onur369's Methods Question Thread :)

[Souljette_93]

having some trouble here..

tan (3x - pi/6) = -1
solve for x..
i get the wrong answer :S

My methods is a bit rusty but I'll give it a go.

So first do tan^-1 on both sides:

3x-pi/6=-pi/4

Add pi/6 to both sides:

3x=-pi/12

Divide by three:

X=-pi/36 +npi

That is a general equation, since you did not provide a
Particular domain

[brightsky]

Another way:

tan(3x - pi/6) = -1
Let 3x - pi/6 = u
tan(u) = -1
u = pi*n-pi/4, where n E Z
so 3x - pi/6 = pi*n - pi/4
3x = pi*n - pi/12 = (12pi*n - pi)/12
x = (12pi*n - pi)/36

[onur369]

Hey guys im stuck with a question, VCAA 2007 Exam 1:

A wine glass is being Þ lled with wine at a rate of 8 cm3/s. The volume, V cm3, of wine in the glass when the depth of wine in the glass is x cm is given by V = 4x 32. Find the rate at which the depth of wine in the glass isincreasing when the depth is 4 cm.


Any quick method for it? Also only 42% got this question correct

[luffy]

Wow - You're doing methods exams really early. Nice work.
- That is a good thing, as long as you know the course properly.

This is an application of chain rule question:















At x = 4,





Therefore, the depth is increasing at a rate of   cm/s when the depth is 4cm.

Hope I helped.

[onur369]

Thanks mate

[onur369]

1/3 πr^2 √(100-r^2 ) How can I derive this to find maximum volume, what should I use, chain rule? I tried chain rule I couldnt do it.

[xZero]

Try product rule, or put r^2 into the sq root then use chain rule

[luffy]

You gotta use chain rule now (I just did it in one step.... sub if you want to see how I got it.
Note: You can also use product rule from the original equation, but I think that would be a bit longer.



For maximums, :







 OR

 OR

You can use your sign table, but obviously r = 0 will not produce a maximum volume and the negative r value is rejected lol.

Therefore for Max Volume,



Sub this r-value back into the volume equation:

At V:











I hope I didn't make an error somewhere at the start.. lol

[onur369]

Its correct, thanks.

[hello_kitty]

This is a silly question, but on the TI-89, when you put in an equation and it says false..i know it means the equation is not true but how would you show your workings?

i.e.y = -1/x + 2

I know the y intercept of the hyperbola is (0,-1/2)
but when it comes to figuring out the x intercept it is 'false'
so what would i put under X int?? Would i just use x = 1 or another point?

Thanks

[onur369]

are you sure bout that ? there is supposed to be no y int, there is a x int which is 1/2

find x int  y=0
0 =-1/x +2 
-2=-1/x
swap x
and -1/-2 thus being 1/2.   
When trying to find the y int we sub in 0 but -1/0 is underfined and since there is no such thing as undefined + 2 it has no y int, its asympote is y=0

[brightsky]

i assume you mean y = -1/(x+2), in which case there are no x-intercepts.

[Respect]

uhh i have question;

if sin(x) = 0.3, cos(a)=0.5, tan(b) = 2.4 and x,a and b are in the first quadrant, find the value of the following.

a.) cos(π-x)
b.) sin(π-a)

π is pi

how u can solve these im not know?

[brightsky]

cos(pi - x) = -cos(x) by symmetry. we know that sin^2(x) + cos^2(x) = 1, so cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.3) = sqrt(0.7).

sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)

EDIT: see below

[jbebbo]

cos(pi - x) = -cos(x) by symmetry. we know that sin^2(x) + cos^2(x) = 1, so cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.3) = sqrt(0.7).

sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)

Don't you mean
cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.09) = sqrt(0.91) therefore cos(pi-x)=-sqrt(0.91)?
similarly
sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.25) = sqrt(0.75)?