Mathematics HL/SL/Studies Help

[adletaY]

If it was an actual IB question then your 2nd question should be fine. There will likely be a followthrough meaning that you found the right answer using your previous wrong answer. All I can help you with

Thanks, I'm pretty sure my teacher uses follow through. Hopefully I can get the marks for the second part, and if she's nice she may give half credit on the first part because at least I found one kind of standard deviation (highly unlikely). Well, I just found out that I divided by 4 instead of 4! on a different problem so I probably dropped 10% on this test for stupid mistakes.

[Drake Glau]

There's also "method" points. Honestly for math barely 1/3 of the points actually come from the answer, a ton more come from you knowing what to do with the numbers So if you did the correct equation or whatever but you failed 2*3=5 you'd get the point for the method but not the answer.

Keep studyin' you'll do fine on the test that counts

[timtamboy63]

Argh, I've forgotten my trig! Help?

[Keel]

Have you tried using your calculus and the formula book? Have you tried using your calculator?

The derivative of sin(x) is cos(x); the chain rule is also involved.

[dessskris]

a)

Find dh/dt

h=8+4sin(πt/6)

dh/dt = (2π/3)cos(πt/6)

When dh/dt=0,

0=(2π/3)cos(πt/6)

0=cos(πt/6)

t=3, 9

h=8+4sin(πt/6)

h=12, 4

b)

h >= 8

8+4sin(πt/6) >= 8

4sin(πt/6) >= 0

sin(πt/6) >= 0

0+12k <= t <= 6+12k

[timtamboy63]

Is there a way to do it without calculus? We haven't covered it yet?

[dessskris]

Part b was done without calculus.

Part a:

h=8+4sin(πt/6)

Max value of height is when sin(πt/6)=1. Min value of height is when sin(πt/6)=-1.

Max height: h=8+4=12

Min height: h=8-4=4

[timtamboy63]

Ah right, with part b, how did you get from the 2nd last step to the last step, and whats k?

[dessskris]

sorry for skipping steps.

k is an integer.

I don't know how to explain it though... I totally suck at explaining can anybody help me explain?

the way I see it is I just look at the graph.

sin(πt/6) >= 0

let x=πt/6, so

sin x >= 0

I work with graphs well, so when sin x is above or on the x axis, I know it's when 0<=x<=π, 2π<=x<=3π, 4π<=x<=5π, etc etc

generalising it:

0+2πk <= x <= π+2πk

substituting back in x=πt/6,

0+2πk <= πt/6 <= π+2πk

0+2k <= t/6 <= 1+2k

0+12k <= t <= 6+12k

[Rigel]

Hiya, can anyone help me with this?

Draw a graph to show what happens in the following jar-water-golf ball situation:

Water is added to an empty jar at a constant rate for two minutes and then one golf ball is added. After one minute another golf ball is added. Two minutes later both golf balls are removed. Half the water is then removed at a constant rate over a two minute period.

Thanks!

[Slovakov]

I think tou should draw a graph of the level of the water in the jar versus time.

It should look like this: (I included no values on Y axis because the are not mentioned in the task, so are not importnat)

Edited May 31, 2011 by Slovakov

[DDuino]

sqrt(pi/4)-arccos x

[dessskris]

√(π/4)-arccosx

since √(π/4) is a constant, we can just ignore it.

arccosx, so you see what x can be, right? -1 ≤ x ≤ 1

therefore the largest possible domain is -1 ≤ x ≤ 1

[DDuino]

how didi you come to the solution that is between -1 to 1?

[dessskris]

to easily explain, plot y=cosx and look at the maximum and minimum values of y reached.

[Ezeh]

Hey there, I'm stuck on this question. I know I have to set y=0 and solve for x, but i've got a negative square root...

Wolfram Alpha says what I have is right, but then it has another answer that i have no clue about....

http://www.wolframalpha.com/input/?i=0%3D3%2B%2820%2F%28x^2-4%29

Here is the question:

Edited June 5, 2011 by Ezeh

[Slovakov]

So, for point i), you need to put 0 for x.

This gives 3 + (20/(-4)) = -2

If you set y=0, you would look for x-intercept, while there is no x-intercept for this function and therefore you get such a weird result.

For point ii), you need to differentiate the function and again put 0 for x.

The derivative is: f'(x)=-40x(x²-4)^-2 if I'm correct.

The result will give you 0

Edited June 5, 2011 by Slovakov

[Mar Le Devine]

Hi

i'm stuck factorizing 4x² + x -5 = 0

i tried factorizing using the box method, but then i'm supposed to factor the numbers in the box, but i just don get it...

can someone help me please?

[Ezeh]

So, for point i), you need to put 0 for x.

This gives 3 + (20/(-4)) = -2

If you set y=0, you would look for x-intercept, while there is no x-intercept for this function and therefore you get such a weird result.

For point ii), you need to differentiate the function and again put 0 for x.

The derivative is: f'(x)=-40x(x²-4)^-2 if I'm correct.

The result will give you 0

Wow, I completely mis-read it, I feel so stupid

Thanks mate

Hi

i'm stuck factorizing 4x² + x -5 = 0

i tried factorizing using the box method, but then i'm supposed to factor the numbers in the box, but i just don get it...

can someone help me please?

I uploaded a solution that I tried to work through. If you have any questions just ask

[Kudzai Mukaratirwa]

great forum thing ^

wow most problems here are with binomial expansion i was just wondering how to solve this........find the area bounded by the curve and the X-axis for:

a.y= 3 cos2(theta) , (theta)=0 and (theta)= 3pie/4

b.y= X⁴ - 2e^x , x=-1 and x=3

non calculator question and p.s. my test is tomoro