they don't have to be, in general x = rcos(t) and y = rsin(t) will parametrize a circle, you could do stuff like, x = rcos(2t) and y = rsin(2t), and that basically means it's traced out twice as fast:) Anyways in this video we were finding vector valued functions, so r(t) = (rcost)i + (rsint)j will always parametrize a circle as long as you let t go from 0 to 2pi. Hope this helps!!!
hmm oh wow, interesting, complete the square maybe? let's try x^2 + x x^2 + x + (1/2)^2 - (1/2)^2 (x + 1/2)^2 - 1/4, now plug back into what you have (x + 1/2)^2 - 1/4 + y^2 = 4 (x + 1/2)^2 + y^2= 17/4 now you have a circle of radius sqrt(17)/2, so set (sqrt(17))/2)cos(t) = x + 1/2, so x = (sqrt(17))/2)cos(t) - 1/2 (sqrt(17)/2)sin(t) = y r(t) = < (sqrt(17))/2)cos(t) - 1/2, (sqrt(17)/2)sin(t)> please please check this, I typed it REALLY fast awesome question:)
@@TheMathSorcerer i made that up, but i was searching how to do this type of problem after I was trying to solve for dx/dt and dy/dt of a given curve. I'm doing calc 3 material currently: using the gradient to find tangent line to a curve in R2, and then finding the tangent plane to a curve in R3. You end up with something like: dot = 0 = dot =0 And I know you can solve for by just creating a perpendicular vector of the gradient. And I know how to do these problems the way we are instructed, but I am trying to see if the dx/dt, dy/dt, (dz/dt in the case of R3 curve) are solvable other ways. But I thought if I can write the given curve as a vector valued function, i could differentiate with respect to t, and solve for dx/dt and dy/dt Sometimes the curve given is in R2 like x^2+y^2 = 4 and that is easy to make a vector valued function of. But thats why I wonder about things like, X^7 +x^2 - y^2 -y = 40 **just giving example of uglier equation** And the given curve can be in R3, 16 = x^2 +y^2+z^2 And a point is specified of course. ill actually post again later with a problem from homework. i just finished the week, want to relax
in 6:41 shouldn't it be y= 2sint - 1? since 1 will turn to negative once it is transposed. please do correct me if I'm wrong.
3:57 Why aren't the cos and sine squared?
they don't have to be, in general x = rcos(t) and y = rsin(t) will parametrize a circle, you could do stuff like, x = rcos(2t) and y = rsin(2t), and that basically means it's traced out twice as fast:) Anyways in this video we were finding vector valued functions, so r(t) = (rcost)i + (rsint)j will always parametrize a circle as long as you let t go from 0 to 2pi. Hope this helps!!!
@@TheMathSorcerer Oh! Thanks, it did help. I'm having difficulty with Calc III, so ye x(
What about finding a vector valued function for something like x^2 + x + y^2 = 4 ?
hmm oh wow, interesting, complete the square maybe? let's try
x^2 + x
x^2 + x + (1/2)^2 - (1/2)^2
(x + 1/2)^2 - 1/4, now plug back into what you have
(x + 1/2)^2 - 1/4 + y^2 = 4
(x + 1/2)^2 + y^2= 17/4
now you have a circle of radius sqrt(17)/2, so set
(sqrt(17))/2)cos(t) = x + 1/2, so x = (sqrt(17))/2)cos(t) - 1/2
(sqrt(17)/2)sin(t) = y
r(t) = < (sqrt(17))/2)cos(t) - 1/2, (sqrt(17)/2)sin(t)>
please please check this, I typed it REALLY fast
awesome question:)
where did you get this question? just curious
@@TheMathSorcerer i made that up, but i was searching how to do this type of problem after I was trying to solve for dx/dt and dy/dt of a given curve.
I'm doing calc 3 material currently:
using the gradient to find tangent line to a curve in R2, and then finding the tangent plane to a curve in R3.
You end up with something like:
dot = 0
= dot =0
And I know you can solve for by just creating a perpendicular vector of the gradient. And I know how to do these problems the way we are instructed, but I am trying to see if the dx/dt, dy/dt, (dz/dt in the case of R3 curve) are solvable other ways.
But I thought if I can write the given curve as a vector valued function, i could differentiate with respect to t, and solve for dx/dt and dy/dt
Sometimes the curve given is in R2 like x^2+y^2 = 4 and that is easy to make a vector valued function of.
But thats why I wonder about things like,
X^7 +x^2 - y^2 -y = 40 **just giving example of uglier equation**
And the given curve can be in R3,
16 = x^2 +y^2+z^2
And a point is specified of course.
ill actually post again later with a problem from homework. i just finished the week, want to relax
@@TheMathSorcerer and yea that looks right
@@tombombadil1351 hardcore!!