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# Mathematics HL/SL/Studies Help

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Sorry for latness Derek. Anyway your question can be solved easily as Desy did. It is not a difficult question, but i think it has a mistake. Because r value will be bigger than 1 this mean you can not caculate the value of the infinite sequence, since it is diverging not converging. Please re-check the question, so we will be able to give a real numerical value. This what i had when i calculated the value of r, maybe i havea mistake, so please check that.

Edited by inm

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Helloo

I just wanted to ask if Maths Studies is accepted if I want to continue with studying pharmacy after IB.

For my HL subjects, I have Chem, Bio and Eco and at SL, I have English A1, Maths Studies SL and French Ab into

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Helloo

I just wanted to ask if Maths Studies is accepted if I want to continue with studying pharmacy after IB.

For my HL subjects, I have Chem, Bio and Eco and at SL, I have English A1, Maths Studies SL and French Ab into

Umm this is actually not the place :/ next time go to the university forum.

Nice you have HL Bio and Chem but Math Studies may make it hard for you to study pharmacy later. I would say at least Math Methods SL is required. A friend who took Math Methods SL and is now studying medicine is even having a hard time with the math, so I suspect you might be struggling very badly in pharmacy.

However it depends on where you want to go. Some unis may accept you but there are some others that will reject you.

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My class did the intro to derivatives while I was sick... now we have a quiz tomorrow, and I dunno how to solve some stuff. Could someone solve one problem for me so I can get an idea of how to do them?

We were given the formula: y ' (x)= lim ((y(x+h)-y(x)) / h)

Then the problem from my math book is:

Use the algebraic/geometric method to find the slope of the tangent to:

y=x^2 at the point where x+3, i.e., (3,9)

Any help would be great

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My class did the intro to derivatives while I was sick... now we have a quiz tomorrow, and I dunno how to solve some stuff. Could someone solve one problem for me so I can get an idea of how to do them?

We were given the formula: y ' (x)= lim ((y(x+h)-y(x)) / h)

Then the problem from my math book is:

Use the algebraic/geometric method to find the slope of the tangent to:

y=x^2 at the point where x+3, i.e., (3,9)

Any help would be great

Well, using a short cut y'=2x so at x=3 the slope will be 6 But that's not how the book wants you to do it =/

y(x)=x2 and y'(x) lim h->0=[y(x+h) - y(x)] / h

From here you can really just plug it all in to find your derivative eventually.

y'(x)=[(x+h)2 - x2] / h

y'(x)=(x2+2xh+h2 - x2) / h

x2 cancel leaving (2xh+h2) / h

2xh / h=2x and h2 / h = h leaving...

y'(x)=2x + h

Since it's a limit as h approaches zero you can really just kind of plug in zero for h now...

h=0 leaving you with just y'(x)=2x

Now that y'(x)=2x and you're looking for the slope at (3,9) you can plug in the 3 to get a slope of 6

Edited by Drake Glau
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y(x)=x2 and y'(x) lim h->0=[y(x+h) - y(x)] / h

From here you can really just plug it all in to find your derivative eventually.

y'(x)=[(x+h)2 - x2] / h

y'(x)=(x2+2xh+h2 - x2) / h

y'(x)=2x + h

Since it's a limit as h approaches zero you can really just kind of plug in zero for h now...

That makes (x2) / h undefined and that end h=0 leaving you with just y'(x)=2x

Now that y'(x)=2x and you're looking for the slope at (3,9) you can plug in the 3 to get a slope of 6

corrected

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I am doing Mathematics HL and can't seem to understand Mathematical Induction!

1. Prove my mathematical induction:

a) 3*5 + 6*6 + 9*7 + 12*8 + ... + 3n(n+4) = [n(n+1)(2n+13)] / 2

b) 1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = [(n^2)(n+1)^2] / 4

and please show working out. i hope you understand the way i typed it!

thanks

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Basically, the method for mathematical induction is:

1. Show true for n=1

2. Assume true for n=k

3. Show true for n=k+1

4. Statement showing that it is true

So...

a) 3*5 + 6*6 + 9*7 + 12*8 + ... + 3n(n+4) = [n(n+1)(2n+13)] / 2

Show true for n=1

3(1)((1)+4)=15 ((1)((1)+1)(2(1)+13))/2=15

LHS=RHS

Assume true for n=k

3*5+6*6+...+3k(k+4)=(k(k+1)(2k+13))/2

Show true for n=k+1

RHS=((k+1)((k+1)+1)(2(k+1)+13))/2

=((k+1)(k+2)(2k+15))/2

LHS=(k(k+1)(2k+13))/2 + k+1

From this point I was unable to find a way to get each side to equal each other (It might be because its been about 4 months since I have attempt mathematical induction or maybe you made a typo, I don't know), but usually you would get each side to be identical and hence showing LHS=RHS

Since true for n=k+1 and true for n+1, then it follows true for n=2, 3,...

b) 1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = [(n^2)(n+1)^2] / 4

If you follow the method from the first example, the next one is quite similar and you should be able to do it, but if you still have trouble just ask again

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I am doing Mathematics HL and can't seem to understand Mathematical Induction!

1. Prove my mathematical induction:

a) 3*5 + 6*6 + 9*7 + 12*8 + ... + 3n(n+4) = [n(n+1)(2n+13)] / 2

b) 1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = [(n^2)(n+1)^2] / 4

and please show working out. i hope you understand the way i typed it!

thanks

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Math HL

how do you integrate:

x arcsin(x)

with respect to x?

∫ x arcsin(x) dx = ??

I tried integrating by parts once and then by substitution twice but ended up with a very weird answer. I wonder if anybody knows how to integrate it like once or twice only instead of thrice. it would be great if they could also write it on a piece of paper and scan it because I guess the solution might be a bit long.. thank you!

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Math HL

how do you integrate:

x arcsin(x)

with respect to x?

∫ x arcsin(x) dx = ??

I tried integrating by parts once and then by substitution twice but ended up with a very weird answer. I wonder if anybody knows how to integrate it like once or twice only instead of thrice. it would be great if they could also write it on a piece of paper and scan it because I guess the solution might be a bit long.. thank you!

My loooong solution:

At some point, I used something our teacher called "the neat trick":

a = b + c - a

2a = b + c

a = 1/2*(b+c)

Wolfram|Alpha's method: http://www.wolframalpha.com/input/?i=integrate+x+arcsin+x

Edited by Gene-Peer
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thank you sooo much! (:

but in the second question, how did

[k^2(k+1)^2] + 4(k+1)(k+1)^2

turn into

(k+1)^2 [k^2 + 4(k+1)] ?

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thank you sooo much! (:

but in the second question, how did

[k^2(k+1)^2] + 4(k+1)(k+1)^2

turn into

(k+1)^2 [k^2 + 4(k+1)] ?

It's simply taking out a common factor. (k+1)^2 is taken out

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Pre-IB student preparing to take Higher Level maths (next year) here.

While I generally enjoy mathematics and can easily pick up new concepts, I find myself frequently struggling with questions where I have to show / prove that a = b or some other equation. My biggest problem is figuring out where to start, since it's difficult for me to organize all the numbers that they give you into a coherent equation (or formula) with which I can prove the mathematical statement presented. Is there some way that I can practice these such questions, or a resource that I can use to help me with these types of problems?

Well,

here's an example of a question that I'm having trouble with.

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You should look at them as "Find a". Ignore the other value for the moment. Then when you're done, you might find yourself with a=b, if not and you've made no mistake, then you'll have a=c where c can be simplified to b. If you can't simplify c, improvise by trying (on a rough paper) to rewrite b to c, and then on your paper just copy the steps in reverse. If you still can't do it, then just write it like:

a=c

=b

At most you'll miss two marks for working, but usually you might get away with it as the examiner might think you skipped the steps due to time constraints.

for the question:

a)

We know that the circle has radius x

So the length of the string used for the circle = circumference of the circle = 2pi*x

There for the perimeter of the rectangle = 32 - 2pi*x

The two "widths" of the rectangle have a combined size/length of 8

Therefore the two "lengths" of the rectangle have a combined size of 32 - 2pi*x - 8 = 24 - 2pi*x

So "one length" = 12 - pi*x

b)

Area of rectangle = length*width = (12 - pi*x)4 = 48 - 4pi*x

Area of circle = pi*x2

Total area = 48 - 4pi*x + pi*x2

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Thanks for helping me with that question! My problem was that I didn't know where to start; the first thought that came to my mind was "So what if the radius of that circle is x? How does that relate to anything?" I never thought that x can be used to find the circumference, and that added to the perimeter of the rectangle would equal the original length of the wire.

I think I need more practice with these types of "show a = b" questions.

A lot of people at my school say that Higher Level maths is difficult and many of them score very low on tests. How challenging is this course? What are the types of questions that are asked on tests?

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Thanks for helping me with that question! My problem was that I didn't know where to start; the first thought that came to my mind was "So what if the radius of that circle is x? How does that relate to anything?" I never thought that x can be used to find the circumference, and that added to the perimeter of the rectangle would equal the original length of the wire.

I think I need more practice with these types of "show a = b" questions.

A lot of people at my school say that Higher Level maths is difficult and many of them score very low on tests. How challenging is this course? What are the types of questions that are asked on tests?

From experience, HL maths is something you will find quite manageable or quite hard, and it just depends on the individual. Saying that, don't think 95% of people struggle with it. Very few that are actually capable to do it do struggle. The problem is it seems all the horror stories seem to stick while anything positive about HL maths seems to be forgotten. About the type of questions, well the best way to know is simply just do past paper after past paper from your teacher to get an idea. Its just like any subject, if you are familiar with the formatting of questions, they are going to make a whole lot more sense.

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help!!

if you have a quadratic expression, like let's say y=ax²+bx+c, where a,b,c are constants, is there any way to express x in terms of y?

my problem is I have an equation:

x²+(y+r+t)²=t² where r and t are constants

I need to express y² in terms of x, r and t, because I need to integrate y² with respect to x.

help, anybody? or is that even possible to be done?

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Well, you can rewrite it as 0 = ax2 + bx + (c-y) and use the quadratic formula.

x = 1/2a*(-b +- sqrt(b2 - 4a(c-y)))

As for your question, it is easier to make y subject of the formula:

x2 + (y+r+t)2 = t2

(y+r+t)2 = t2 - x2

y+r+t = +- sqrt(t2 - x2)

y = sqrt(t2 - x2) - r - t

or

y = -sqrt(t2 - x2) - r - t

Assuming you want to find that integral for the volume of revolution about the x-axis, the function you have is a circle with center (0,-(r+t)) and radius t. If you rotate it about the x-axis you form a shape that is similar to folding a cylinder of length 2pi(r+t) and radius t. This shape is known as a torus (or a doughnut, lol).

The volume of this cylinder = base area*length = pi*t2*2pi(r+t) or 2pi2t2(r+t)

This will work only if r + t > t, otherwise, you won't form a torus. This means r has to be positive.

You might want to look into Pappus' Centroid Theorem

Edited by Gene-Peer
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ohhhhh haha thank you! I didn't see that I expanded (y+r+t)² and was stuck

yup, the original question was: find the formula of the volume of a doughnut. teacher wants us to use integration

2t=the thickness of the doughnut

r=the radius of the hole of the doughnut

ok now since

y = ±√(t²-x²) - (r+t)

doesn't that mean that

y² = t² - x² ±2(r+t) + (r+t)²

?

what should I do to the ± sign? can I just take its positive value and then integrate it or just leave it and integrate it as it is?

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Well, it's actually y2 = t2 - x2 ±2(r+t)sqrt(t2-x2) + (r+t)2

The plus-minus sign means you have two curves, i.e. the top half of the circle and the bottom half of the circle. To find the volume between the two curves you'll have to integrate one curve and subtract the integral of the two curves. Just like you would if you were finding the area of the circle with integration.

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oh yeah I missed that out thanks!

can I just integrate it with the ± and divide it by 2 then?

theoretically, according to Pappus's centroid theorem, the volume is

V=2π²r²(r+2t)

How do I get π² from integration, though? Because V=π∫y²dx (definite integral from -h to h)

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V = π∫(t² - x² + 2(r+t)√(t²-x²) + (r+t)²)dx - π∫(t² - x² - 2(r+t√(t²-x²)) + (r+t)²)dx, interval [-t, t]

Since both integrals have the same interval, we can combine them giving

V = π∫(4(r+t)√(t²-x²))dx

V = 4π(r+t)∫√(t²-x²)dx

∫√(t²-x²)dx = x√(t²-x²) + ∫x²/√(t²-x²)dx

Using methods extremely similar to my earlier "loooong solution" we can integrate the second integrand to get:

∫√(t²-x²)dx = 1/2*(x√(t²-x² + t²arcsin(x/t)) + c

integrate this from -t to t, gives -1/2*πt²

V = 4π(r+t)* -1/2*πt²

V = -2π²t²(r+t)

Volume can't be negative so you ignore it, V = 2π²t²(r+t). This is the answer I gave using the cylinder method.

Edited by Desy ♫
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OMG... that is very complicated I hate integration now, but we are going to take series&differential equations option

btw,

∫√(t²-x²)dx = x √(t²-x²) -2 ∫x²/ √(t²-x²)dx

Integration by parts,

u=√(t²-x²)

du= -2x/√(t²-x²)

dv=dx

v=x

therefore ∫√(t²-x²)dx = x√(t²-x²) - 2 ∫x²/ √(t²-x²)dx

Thank you btw!

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Don't worry, you'll never find integration like that in the exams...

u = (t²-x²)^(1/2)

du/dx = 1/2*(t²-x²)^(-1/2)*(-2x) = -x/√(t²-x²) The twoes cancel.

∫1*√(t²-x²)dx = x*√(t²-x²) - ∫x * -x/ √(t²-x²)dx = x√(t²-x²) + ∫x²/√(t²-x²)dx

You're welcome. This was really fun to do. There's probably an easier & faster way to integrate √(t²-x²). I'll post it if I find it. This long method was more tempting as I'd already done something very similar earlier.

Edited by Gene-Peer
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