find the point on the curve r(t)=(sin(3t))i+(cos(3t))j+(4t)k at a distance 5pi/12 units along the curve from the point (0,-1,4pi/3) in the direction of increasing arc length

Question

anonymous · Answer

i need verification..

anonymous · Answer

r'(t)=(3cos(3t))i-(3sin(3t))j+4k
|r'(t)|=5
S=$$\int\limits_{t(0)}^{t}5d \tau$$

anonymous · Answer

and t(0)= pi/3 so that it satifies original point (0,-1,4pi/3)

anonymous · Answer

i think $$5\tau-5(\pi/3)$$ has to be set to equal 5pi/12..?

mathmate · Answer

The integral is correct, but it should be integrated with respect to t.
After integration, it would be
5t] from pi/3 to t1 = (5/12)pi
Solve for t1.

anonymous · Answer

right, i just didn't want to confuse myself so just added another variable. so after solving, i get 5pi/12 which then gets plugged into r(5pi/12)=(sin(3*5pi/12))i+(cos(3*5pi/12))j+4(5pi/12)k...?

mathmate · Answer

All good, great job!

anonymous · Answer

ok great, finally got something right, thank you

mathmate · Answer

You're welcome!