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@Mark both! any binomial expansion should never miss out the nCr!


i need help with this question on vectors

it says: Gina drives along a straight highway for 4.2km in a north-westerly direction and then along another road for 5.3km in a north-easterly direction. Find her displacement from her starting position

Thanks

you see, Gina is going 45° left-upwards and then 45° right-upwards forming a right angle. so you have a right angle with 4.2 and 5.3 as the length of the sides. you are asked the length of the hypotenuse :o which can be found using the pythagoras theorem. then the answer is that hypotenuse length and the direction is upwards :drake:

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i need help with this question on vectors

it says: Gina drives along a straight highway for 4.2km in a north-westerly direction and then along another road for 5.3km in a north-easterly direction. Find her displacement from her starting position

Thanks

This should help you out:

vec.png

EDIT: Desy already beat me to the answer. :drake:

Edited by chrypton
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Hahaha chrypton :P

but now idk what that bearing means so please do help her :drake: I haven't learned vectors in IB

@Mark I'll just take your HW as an example yeah

b) (2+1/2x)^4

First look at the pascal triangle, the coefficients are 1 4 6 4 1 so that's your nCr in this case

(2+1/2x)4

= 1*24 + 4*23*(1/2x) + 6*22*(1/2x)2 + 4*2*(1/2x)3 + 1*(1/2x)4

Then..just calculate those :o I'm too lazy to do that haha. Very easy, right?

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For part 2, when you find the coefficients, you expand them as I've shown before and using pascals triange you can work out for example the 4th coeffeicient. Or you wouldn't even need to draw out pascals triangle because you can work out the coefficients easily by knowing that the difference between the first coefficient and the second is *5/1 for a), below and all you need to do is remember that the '5' decreases whilst the 1 increases till 5.

a) (7-4x)^5 [x4]

I probably confused you even more...but that's basically the process. You wouldn't even need a calculator.

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Hahaha chrypton :P

but now idk what that bearing means so please do help her :drake: I haven't learned vectors in IB

@Mark I'll just take your HW as an example yeah

b) (2+1/2x)^4

First look at the pascal triangle, the coefficients are 1 4 6 4 1 so that's your nCr in this case

(2+1/2x)4

= 1*24 + 4*23*(1/2x) + 6*22*(1/2x)2 + 4*2*(1/2x)3 + 1*(1/2x)4

Then..just calculate those :o I'm too lazy to do that haha. Very easy, right?

ahaha, when I get to that question I will see if it IS that easy :D Btw i just wanted to make sure, even when they ask you to just Expand it, I still have to use that nCr process? =/

For part 2, when you find the coefficients, you expand them as I've shown before and using pascals triange you can work out for example the 4th coeffeicient. Or you wouldn't even need to draw out pascals triangle because you can work out the coefficients easily by knowing that the difference between the first coefficient and the second is *5/1 for a), below and all you need to do is remember that the '5' decreases whilst the 1 increases till 5.

a) (7-4x)^5 [x4]

I probably confused you even more...but that's basically the process. You wouldn't even need a calculator.

thanks dude! But, i have a question, when we are asked to just EXPAND, do we need to do the nCr also? Or JUST expand it?

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Hey, one of the users, fan, is having trouble posting, so I'm posting this math question for him. Thanks:

1. integrate Sin^3(x) / cos^5 (x) by substitution--> Thank you Gene-Peer!

2. integrate csc^3(2x)cot(2x) by substitution (okay, he solved this one)

Any immediate help would be awesome :drake:

Edited by ~Julie~
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binomial expansion needs the nCr. it is a MUST.

so i do the

√4.3²+5.3² and that gives me 6.76

the book says the answer is 67.6km from her starting point at a bearing of 006.60°

idk how to get those answers

The 67.6 km is probably a typo. The answer should logically be 6.76 km :o

Ok I can't draw this thing now so let me just briefly describe this thing.

Gina starts from point A. She goes 4.3 km northwest to point B. Then she goes 5.3 km north east to point C. Now you see a triangle ABC. The angle B is a right angle. Let angle A be θ.

The displacement is yes √4.3²+5.3² :drake: and then you find angle θ.

tan θ = 5.3/4.3

you find θ

Then the bearing of displacement is θ - 45° :P

I hope that's correct. I haven't learned vectors in IB. The last time I dealt with it is like...8 months ago lol

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Hey, one of the users, fan, is having trouble posting, so I'm posting this math question for him. Thanks:

1. integrate Sin^3(x) / cos^5 (x) by substitution

2. integrate csc^3(2x)cot(2x) by substitution

Any immediate help would be awesome :o

For 1)

let u = cos(x)

du/dx = -sin(x)

dx = -du/sin(x)

Using II as the integration sign.

|| sin3(x)/cos5(x) dx = || - sin2(x)/u5 du = || (cos2(x) - 1)/u5 du = || (u2 - 1)/u5 du = || u-3 - u-5 du ... finish off :drake:

For 2)

csc3(x)cot(x) = 1/sin3(x) * cos(x)/sin(x) = cos(x)/sin4(x)

u = sin(x)

du/dx = cos(x)

dx = du/cos(x)

|| cos(x)/sin4(x) dx = || 1/u4 du = || u-4 du ...

Edited by Gene-Peer
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Hey guys! :)

I just need some clear instructions on Binomial Expansions, 'cause I couldn't find a way around it. Here is one question:

Find the coefficient of the term indicated in square barckets in the expansion of each of these expressions

(3 + 2x)^7 [x^5] the answer is 44 800

I tried everything I knew, I know you have to expand it to get the coefficient of x^3 then you have to power them up to its respective exponent, then add on the values that you'd get from the pascal triangle...But I ended up with a bigger answer than normal...

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Hey guys! :)

I just need some clear instructions on Binomial Expansions, 'cause I couldn't find a way around it. Here is one question:

Find the coefficient of the term indicated in square barckets in the expansion of each of these expressions

(3 + 2x)^7 [x^5] the answer is 44 800

I tried everything I knew, I know you have to expand it to get the coefficient of x^3 then you have to power them up to its respective exponent, then add on the values that you'd get from the pascal triangle...But I ended up with a bigger answer than normal...

Aww Mark, why do you still not get it? :)

(3 + 2x)7 = 37 + 7*36*(2x) + 21*35*(2x)2 + 35*34*(2x)3 + 35*33*(2x)4 + 21*32*(2x)5 + ...

= ... + 189*(2x)5 + ...

= ... + 189*25*x5 + ...

= ... + 6048 x5 + ...

So the answer is 6048

Oh wait, how come my answer is wrong?! Haha sorry then idk, perhaps somebody else could help and point out my mistake :)

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Hey guys! :)

I just need some clear instructions on Binomial Expansions, 'cause I couldn't find a way around it. Here is one question:

Find the coefficient of the term indicated in square barckets in the expansion of each of these expressions

(3 + 2x)^7 [x^5] the answer is 44 800

I tried everything I knew, I know you have to expand it to get the coefficient of x^3 then you have to power them up to its respective exponent, then add on the values that you'd get from the pascal triangle...But I ended up with a bigger answer than normal...

Aww Mark, why do you still not get it? :)

(3 + 2x)7 = 37 + 7*36*(2x) + 21*35*(2x)2 + 35*34*(2x)3 + 35*33*(2x)4 + 21*32*(2x)5 + ...

= ... + 189*(2x)5 + ...

= ... + 189*25*x5 + ...

= ... + 6048 x5 + ...

So the answer is 6048

Oh wait, how come my answer is wrong?! Haha sorry then idk, perhaps somebody else could help and point out my mistake :)

Your answer's correct!! Btw, thanks for the reply. My moment of stupidity!!

Edited by Derek
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Hey guys! :)

I just need some clear instructions on Binomial Expansions, 'cause I couldn't find a way around it. Here is one question:

Find the coefficient of the term indicated in square barckets in the expansion of each of these expressions

(3 + 2x)^7 [x^5] the answer is 44 800

I tried everything I knew, I know you have to expand it to get the coefficient of x^3 then you have to power them up to its respective exponent, then add on the values that you'd get from the pascal triangle...But I ended up with a bigger answer than normal...

Aww Mark, why do you still not get it? :)

(3 + 2x)7 = 37 + 7*36*(2x) + 21*35*(2x)2 + 35*34*(2x)3 + 35*33*(2x)4 + 21*32*(2x)5 + ...

= ... + 189*(2x)5 + ...

= ... + 189*25*x5 + ...

= ... + 6048 x5 + ...

So the answer is 6048

Oh wait, how come my answer is wrong?! Haha sorry then idk, perhaps somebody else could help and point out my mistake :)

I am sorry,Desy...It's just, my mind complicates maths solutions first before understanding it completely. :) but thanks a lot :) I wish i could repay you.


Does any one here knows how to use MS Excel by any chance? :)

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@Derek so the answer key given by Mark is wrong?

@Mark no problem, one day you should help me in TOK or Chem :)

Does any one here knows how to use MS Excel by any chance? :)

For graphing? Yessh my physics teacher taught us how to plot stuffs for our lab reports :) anything I could do to help?

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