Question

Calculus Trajectory Question (Comment Below)

Answer 1

What value of T should be used for the trajectory below that starts at (1,0) and ends at (0,1)?

\[r(t) = <\cos ^{3} (u(t)),\sin ^{3} (u(t))>\]

Using this:

\[S(t) = 3u'(t)\sin(u(t))\cos(u(t)) = V_{o}\]

Solving that down we find u(t) which is below which has two solutions because integration you can use either sin or cos substitution above:

\[u(t) = \cos^{-1} \left( 1-\frac{ 2V _{o}t }{ 3 } \right)\]or\[u(t) = \sin^{-1} \left( \sqrt{\frac{ 2V _{o}t }{ 3 }} \right)\]

So again in this question we are looking for what the values of t are when the trajectory starts at (1,0) and ends at (0,1).

Answer 2

@oldrin.bataku @dumbcow

Any advice?

Answer 3

looks like an initial value problem

Answer 4

(1,0) = cos(0), sin(0)

(0,1) = cos(pi/2), sin(pi/2)

Answer 5

\[r(t) = \cos ^{3} u(t)~X+\sin ^{3} u(t)~Y\]
the velocity vector is
\[r'(t) = -3u'\cos ^{2} u(t)~\sin u(t)~X+3u'\sin ^{2} u(t)~\cos u(t)~Y\]

Answer 6

i think the cos solution should be inside sqrt
\[u(t) = \cos^{-1} \sqrt{1-\frac{2V_ot}{3}}\]

Answer 7

because
\[\sin^{2} u + \cos^{2} u = 1\]
right

Answer 8

True

Answer 9

so
\[r(t) =< (1-\frac{2V_o t}{3})^{3/2} , (\frac{2V_o t}{3})^{3/2}>\]
set initial endpoint (1,0)
t = 0

Answer 10

ends at (0,1)
solve
\[\frac{2V_o t}{3} = 1\]

Answer 11

So \[t = \frac{ 3 }{ 2V _{o} }\]

Answer 12

right

Answer 13

so between 0 and that then for t values?

Answer 14

yep

Answer 15

Ahh ok, thank you :)

Answer 16

yw