[Math] Help!?

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10-15-2009, 09:32 PM
3Garrett3

[Math] Help!?

New question guys. Sorry:

Bump, sorry guys. I have another. Even brain didn't tell me the answer in 30 seconds :O

Here it is. Anyone help?

http://i35.tinypic.com/r8h20z.png

I need help with part d.
10-15-2009, 09:37 PM
Awkwardsaw

we just finished matracis in alg 2, but iv never seen something like this :o what class are you taking?
10-15-2009, 09:41 PM
3Garrett3

Quote:

Originally Posted by Awkwardsaw

we just finished matracis in alg 2, but iv never seen something like this :o what class are you taking?

IB Math 12. We're doing precalculus. The section is vectors.
10-15-2009, 09:57 PM
Runaway

I'm doing precalc also :)
unfortunately, we have not ventured into the world of vectors yet...

EDIT: This could be what you did wrong, but I don't really know.

Quote:

Originally Posted by 3Garrett3

Now, seeing as you have 37.7km, and 13km/h, the math seems simple. 37.7/13 and you get the time, right? I get 2.9 hours, which puts it at 8:54am.

That would be the calculation to get directly to the port, correct?

don't you want to find when the ship will be "due north" of the port, not arriving at the port?
10-15-2009, 10:03 PM
3Garrett3

Quote:

Originally Posted by Runaway Cop

I'm doing precalc also :)
unfortunately, we have not ventured into the world of vectors yet...

EDIT: This could be what you did wrong, but I don't really know.

That would be the calculation to get directly to the port, correct?

don't you want to find when the ship will be "due north" of the port, not arriving at the port?

Yeah, I misread it. I still don't know how to do it though..
10-15-2009, 10:20 PM
Nava2

I've never seen that notation before. I know basic vectors, but I don't understand why you have them listed like matrices..
10-15-2009, 10:25 PM
3Garrett3

Quote:

Originally Posted by Nava2

I've never seen that notation before. I know basic vectors, but I don't understand why you have them listed like matrices..

Its called "Vector notation" in the IB world. A matrix in IB is actually listed in normal brackets.

Basically, it's the same as saying

"The vector is at (-20,32) and it moves (12, -5) every hour, represented as t"
"What is t when the ship is directly north of the port (0,0)"
10-15-2009, 10:26 PM
Runaway

Ok, I got it.

It's much more simple than you'd think.
Take only the top two numbers of the matrices.

r = -20 + 12t
so this means you are 20km west, and can go 12 km/hour east.
20km / 12km/hour = 1.66 hours
1.66 = 1 hour 40 minutes
6:00 am + 1:40 = 7:40 am

:)
10-15-2009, 10:27 PM
3Garrett3

Quote:

Originally Posted by Runaway Cop

Ok, I got it.

It's much more simple than you'd think.
Take only the top two numbers of the matrices.

r = -20 + 12t
so this means you are 20km west, and can go 12 km/hour east.
20km / 12km/hour = 1.66 hours
1.66 = 1 hour 40 minutes
6:00 am + 1:40 = 7:40 am

:)

I kinda overthought it a little XD

Can I add you on msn? I have another problem too with a different question. I've litterally been stuck on it for 30 minutes

E: Haha, forgot that I had you added XD
10-15-2009, 10:51 PM
3Garrett3

Bump, sorry guys. I have another. Even brain didn't tell me the answer in 30 seconds :O

Here it is. Anyone help?

http://i35.tinypic.com/r8h20z.png

I need help with part d.

E: Image broken?

E: Image fixed.
10-15-2009, 11:01 PM
Runaway

Image is still broken I think.
I don't see any of the question parts :o
10-15-2009, 11:09 PM
3Garrett3

Quote:

Originally Posted by Runaway Cop

Image is still broken I think.
I don't see any of the question parts :o

I see it? :\
10-15-2009, 11:09 PM
Brain

I only see this picture: http://i38.tinypic.com/ealdgy.jpg
no questions :p
10-15-2009, 11:13 PM
3Garrett3

Quote:

Originally Posted by Brain

I only see this picture: http://i38.tinypic.com/ealdgy.jpg
no questions :p

Linked the wrong one XD

My brain is so fried by this I didnt even notice.
10-15-2009, 11:20 PM
Runaway

I tried it out and double-checked it, keep getting 10:36am... (even graphed it O_o)

apparently the correct answer is 10:12am though.
10-15-2009, 11:57 PM
3Garrett3

Is this even possible guys? Anyone? :\
10-16-2009, 07:07 AM
nielsie95

Did you draw the lines? If I draw them, they collide. I forgot how to do it with vectors, so I turned them into y=ax+b lines:

A: y=-2x+13
B: y=.5x-8.5

A=B -> x= 8.6

6 + 8.6 = 14.6h = 14:36

EDIT: Hold on.. They start on different points...

A: y=13 - 2(x + 4)
B: y=0.5(x + 1) - 8.5

A=B -> x=5.2

6 + 5.2 = 11.2h = 11:12

Hmm.. Both don't seem right though, I don't know =\
10-16-2009, 10:01 AM
3Garrett3

Quote:

Originally Posted by nielsie95

Did you draw the lines? If I draw them, they collide. I forgot how to do it with vectors, so I turned them into y=ax+b lines:

A: y=-2x+13
B: y=.5x-8.5

A=B -> x= 8.6

6 + 8.6 = 14.6h = 14:36

EDIT: Hold on.. They start on different points...

A: y=13 - 2(x + 4)
B: y=0.5(x + 1) - 8.5

A=B -> x=5.2

6 + 5.2 = 11.2h = 11:12

Hmm.. Both doesn't seem right though, I don't know =\

They travel at different rates.

Runaway Cop tried graphs, but realized that it isn't necessarily the intersection, because they aren't actually travelling at the same constant rate, or from the spots on the graph which would allow them to collide.

I will find out how to do it today, and I'll post the answers here :)
10-16-2009, 03:10 PM
Nava2

Yacht A:
[4 + t, 5 - 2t]

Yacht B:
[1 + 2t, -8 + t]

d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
d = sqrt((1 + 2t - (4 + t))^2 + (-8 + t - (5 - 2t))^2)
d = sqrt((t - 3)^2 + (-(13 + t))^2)
d'=.5((t-3)^2 + (-(13 + t))^2)^(-.5)(2(t-3)(1) + 2(-t -13)(1))
d'=(t - 3 - t - 13)/((t-3)^2 + (-(13 + t))^2)

Find t when d' = 0.

nvm.. if I buggered up, someone correct me but I think thats how you would do it.. I ended up with a constant on top which seems wrong..
10-16-2009, 05:17 PM
3Garrett3

Quote:

Originally Posted by Nava2

Yacht A:
[4 + t, 5 - 2t]

Yacht B:
[1 + 2t, -8 + t]

d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
d = sqrt((1 + 2t - (4 + t))^2 + (-8 + t - (5 - 2t))^2)
d = sqrt((t - 3)^2 + (-(13 + t))^2)
d'=.5((t-3)^2 + (-(13 + t))^2)^(-.5)(2(t-3)(1) + 2(-t -13)(1))
d'=(t - 3 - t - 13)/((t-3)^2 + (-(13 + t))^2)

Find t when d' = 0.

nvm.. if I buggered up, someone correct me but I think thats how you would do it.. I ended up with a constant on top which seems wrong..

Haha thanks for the try. XD

I had a test on it today so the teacher wouldn't tell me how to do it. Needless to say that since two questions out of 6 were like the question I showed, it took me an hour to do it, and I most likely failed.

Anyone who could show this? XD