Mathematics HL/SL/Studies Help

[tank85]

point a and b have coordinates (4,1) and (2, -5) respectively. Find a vector queation for the line which passes through the point a and is perpendicular to AB

Can't figure out vectors

[The Economist]

well, first of all find the equation of the line passing through A and B which is:

r=(4,1)+λ(-2,-6)

and then simply find a vector perpendicular to (-2,-6) (which means that the dot product has to be zero). So a random perpendicular vector would be (-3,1)

so a line passing through A and perpendicular to AB is:

r=(4,1)+μ(-3,1)

Edit: I just noticed that you didn't even need the eq. of the line passing through AB so ignore that

AB can be found by: (2,-5)-(4,1)=(-2,-6)

[dessskris]

guys one quick question,

in a cylinder, what do you call the circles on the top and bottom? also, what do you call the other surface thing?

I'm used to studying maths in indo so I only know what it's called in indo.. we didn't learn cylinder again in IGCSE and IB because we're assumed to have been very familiar with it. which I am but I'm not familiar with the English terminology.

just want to know what the surfaces are called. thank you!

barrel!!but that is more engineering terminology not mathematical...in math in our school they ues tub/side/face idk

no... barrel is the whole cylinder, right?

so just call the curved side as the "face" or "side"?

if anyone else has something to say about this please do, I really really need to know what it's called in English. it's for my EE.

[timtamboy63]

I'm not sure there is specific terminology.

What I would do, is indicate that for the rest of your extended essay, you will be calling the circles on the top and the bottom the "bases" of the cylinder, and the 'other surface thing' the 'barrel' of the cylinder. I'd recommend you use a diagram so the market knows exactly what you're talking about. Good luck!

[DeBrogliez]

Hello everyone,

This is my first topic here

And I need your help with my GDC (Texas TI-84 Plus Silver Edition)

Let's say I have two functions:

1) Y = 0.2sin(10x)

2) Y = -0.1

I want to find the slope of the tangent where these to functions intersect, for the function Y = 0.2sin(10x).

What I do is plot them together then find the intersection (2nd --> Trace --> 5).

Then I write the solution on on a piece of paper, then go to (2nd --> Trace --> 6) and type that solution in the calculator again to find the slope.

This way works for me. However, sometimes the solution is not a whole number, and writing the solution on a piece of paper then re-copying it requires a lot of time and I might do mistakes.

Is there a much simpler way where I can tell the calculator to find the slope of the tangent to the point where those to functions intersect?

Thank you

[Drake Glau]

0.2sin(10x)=-0.1

solve for x to get your x value.

Find derivative of the sin function given, plug in x, obtain slope.

-0.1/0.2=sin10x

sin^-1(-0.1/0.2)=10x

[sin^-1(-0.1/0.2)]/10=x

f'(x)=2cos(10x)

Edited November 20, 2011 by Drake Glau

[LMaxwell]

I'm having a real bad time doing this binomial question...

Expand (3r+2)(2r+3)^3.

The answer should be: 24r^4 + 124r^3 + 234r^2 + 189r + 54.

Any help would be much appreciated.

Thanks

Edited April 2, 2012 by brofessional

[Drake Glau]

Just set it all up like

(3r+2)(2r+3)(2r+3)(2r+3)

And start doing all the math

FOIL the first two and last two separately giving you two 2nd degree polynomials and then when you multiply them together you just distribute each of the terms from the first polynomial into each of the terms of the 2nd.

Course there might be an easier way, but I don't know it. This is just long and tedious as far as I know...

(3r+2)(2r+3)(2r+3)(2r+3)

(6r²+13r+6)(4r²+12r+9)

Distribute first term, 6r²:

24r⁴+72r³+54r²+(13r+6)(4r²+12r+9)

Distribute second term, 13r:

24r⁴+72r³+54r²+52r³+156r²+117r+(6)(4r²+12r+9)

Now distribute the 6 in:

24r⁴+72r³+54r²+52r³+156r²+117r+24r²+72r+54

Combine like terms will give you:

24r⁴+124r³+234r²+189r+54

Edited November 21, 2011 by Drake Glau

[Guest Animistic Anaemia]

you should first expand the powered part, giving

8r^3+36r^2+54r+27

then, you multiply this by the other part of the equation, giving the final result you expect

@Drake

the easiest way is to use Pascal's triangle and to replace 2r by a and 3 by b, giving a^3 + 3a^2+3ab^2+b^3 and then you can calculate it all simply

Edited November 21, 2011 by Animistic Anaemia

[ILM]

This can be expanded by the following method:

i) Binomial theorem can be used to expand (2r+3)^3:

1 (2r)³ (3)⁰ + 3(2r)² (3) + 3 (2r) (3)² + (3)³ = 8r³+36r²+54 r + 27

ii) multiply it by 3r+2

= 24r⁴+ 124r³ + 234r²+189r +54

any further questions are welcomed

[LMaxwell]

thanks for the quick replies people

I got 24r⁴+ 124r³ + 234r²+189r +54

but whenever I times by 3r+2 I get a different result...

EDIT: I got it now

Edited April 2, 2012 by brofessional

[LMaxwell]

More questions, sorry

1) Given that (3+(square root 7))^3 = p+q(square root 7) where p and q are integers, find

a) p b) q

2) Find the term containing x^10 in the expansion of (5+2x^2)^7.

3) Consider the expansion of (3x^2 - (1/x))^9.

Find the constant term in this expansion.

Thanks again for any help

[Procrastination]

2)

In order to find a term with x^10 you must multiply something by two that gives this value... I guess it is 5

The exponent of 2x^2 must be 5 and the one of 5 must be two in order to sum up 7.

So it'd be:

(7 combined 5) (5)^2 (2x^2)^5

(21)(25)(32x^10)

16800x^10

Edited November 22, 2011 by Procrastination

[CkyBlue]

1.

(3+sqrt7)^3=p+q*sqrt7

27+9*sqrt7+21+7*sqrt7=p+q*sqrt7

48+16*sqrt7=p+q*sqrt7

given that p and q are integers,

p=48, q=16

2. (5+2x^2)^7= Find x^10 term

You're familiar with binomial expansion right? 21 is the coefficient derived from Pascal's triangle

(5^2)[(2x^2)^5]*21

=16800x^10

3. In that expression, you see terms with overall positive and negative exponents. The key is to find one of the terms where there is a constant, meaning there is no variable, making the exponent 0.

The corresponding term is

{[(3x^2)^3]*[-1/x]^6}*84

See how their exponents cancel?

[(27x^6)/(-1/x^6)]84

=2268

Hope I got that right... Haven't touched math since May lol

[LMaxwell]

1.

(3+sqrt7)^3=p+q*sqrt7

27+9*sqrt7+21+7*sqrt7=p+q*sqrt7

48+16*sqrt7=p+q*sqrt7

given that p and q are integers,

p=48, q=16

2. (5+2x^2)^7= Find x^10 term

You're familiar with binomial expansion right? 21 is the coefficient derived from Pascal's triangle

(5^2)[(2x^2)^5]*21

=16800x^10

3. In that expression, you see terms with overall positive and negative exponents. The key is to find one of the terms where there is a constant, meaning there is no variable, making the exponent 0.

The corresponding term is

{[(3x^2)^3]*[-1/x]^6}*84

See how their exponents cancel?

[(27x^6)/(-1/x^6)]84

=2268

Hope I got that right... Haven't touched math since May lol

Your answer to question 1) isn't right Capt'n Marth but it's fine, I worked it out. Thanks for all the help guys/girls

Edited April 2, 2012 by brofessional

[CkyBlue]

1.

(3+sqrt7)^3=p+q*sqrt7

27+9*sqrt7+21+7*sqrt7=p+q*sqrt7

48+16*sqrt7=p+q*sqrt7

given that p and q are integers,

p=48, q=16

2. (5+2x^2)^7= Find x^10 term

You're familiar with binomial expansion right? 21 is the coefficient derived from Pascal's triangle

(5^2)[(2x^2)^5]*21

=16800x^10

3. In that expression, you see terms with overall positive and negative exponents. The key is to find one of the terms where there is a constant, meaning there is no variable, making the exponent 0.

The corresponding term is

{[(3x^2)^3]*[-1/x]^6}*84

See how their exponents cancel?

[(27x^6)/(-1/x^6)]84

=2268

Hope I got that right... Haven't touched math since May lol

Your answer to question 1) isn't right Capt'n Marth but it's fine, I worked it out. Thanks for all the help guys/girls (I've got a test today)

Argh, I forgot to add the coefficients. I hope you took the idea from it though ! Sorry about that...

[ib_is_a_bitch]

I don't understand simple interest,

question : while in Britain, Bobby decides to put this money in a bank that pays 6% simple interest per annum, and he gets a part-time job to cover his expenses. Bobby remains in Britain for six months.

Calculate how much interest he receives for the six months

WHAT DO I DO>?!?!?! HELP

[dessskris]

the interest for one year i.e. 12 months is 6%

the interest for 6 months is...?

[+Agastya]

Assuming that an annum interest will yield 3% interest after six months (which I'm not sure is the case), the answer is 3%.

[CkyBlue]

Yeah^ that sounds a little superficial. Technically he doesn't get anything, as the interest is compounded annually. Is there more information?