Question

Cal 3 Angle of Intersection

Answer 1

^^^image

Answer 2

k so you should be able to find the derivative of both vectors

then do a dot b' + a' dot b
then divide by the magnitudes of both right?

Answer 3

oh and given (0,0,0) this would correspond to t=0 correct?

Answer 4

i got \[\cos \theta = (<6,4,0>\dot<0,0,0>+<0,0,0>\dot<9,9,2>) / (\sqrt{52} * \sqrt{166})\]

Answer 5

which would be inverse cos = 0

Answer 6

theta = 90 degrees

Answer 7

i will be your fan if you can help me with this :)

Answer 8

ok wait i'm going about this all the wrong way... no need for derivative

Answer 9

however it should still be t=0, which would give <0,0,0> dot <0,0,0> div mag of both. equating to the same thing no?

Answer 10

or would you use t=1

Answer 11

the derivative r'(t)=<6,4,-14t>
so r'(0)=<6,4,0>

Answer 12

using t=1 i got 128 degrees (rounded up)

Answer 13

the derivative of s'(t)=<9cos(9t),9cos(9t),2>
s'(0)=<9,9,2>
we can use the formula\[\vec r\cdot\vec s=|r||s|\cos\theta\]and we want theta, so that seems to be the way to go
do the dot product

Answer 14

the derivative is*

Answer 15

see that's what i was thinking at first also but it would just be r dot s not dx/dy(r dot s) right?

Answer 16

where are y and x?
this is parameterized in t...

Answer 17

yeah, you should be able to find t through the given point (origin)

Answer 18

er hrm, but t would be 0 which is a problem i guess cuz you can't just plug them into each vector without getting 2 zero vectors

Answer 19

oh I see, you mean the origin is not t=0 ?
it is the origin of x y z?

Answer 20

yeah but at the origin t = 0 because they are paremetrized so you could do say x=6t... => 0 = 6t.... => t = 0

Answer 21

but they are...

Answer 22

right, so what's the problem with finding the vectors r'(0) and s'(0) ?

Answer 23

so i was thinking you could just set t = 1, find two vectors by plugging in t=1, then use the cos theta = adotb/mag of both

Answer 24

not sure if that would be accurate tho

Answer 25

because they give 2 zero vectors

Answer 26

why would you use t=1 ?

Answer 27

and i tried inverse cos(0) = 90 but the prog didn't like it :(

Answer 28

they don't intersect at t=1

Answer 29

not sure lol, just figuring it would give you two vectors that you could use the formula

Answer 30

they intersect at t=0
I'm pretty sure my way is right, what makes you think it's wrong?

Answer 31

i think that's the way i originally did it
wait were you doing
a dot b' + a' dot b

Answer 32

then divide by the mag of both at t = 0

Answer 33

no, I just dotted the individual derivatives...

Answer 34

\[r'\cdot s'=\cdots\]

Answer 35

oh hrm so you are saying dot the individual derivatives and divide by mags all of which using t = 0?

Answer 36

<6,4,0> dot <9,9,2> / sqrt 52 * sqrt 166

Answer 37

yeah, what makes you think you need a product rule thing?

Answer 38

yeah kinda messed up with that, was looking at the book and i didn't realize it had the derivative in front of it

Answer 39

So my theory is that the tangent vectors are in the direction of the curve, and that's the derivative (individually)
so if we use those vectors at the origin in the formula above, that should work.
It looks like you're on your last chance though, I don't wanna be responsible if you get it wrong, so I want to see if you agree.

Answer 40

ahh yeahhhh that makes sense

2 more chances :-0

Answer 41

good!

Answer 42

hrm coming up out of bounds with this inverse cos tho

Answer 43

ah whoops accidently multiplied when i should have added
answer i got is 14

Answer 44

no I don't...
that sounds right

Answer 45

saaaawweeet 14 was right
you are awesome
Thank you soo much :)

Answer 46

very welcome :D
I'm glad my theory panned out

Answer 47

haha glad your theory did too

now if i could just somehow get a copy of your brain and install it into my own like in the matrix... i'd be set for this test

Answer 48

some fine day, until then study!

Answer 49

haha will do
thanks again