Question

Let r(t) = (24 sin(16t), -4, 24 cos(16t)). Compute d/dt(1/2r'(t) * r"(t))

Answer 1

r'(t)=(16(24)cos(16t),0,-16(24)sin(16t))
r''(t)=(-16^2(24)sin(16t),0,-16^2(24)cos(16t))
Let w=16^2(24) to keep our sanity
r''(t)=(-wsin(16t),0,-wcos(16t))

now before i start the d/dt(1/2)(r'(t)*(r''(t)) is the r'(t) in the denominator or the numerator?

Answer 2

It's actually 1/2 * r'(t).

Answer 3

ok then let z=16(24)
then r'(t) dot r''(t)=
((z)cos(16t),0,-(z)sin(16t)) *
(-(w)sin(16t),0,-(w)cos(16t))=
((-zw)cos(16t)sin(16t)+(zw)cos(16t)sin(16t))
d/dt((r'(t)*r''(t))=
Im going to pull out the zw because i already am hating how its looking
zw(d/dt(-cos(16t)sin(16t)+sin(16t)cos(16t))

to me thats looking like 0

Answer 4

which means they are orthogonal (⊥) to eachother

Answer 5

Even with the 1/2 next to r'(t), will it still equal to 0?

Answer 6

yes because you can pull the 1/2 out and dot them first then add in the 1/2 after. wz/2 can be pulled out.

is the answer not 0?

Answer 7

Actually I got 0 as well when I was working on this. So I guess it should be right. But I'm still checking on it.

Answer 8

yeah well its an UGLY problem in deed. well if you took away all coefficents and were left with
(sin(t),-4,cos(t))=r(t)
r'(t)=(cos(t),0,(-sin(t)) which is ⊥ to r(t)
r'"(t)=(-sin(t),0,-cos(t)) which is ⊥ to r'(t)

Answer 9

r'''(t)=(-cos(t),0,sin(t)) which is ⊥ to r''(t)
pattern

Answer 10

lol, it was annoying and I pretty much got the same as you have. I've asked a couple of my friends about this and they said they all ended up with 0 as well. So maybe that is right.

Answer 11

Anyways, thanks again for helping me. :)

Answer 12

welcome