Mathematics HL/SL/Studies Help

[nametaken]

Could someone help me with this?

Differentiate using first principles:

f(x)=x^2-4x+9

I know that differentiated, it is 2x-4, but when I go through it, it just doesn't work.

My steps-

(x+h)^2-4(x+h)+9

(x+h)^2-4(x+h)+9=x^2-4x+9

x^2+2hx+h^2-4x-4h+9-x^2-4x+9

2hx+h^2-8x-4h+18

2hx+h^2-8x-4h+18/h = 2x+h-8x+18

As h approaches 0, -6x+18

I've gone very wrong here, can someone help me out?

Thanks!

Edited March 6, 2011 by nametaken

[adletaY]

You just made a computational error in the third line.

x^2 + 2xh + h^2 - 4x - 4h + 9 - (x^2 - 4x + 9)

x^2 + 2xh + h^2 - 4x - 4h + 9 - x^2 + 4x - 9

2xh + h^2 - 4h

Divide h out

2x + h - 4

Limit h => 0

2x - 4

Edited March 6, 2011 by adletaY

[dessskris]

[kiwi.at.heart]

I'm still really struggling with integration with substitution and I was wondering if anyone could help me out.

a) By substitution, integrate x(1/2x+1)^1/2

b) By substitution, integrate (x^3)/((x+2)^2)

I just can't seem to get my head around it at all

Edited March 6, 2011 by Kiwiatheart

[dessskris]

I can't seem to solve the first one either what is it actually?

The one above or below?

For the second one:

[fan]

check the PDF that I have posted below. Desy got a different answer for the second one though.. so we need a third person to check who is correct.

question.pdf

Edited March 6, 2011 by fan

[kiwi.at.heart]

Its the second one. Can anyone tell I really do not like substitution? Can anyone actually explain it to me, because I just happened to be away when we got taught it and my textbook is really crap at explaining it.

[Keel]

I can't seem to solve the first one either what is it actually?

The one above or below?

For the second one:

Thats correct, you but you need to sub back in the (x+2) for each u.

[dessskris]

You usually integrate a function by substitution if you cannot integrate it directly and if you see that the function consists of 2 functions and one is the differentiation of the other.

For example:

∫ 3x²(x³+1)⁴ dx

as you can see, there are 2 functions, and when you differentiate x³+1, you get 3x²

so you substitute u=x³+1

du=3x² dx

As you can see, u⁴ du is actually equal to the function you wanted to integrate. So it's

∫ u⁴ du

= u⁵/5 + c

Then you substitute back u=x³+1 and you get

(x³+1)⁵/5 + c

Another example would be

∫ (3x+1)²√(x³+x) dx

Let u=x³+x

du=(3x+1) dx

So you have

∫ √u du

= 2/3 u^3/2 + c

= 2/3 √(x³+x)³ + c

Yet another example:

∫ x/√(x-3) dx

u=√(x-3)

du= 1/(2√(x-3)) dx

u²=x-3

x=u²+3

So you have

∫ 2(u²+3) du

= ∫ (2u²+6) du

= 6u + 2u³/3 + c

= 6√(x-3) + 2√(x-3)³/3 + c

As for your first question, I don't see any of the functions as the differentiation of the other it can still be done though, but not by me . I hope somebody would help you out!

[nametaken]

How would I work this out?

Calculate the area enclosed by the two functions:

1) y=x^2 and y=x

My steps:

x^2=x

x^2-x=0

x(x-1)=0

x=0, x=1

Then the integral sign with limits 1,0: 1/3x^3

(1-1/3)-(0)

But in the book, it says that the answer is 1/6.

Any help?

Thanks!

[kiwi.at.heart]

How would I work this out?

Calculate the area enclosed by the two functions:

1) y=x^2 and y=x

My steps:

x^2=x

x^2-x=0

x(x-1)=0

x=0, x=1

Then the integral sign with limits 1,0: 1/3x^3

(1-1/3)-(0)

But in the book, it says that the answer is 1/6.

Any help?

Thanks!

basically you integrate x - x^2 with the limits 0,1

[x^2/2-x^3/3]0,1

=1/2-1/3

=1/6

Hope you understand that, if you don't, just ask me to explain

[BlackHawk]

Hello,

I am struggling with my maths homework. Any help, please?

2 exercise f only and 3 a only

THANK YOU!

[fan]

please state the question as I do not know what book you are referring to

[BlackHawk]

11 Exercise

5 exercise

2 exercise f only and 3 a only

Edited March 6, 2011 by BlackHawk

[dessskris]

2f)

sinθ

3a)

- cosθ

---

11a)

Let A' be the midpoint of BC, and AA'=3 cm

BA'=1.5 cm

Use pythagoras theorem to find AB, using the triangle AA'B

11b)

Area=θr²/2 where θ is angle BAC in radians

Find θ/2 first, using cos, hypotenuse and adjacent sides of triangle AA'B

Then find θ

r is AB

Then get the area

11c)

s (arc) BC is equal to θr

Perimeter=s+2r

Perimeter=(θ+2)r

---

5a)

as I mentioned in 11c, s=θr

so p=θr and θ=p/r

5b)

Area=θr²/2

Area=p/r * r²/2

the r cancels out

Area=pr/2

5c)

Area=pr/2

Least possible value of the Area is when

p=5.250 cm and r=4.650 cm

5d)

θ=p/r

min p=5.250 cm

max p=5.349 cm or maybe 5.34 cm

min r=4.650 cm

max r=4.749 cm or maybe 4.74 cm

max θ=max p/min r

min θ=min p/max r

---

However I am a bit unsure on questions 2 and 3

Any doubt or any need of further explanation just ask

Edited March 6, 2011 by Desy ♫

[BlackHawk]

Help getting the drawing. I don't need the solution

[Drake Glau]

Help getting the drawing. I don't need the solution

It's done with paint, bare with it

[BlackHawk]

I missed some lessons because of illness and I had an impression that these HW will be easy. Could anyone help me cuz I'll be in a big trouble latter...

I would be very thankful if you help me finish them ;(

[Drake Glau]

#11 can be done by the rule of sines where A/sin(a)=B/sin(b)=C/sin©

So it would be 5sqrt2/5sqrt8=x/sin30

Ratios can be cross multiplied so x=[(5sqrt2)*(sin30)]/5sqrt8

#2 Right angled triangle so you have angles of 90, 45 and 45.

Supposedly sin(90-x)=cos(x) and cos(90-x)=sin(x)

If you are using acute angles (45 in this case) you would just plug in and see if cos(90-45)=sin(45) and sin(90-45)=cos(45)

sin(45)=cos(45)? According to the unit circle it does

Same with the tangent one. Plug in your angle and prove that tan(90-45)=1/tan(45)

Edited March 6, 2011 by Drake Glau

[BlackHawk]

2f)

sinθ

3a)

- cosθ

However I am a bit unsure on questions 2 and 3

Any doubt or any need of further explanation just ask

Maybe, someone could comment on these, I still don't get it