2 unit maths revision (1 Viewer)

[cutemouse]

It's 60 degreez.



Cross multiply x=


Alternatively, grab a protracter and hold it against your clock and measure the angle when it's 2:00.

Okay my bad. I thought I wrote 2:15. Derp.

I should really start using pen and paper for maths...

 

[AAEldar]

If we square x, don't we square
Wouldn't it be (Don't go harsh on me)

Nope, what he's doing is putting the under the square root sign. So . It's like making a common denominator on a fraction, you're just multiplying by in that case, and here you're putting the under the square root sign by making it .

 

[SpiralFlex]

What question are we on now?

 

[Bored Of Fail1]

NEW QUESTION:

Use simpsons rule with 4 sub intervals to estimate the area enclosed by the curve y= e^(x) +1 , and the lines y=1.5 , y=3.5 and x=0.

give answer to two decimal places

 

[hscishard]

I've noticed that noone answers my questions
What the hell?

 

[hscishard]

x=ln(y-1)
Then simpsons rule

 

[Bored Of Fail1]

x=ln(y-1)
Then simpsons rule

is it really? XD

yet another person tricked

 

[hscishard]

?

 

[Bored Of Fail1]

?

that would answer a similar question, but not the one I asked

 

[hscishard]

that would answer a similar question, but not the one I asked

0.45 units^2?

 

[cutemouse]

You need to split the area into two integrals. Take one of them with abs value maybe.

 

[HSCAREA]

lmao. Think.

60 deg would be 4:00

no it wouldn't, 4 o'clock would be 120 degrees

 

[Bored Of Fail1]

You need to split the area into two integrals. Take one of them with abs value maybe.

yeh but looking back at it , there wouldnt be a way whereby you could use 5 function values and still isolate the "negative" part of the integral and take its absolute value. thats what happens when you make questions up on the spot

 

[Squishxmishyx]

Nope, what he's doing is putting the under the square root sign. So . It's like making a common denominator on a fraction, you're just multiplying by in that case, and here you're putting the under the square root sign by making it .

THANKS BRO!! <3 <3 <3

Someone do this... preferably before tonight =]

Question.
a) The sum of the radii of two circles is 100cm. If one of the circles has a radius of x cm, show that the sum of the areas of the two circles is given by


b) Find the value of x for which A is least.

 

[Bored Of Fail1]

A= 2 pi ( x^2 -100x +5000 )

dA/dx = 2 pi ( 2x -100 ) [ note the constants remain out front, and we differentiate the inside ]

set equal to zero

so 2x-100 =0

x=50

now the second derivative is 2 pi ( 2 ) = 4 pi, which is positive, so the graph is concave down, and thus gives a minium value

 

[Bored Of Fail1]

for part a.

let radii be x and y

therefore x+y = 100

and the Area of the two circles = pi x^2 + pi y^2 = pi ( x^2 +y^2 )

now rearrage the first eqn for y, ie y=100-x, and sub into the above

A= pi ( x^2 + ( 100-x)^2 )
expand

A= pi [ x^2 + 10000-200x +x^2 ]

factor out a factor of 2

so A = 2pi [ x^2 -100x +5000]

 

[SpiralFlex]

Let the radius be such that the other radius is

Now,











Working on second question now. Too late, Bored beat me to it.

 

[Squishxmishyx]

Thank you Matthew Goodwin, you're my hero!
And SpiralFlex, are you doing maths accelerated o.o.
How come you know how to do this...

I'll continue the questions so I don't seem like a whore.
The distance between the sun and the earth is approximately . The sun subtends an angle of approximately at a point on the earth. Calculate the approximate diameter of the sun.

 

[Bored Of Fail1]

Thank you Matthew Goodwin, you're my hero!
And SpiralFlex, are you doing maths accelerated o.o.
How come you know how to do this...

I'll continue the questions so I don't seem like a whore.
The distance between the sun and the earth is approximately . The sun subtends an angle of approximately at a point on the earth. Calculate the approximate diameter of the sun.

isnt that a worked example in cambridge??

its just using the arc length formula

l= r theta


where theta is in radians

 

[Riproot]

its just using the arc length formula

l= r theta


where theta is in radians

THAT'S WHAT I'M UP TO!!!
in Fitzpatrick of course.