought process to put it together. I know the equation of a plane curve r(t)=x(t)i+y(t)j and everything. But its just not making sense to me. I would really like to understand how the book got this answer, I know there are many answers but if I can figure this one out i might be able to see other ways. If you know a good vid for this stuff please link it.[PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/IMAG0094.jpg[/PLAIN]

 

[Simon Bridge]

Well,

c1: (x,y) = (t,0) : 0 < t ≤ 3
c2: (x,y) = (3,t-3) : 3 < t ≤ 6
c3: ... you should be able to figure the rest.

Then you can integrate from t=0 to t=12.

This is pretty much what the book says - but they use unit vectors.
To understand it, put a dot on the graph at the t=0 position, then watch where it goes as t gets bigger.

 

[Mark44]

I am having a real tough time trying to build a parametrization of the path c. its a really simple one but my professor just briefly went over it, and I can't find the thought process to put it together. I know the equation of a plane curve r(t)=x(t)i+y(t)j and everything. But its just not making sense to me. I would really like to understand how the book got this answer, I know there are many answers but if I can figure this one out i might be able to see other ways. If you know a good vid for this stuff please link it.

Since you already have the answer in the book, maybe it would make more sense to see why their parameterization works.

For the first leg, r(t) = ti, 0 ≤ t ≤ 3
Each value of t will get you to a point on the x-axis between 0 and 3. When t = 3, you're at the point (3, 0).

For the second leg, all the x values are 3, which explains the 3i component.
For the vertical component we want to map the t values from 3 to 6 to the values 0 to 3, so we subtract 3. This gives us r(t) = 3i + (t - 3)j, 3 ≤ t ≤ 6.
When t = 3, we're at the point (3, 0). When t = 4.5, we're at the point (3, 1.5). When t = 6, we're at (3, 3).

For the third leg, the y-value stays fixed at 3, but the x-value has to go from 3 down to 0, while the t values range from 6 to 9. If we were mapping [6, 9] to [0, 3] we could use t - 6, but we want to end up with [3, 0], so t - 6 won't work. Switching the order from t - 6 to 6 - t maps the t values in [6, 9] to x values of 0 to -3, so they're off. If we adjust by adding 3 to 6 - t, we get 9 - t, and this correctly maps t values between 6 and 9 to x values from 3 down to 0. So r(t) = (12 - t)i + 0j, 6 ≤ t ≤ 9.

The last leg keeps x fixed at 0, but maps t values from 9 to 12 onto y values from 3 down to 0. I leave it for you to figure out a formula for this part.

 

[whynot314]

ok I think I kind of got it now, I tried this one. And found a piecewise, does this one work?
[PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/IMAG0095.jpg[/PLAIN]

 

[Mark44]

Test it and see. You have c₁(t) = 1 - t for 0 <= t <= 2. You need to include i in your formula.

Does c₁(0) give you the point (1, 0)?
Does c₁(2) give you the point (-1, 0)?

Same thing for the other two legs.

 

[happysauce]

I think you made a mistake on the second one. C1 ends at -i but C2 starts you at -i + j

 

[whynot314]

k found my mistake on C2 and believe it to be (t-2)j

so I now have c1 r(t)=(1-t)i 0≤t≤2
C2 r(t)=(t-3)i +(t-2)j 2≤t≤3
c3 r(t)=(t-3)i +(4-t)j 3≤t≤4

I think this is correct.