Mathematics HL/SL/Studies Help

[genepeer]

Haha, never mind... I wrote the question before working it out on word, smh. Forgot to remove it. Well, where you went wrong was substituting u and x. To answer your question, (So I can say that it's ∫-√(t²-x²) dx, right?) No. Which why you were getting V=0

Anyway, I've got all the time in the world to waste, and I enjoy math

Edited February 19, 2011 by Gene-Peer

[VSmithIB]

Hi Guys,

Just wondering if any of you could answer a quick question for me about my maths studies ia?

Are we supposed to refer to ourselves throughout the project as in I did, next I??

Thank you!

[nametaken]

Hi Guys,

Just wondering if any of you could answer a quick question for me about my maths studies ia?

Are we supposed to refer to ourselves throughout the project as in I did, next I??

Thank you!

Some people say that third person sounds more professional, but for my maths IA, I used first person. Don't overuse 'did' as in 'I did this, then I did that'.

[kiwi.at.heart]

I'm still getting really confused with Integration by substitution and this question is really giving me trouble:

(ln x)^2

Any help would really be appreciated

[nametaken]

I'm still getting really confused with Integration by substitution and this question is really giving me trouble:

(ln x)^2

Any help would really be appreciated

Ok...I'll give you an example with another question:

Integrate:

(a) -sin x cos^2x

For this question you need to imagine f(x) = cos x and n = 2. Therefore [f(x)]^2 = cos^2x and f "(x) = -sin x. Therefore, since n = 2, the answer is simply (cos^3x)/ 3 + c

Follow these rules:

Edited February 28, 2011 by nametaken

[Keel]

I'm still getting really confused with Integration by substitution and this question is really giving me trouble:

(ln x)^2

Any help would really be appreciated

I wouldn't use substitution for this, I would use intergration by parts.

Have a look:

In this case you will have to apply 'by parts' twice as you can see, you get another integral of lnx which is not good. repeating by parts solves this.

EDIT: AS YOU CAN SEE I FORGOT TO ADD +C AT THE END!!!

Edited February 28, 2011 by Keel

[dessskris]

So...yesterday I was absent. and we started on differential equations. we got a HW and idk how to solve it I tried reading the explanation in the textbook but I can't seem to get the answer that's written at the back of the book

A curve passes through the origin and satisfies the differential equation dy/dx = a+y, where a>0. Find an expression for y in terms of x.

(Mathematics for the IB Diploma Higher Level 2, Cambridge, page 311, Ex 23B, qn 2)

The answer written at the back of the book is:

y = a(e^x-1)

My answer is:

y = e^x/a - a

My rough working:

dx/dy = 1/(a+y)

x = ln(a+y) + c

x = ln(a+y) + ln(a)

y = e^x/a - a

Where did I go wrong? I regret skipping school yesterday

[Keel]

x = ln(a+y) + c

x = ln(a+y) + ln(a)

c = -ln(a)

Would that help?

[genepeer]

Well that would solve the question but you'll lose marks for wrong method. You should know that solving a differential equation gives a family of equations. So the question has two steps: 1) Solving the differential equation and 2) finding which one of this family of equations passes through the origin.

dx/dy = 1/(a+y)

x = ln(a+y) + c₁

x = ln(a+y) + ln(c₂)

y = e^x/c₂ - a

y = c₃e^x - a, here I can drop the subscripts and just write:

y = ce^x - a

This is the family of equation that satisfies that differential equation.

At the origin, x=0 & y=0. So

0 = ce⁰ - a

c = a

Therefore, the solution is y = ae^x - a

PS: Subscripts help to show a change. Otherwise, the working would imply c = ln( c ) which is not right.

Edited March 1, 2011 by Gene-Peer

[nametaken]

Just a quick question. How do you work out the mean of a population from a sample of discrete data? The data is as follows:

66, 72, 65, 70, 69, 73, 65, 71, 75

It's just that for some reason, the way to work out the mean isn't told in my textbook.

Thanks.

[genepeer]

Add and divide by number of terms.

[kiwi.at.heart]

Just a quick question. How do you work out the mean of a population from a sample of discrete data? The data is as follows:

66, 72, 65, 70, 69, 73, 65, 71, 75

It's just that for some reason, the way to work out the mean isn't told in my textbook.

Thanks.

Unless you answer isn't 69.6 the chances are there is simply a typo in the textbook.

[dessskris]

Well that would solve the question but you'll lose marks for wrong method. You should know that solving a differential equation gives a family of equations. So the question has two steps: 1) Solving the differential equation and 2) finding which one of this family of equations passes through the origin.

dx/dy = 1/(a+y)

x = ln(a+y) + c₁

x = ln(a+y) + ln(c₂)

y = e^x/c₂ - a

y = c₃e^x - a, here I can drop the subscripts and just write:

y = ce^x - a

This is the family of equation that satisfies that differential equation.

At the origin, x=0 & y=0. So

0 = ce⁰ - a

c = a

Therefore, the solution is y = ae^x - a

PS: Subscripts help to show a change. Otherwise, the working would imply c = ln( c ) which is not right.

umm ok... but we discussed it in class with our teacher today, and after we integrate and get the expression of x in terms of y, my teacher told us to substitute (0, 0) straightaway to find c and then express y in terms of x. which method should I follow then?

I got another question. This is from a worksheet so I don't get to know the correct answer

Solve this differential equation, giving your answer in the form of y=ƒ(x)

dy/dx = e^y-x

ƒ(0)=ln(2)

[3 marks]

My working:

dy/dx = e^y/e^x

e^-y dy = e^-x dx

∫ e^-y dy = ∫ e^-x dx

-e^-y = -e^-x + c (?)

should I put a c in here? if not then:

-e^-y = -e^-x

-y = -x

y = x+c

or if I must put a c there then:

-e^-y = -e^-x + c₁

...

How do I proceed?

or if I can just substitute (0, ln(2)) then:

-e^-y = -e^-x + c

-e^-ln(2) = -e⁰ + c

-½ = -1 + c

c = ½

Therefore,

-e^-y = -e^-x + ½

but how do I express y in terms of x?

---

Another question..

I have integrated the function and got:

1/y = ½ (½sin(2x) + x) + c

This curve passes through (0, 1).

If I express y in terms of x first and then find c,

y = 2/(½sin(2x) + x) + c₂

Then substitute in (0, 1):

1 = 2/0 + c

1 = ∞ + c ???

But if I find c first and then express y in terms of x,

1/y = ½ (½sin(2x) + x) + c

1 = 0 + c

c = 1

1/y = ½ (½sin(2x) + x) + 1

1/y = ½ (½sin(2x) + x + 2)

y = 2/(½sin(2x) + x + 2)

So how, Gene-Peer? I am confused now

Help... anybody...

[genepeer]

It doesn't matter when you substitute for c. It depends on your preferences. However, if the question is "Find the family of equations that satisfy this diff. equation", you have no choice but to continue using c.

1)

Yes, the "+c" must always be there after integrating...

-e^-y = -e^-x + 1/2

e^-y = e^-x - 1/2

-y = ln(e^-x - 1/2)

y = -ln(e^-x - 1/2)

I can't simplify any further but even if I could, it's not necessary unless the question requires it.

2)

You put the brackets in the wrong positions. It should be

1/y = ½ (½sin(2x) + x) + c₁

y = 1/(½(½sin(2x) + x) + c₁)

y = 1/(½(½sin(2x) + x) + ½c₂)

y = 2/(½sin(2x) + x + c₂)

y = 2/(½sin(2x) + x + c)

Now I substitute:

1 = 2/(0+c)

c = 2

y = 2/(½sin(2x) + x + 2), the same as you got when you substituted for c earlier.

[dessskris]

Ohhh so it's just because of the 'family of equations', that's why I needed to find y in terms of x first (the general expression) and then find c, right?

I'm getting it now...

One stupid question:

P is the point where x=1 on the curve whose equation is y=1+ln(x).

Find the equation of the normal to the curve at P. (I think I got it: y=2-x)

Find the area of the region enclosed by the normal, the curve and the coordinate axes.

I'm having a problem with the 2nd part. The coordinate axes are x-axis and y-axis, right? If that is so, I don't know which area the question is referring to. I tried plotting y=1+ln(x) and y=2-x on the same set of axes but looking at the area bounded by them and the coordinate axes..... I don't know which it is

Would anybody plot it, shade the area that I need to find and put a screenshot here? haha thank you!!

[genepeer]

Yeah, sometimes the question is even separated into a) & b) where in a) you must find the family of equations and b) you then substitute to get c. Anyway,

GeoGeobra FTW!! Though I shaded with MSPaint, lol

Edit: Oops, never mind those labels "a" & "b".

Edited March 3, 2011 by Gene-Peer

[nametaken]

Simple polynomial question.

The same remainder is found when 2x^3+kx^2+6x+31 and x^4-3x^2-7x+5 are divided by x+2. Find k.

It says in the book that the answer is 5. I did synthetic division and got the answer as -1. How would you work it out then?

Thanks for your help!!

[dessskris]

[nametaken]

Anyone got a tip on how to remember all the special angles without the use of triangles?

[kiwi.at.heart]

Anyone got a tip on how to remember all the special angles without the use of triangles?

I honestly can't think of an easier way that the triangles. The easiest way to remember them is just remembering the smallest side is opposite the smallest angle and then just drawing them again and again. At the top of nearly every page of my workbook I have the exact triangle drawn for quick reference and I don't even have to think about how to draw them now.