Mathematics HL/SL/Studies Help

[LMaxwell]

3(a) Use your calculator to check that

20

∑ i^2 = 2870

i=1

Obviously I could check it by doing: (1^2)+(2^2)+(3^2)+...+(20^2), but I doubt that is the correct way.

Could someone please help me and tell me the correct method?

Thanks.

[HoolaBaloola]

It's the correct way how you listed the sequence.

Use the Arithmetic Sum Rule to prove that the sequence adds up to 2870.

EDIT: Do not manually add up the sum of the sequence! It's way too time consuming, and on an exam, you can't afford to waste time. Simply use the equation.

Edited October 3, 2011 by HoolaBaloola

[LMaxwell]

It's the correct way how you listed the sequence.

Use the Arithmetic Sum Rule to prove that the sequence adds up to 2870.

EDIT: Do not manually add up the sum of the sequence! It's way too time consuming, and on an exam, you can't afford to waste time. Simply use the equation.

It doesn't seem to work...

n/2 (U1+Un)

20/2 (1+400)

10 x 401

= 4010

[Hexa]

If you have a TI-84 with the newest software version, you can press Alpha and then one of the function keys at the top to bring up a drop down that has sigma notation as one of the options.

[CkyBlue]

She placed a small negative sign in front of the equal sign.

Idk about the sqrt....

[Guest hellokitty818]

How do you find dy/dx of y=x^3-x/x^2??

[Lero]

How do you find dy/dx of y=x^3-x/x^2??

You'd need so add some brackets in there or else I'd have no idea how to interpret that on the computer.

[dessskris]

Yeah Desy you're right, the answear key says that's the correct answear. 1. I'm shocked, how do you type the square root? 2. In the second exercise, why did you erased the /(-1)?

1. copy it from Ms.Word or if you use Windows 7, you can go to Start>All Programs>Accessories>System Tools>Characters Map

2. what Captain Marth said

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@hellokitty818, what Hus said.

[HoolaBaloola]

Oh, this questions obviously needs the calculator then. It's niether geometric nor arithmetic (it has no common ratio and the difference not equal).

I have an fx-9860GII, by clicking F4-F6-F2, sigma can be used, inputting the numbers and np!

Edited October 4, 2011 by HoolaBaloola

[scarlettjazz]

Hi,

I'm really stuck on this question. I've print screened it from the text book here

http://i1187.photobucket.com/albums/z393/scarlettjazz1/mathsquestion.jpg

if you have the HL textbook then you can find the Q on pg 698, it's Q1.

Thanks to anyone who can help me!

[adletaY]

This one is tricky.

So, to get the expression for A, you have two parts of the shape: the sector and the rectangle minus a chunk.

The sector is fairly straightforward, use A = 1/2(θ)®^2

A = (1/2)(θ)(10)^2

A = 1/2(θ)(100)

A = 50θ

Now comes the good part, finding the rectangle minus a chunk.

So, pretend the segment does not exist and visualise it.

You have three triangles, which I have labelled 1, 2, and 3.

1 and 2 are congruent.

To find the area of 1, you use the angle p, which is (180-θ).

Area = (1/2)(10)(10)(sin(p))

Area = 50sin(p)

Since you have congruent triangles 1 and 2, you multiply this area by 2 to leave 100sin(p)

So, you have this in terms of p, but we need θ.

Well, p = 180-θ . . .

100sin(180-θ) =

100[sin180cosθ - cos180sinθ]= [using difference formula for sin]

100[0 - (-sinθ)]= [sin180 = 0 and cos180 = -1]

100sinθ

Now, on to triangle 3.

For this triangle, you use the same formula, but you have θ so there is no need to simplify.

Area = (1/2)(10)(10)(sinθ)

Area = 50sinθ

Adding these up leaves 150sinθ

Summing that with the 50θ area of the sector leaves:

A = 50θ + 150sinθ

Which can be factored out to give the required:

A = 50(θ + 3sinθ)

Now, to find the maximum area you find A':

A' = 50 + 150cosθ

Set it equal to 0

0 = 50 + 150cosθ

-50 = 150cosθ

-(1/3) = cosθ

θ = 109.5 degrees. (I hope )

EDIT: This is the correct answer according to the H&H book.

Hope that helps!

Edited October 4, 2011 by adletaY

[dessskris]

yeah, use the sigma notation in your calculator.

[scarlettjazz]

Well, p = 180-θ . . .

100sin(180-θ) =

100[sin180cosθ - cos180sinθ]= [using difference formula for sin]

100[0 - (-sinθ)]= [sin180 = 0 and cos180 = -1]

100sinθ

Wow, thanks so much for your help!! Really, thank you for putting in the time, I appreciate it.

Can you please explain this bit to me:

100sin(180-θ) =

100[sin180cosθ - cos180sinθ]= [using difference formula for sin]

100[0 - (-sinθ)]= [sin180 = 0 and cos180 = -1]

Maybe I haven't learnt it yet? How did you get from sin(180-θ) to [sin180cosθ - cos180sinθ]?

Thanks!!

[dessskris]

it's the compound angle formula (also in the data booklet). it states that:

sin(A+B) = sinA cosB + sinB cosA

[adletaY]

Wow, thanks so much for your help!! Really, thank you for putting in the time, I appreciate it.

Can you please explain this bit to me:

100sin(180-θ) =

100[sin180cosθ - cos180sinθ]= [using difference formula for sin]

100[0 - (-sinθ)]= [sin180 = 0 and cos180 = -1]

Maybe I haven't learnt it yet? How did you get from sin(180-θ) to [sin180cosθ - cos180sinθ]?

Thanks!!

No problem, I love helping others with maths, it keeps my on my toes too.

Desy pretty much explained the identity you need, except it would be with minus signs. Some advice for if you aren't sure how to simplify something: look in the data booklet. If you can whittle a problem down to sin(a-b) or d(cosx)/dx, or integral of 1/(x^2 + a^2) or something, look in the formula packet to see if there is something that fits the situation you see that you can use to help you.

Was this a problem your teacher gave you? Without learning trig identities, it seems kind of unfair to ask you to do this, though you may be able to use a graph or other knowledge to see that sin(180-x) = x.

Anyway, glad I helped, ask again anytime, there's a bunch of people on this website who are really helpful with this stuff.

[Drake Glau]

How do you find dy/dx of y=x^3-x/x^2??

I'm guessing it's y=[(x^3)-x]/(x^2)

Use the quotient rule

f(x)=(x^3)-x

g(x)=(x^2)

y'=[g(x)*f'(x)-g'(x)*f(x)]/g(x)^2

Or if you don't want to do that you can make x^2 become x^-2 and then use the product rule of

y'=f'(x)g(x)+g'(x)f(x)

Either way should work

I'd work it out but my laptop is going to die in about 1ish minute...

[CkyBlue]

^^ or you could simplify it to x-x^(-1), whose derivative is 1+x^(-2)

[wazii]

Ok so there are 2 tangent and normal questions I can't wrap my head around and I can't find any examples in my maths book on how to solve them either. Please help out a fellow IB student if you can. They are SL questions

1. Find the equation of the tangent to the curve y= x^2 - 2x that is perpendicular to the line x-2y = 1 [ solution is y = -2x ]

2. THe normal to the curve y= ax^(0.5) + bx at the point x= 1 has a slope of 1 and intersects the y-axis at (0.-4). Find the value of a and b

( solution a= -4 , b = 1 )

Thanks Heaps

[Lero]

1. The line x - 2y = 1, can simply be re-written for y in terms of x. Therefore, y = (1/2)x + (1/2).

From this we can deduce that the slope of this line is 1/2. Therefore its perpendicular is the negative reciprocal, which is -2.

Therefore we need to find the point on y = x^(2) - 2x at which the slope of the tangent is -2.

Find the derivative of this equation, and equate it to -2, therefore: y' = 2x - 2, so 2x - 2 = -2, 2x = 0, x = 0.

Now when x = o, y = 0 also. Therefore the line passes through the origin and has a slope -2, so y = -2x.

2. Firstly I'd find the corresponding y-coordinate when x = 1, this is simply y = a + b. (plug in equation)

Now if the normal has a slope of 1, then the tangent has a slope of -1 (negative reciprocal). So take the derivative with respect to x of the equation, y' = a/(2x^(1/2)) + b.

When x = 1, the slope is -1, plug this value of x into the equation, therefore a/2 + b = -1. This is one equation involving a and b.

To get the second we simply find the equation of the normal in terms of a and b. Using the equation of a line formula: y - (a + b) = 1( x - 1).

Manipulate the equation and you should get a + b = -3.

Combine the two equations simultaneously and solve for each variable and you should find your answer .

[dessskris]

1)

since it's perpendicular to x-2y=1, remember m1*m2=-1?

x-2y=1

2y=x-1

y=0.5x-0.5

the gradient of the tangent you're looking for= -1/0.5 = -2

this line has the equation y=-2x+c

since it's a tangent to the curve y=x²-2x, it passes through the curve at one point. but which point is this? let's try to find the gradient of the tangent to the curve at ANY point using differentiation.

dy/dx=2x-2

at that point, we've previously found that the gradient of the tangent is -2.

-2=2x-2

x=0

so at x=0, the curve gives y=x²-2x=0.

since the line is tangent to the curve, the line passes through this point also.

y=-2x+c

0=0+c

c=0

so the equation of the line is y=-2x

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2)

let's find the gradient of the tangent to the curve first

y=a√x +bx

dy/dx= 0.5a/√x +b

at x=1, this gradient is = 0.5a+b

the gradient of the normal is hence = -1/(0.5a+b)

the question says this is one, so

1 = -1/(0.5a+b)

0.5a+b = -1

the equation of the normal is y=x+c

this intersects the y axis at -4, so c=-4, the equation is y=x-4

when x=1, y=-3 <-- the curve also passes through this point, so

y=a√x +bx

-3=a+b

now you have two equations:

0.5a+b = -1

-3=a+b

you can use whatever method you like to solve them... elimination, substitution, graphing (finding intersection). but since it's simple let's use substitution

b=-a-3

0.5a+b = -1

0.5a-a-3 = -1

-0.5a=2

a=-4

b=1

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ahaha damn tbh they're difficult, as in they're not very straightforward and I kinda had to think how to play with them in order to get the answers. where did you get these questions from?