Post Your Problem Here (1 Viewer)

[Xayma]

Subtended=created.

Hmm maths in focus, to have an idea of what it will look like go to the next exercise and look at the diagram, it is doing areas there, but the angle subtended at the centre by the chord is theta.

 

[stoydgen]

winston - "Is this under Trignometric Functions?

If so... isn't there a formula for findin the length of the arc or something? something about theta = l/ r or something..."

The confusing thing is the fact that im finding the length of the arc cut off by the chord aye. The formula for the length of an arc is
l=r*theta
Its just the chord thing that got me confused. All the Q's before it weren't so hard lol, then its just like "WHACK" here's one u cant do

Can anyone figure it out? Coz i can't...

 

[CM_Tutor]

Originally posted by stoydgen
"A circle has a chord Of 25mm with an angle of (pi/6) subtended at the centre. Find to 1 decimal place, the length of the Arc cut off by the chord."

And can someone please tell me what "subtended" actually means?

Ok, draw a circle, mark the centre as O. Draw in any chord (we'll call it AB) that isn't a diameter, and join AO and BO. The angle subtended by the chord AB is angle AOB. We know that AO = OB (equal radii), so call them both R. The arc length that you seek is the arc length of AB, with is R * pi / 6, so all we need to do is find R. Now, since we know that AB = 25 mm, we can apply the cosine rule in triangle AOB to find R, in which case you must solve:

cos(pi / 6) = (R² + R² - 25²) / (2 * R * R)

or you can find angle OAB (which is 5 * pi / 12) and then use the sine rule, in which case you must solve:

R / sin(5 * pi / 12) = 25 / sin(pi / 6)

 

[Xayma]

Or you could state perpendicular line from chord to center bisects it.

As is an isos triangle (radii equal) bisects triangle therefore sin pi/12=12.5/R

then use l=Rsin theta (the only problem being is you have to remember that theta is pi/6 not pi/12)

 

[velox]

how do u do this
for what values of m does the quadratic equation have one root?

(5m-3)x^2 - 4mx + m + 1 = 0

 

[Xayma]

When the discriminant=0

ie when b²-4ac=0
so.

16m²-4(5m-3)(m+1)=0
4m²-(5m²+2m-3)=0
4m²-5m²-2m+3=0
m²+2m-3=0
(m+3)(m-1)=0

Therefore m=-3, m=1.

 

[Seraph]

hmm how would i find the smallest term of a geometric sequence , 0.5 , 1.5 , 4.5 which is greater than > 300 ???
ive tried to use the Nth term for a GP sequence to find it but i could not change the bases!!!

 

[CM_Tutor]

GP has first term 0.5, common ratio 3, and so the nth term is T_n = 0.5 * 3^n-1.

We seek the smallest T_n, such that T_n > 300
0.5 * 3^n-1 > 300
3^n-1 > 600
ln 3^n-1 > ln 600
(n - 1)ln 3 > ln 600
n - 1 > ln 600 / ln 3
n > 1 + ln 600 / ln 3
n > 6.8227...

So, all terms with n => 7 satisfy T_n > 300, and the smallest of them is:
T₇ = 0.5 * 3^7-1 = 364.5

 

[Seraph]

excuse my ignorance but what is ln?

 

[CrashOveride]

ln just denotes log with base e

 

[CM_Tutor]

The whole thing could be done in logs with base 10, if you prefer:

We seek the smallest T_n, such that T_n > 300
0.5 * 3^n-1 > 300
3^n-1 > 600
log₁₀3^n-1 > log₁₀600
(n - 1)log₁₀3 > log₁₀600
n - 1 > log₁₀600 / log₁₀3
n > 1 + log₁₀600 / log₁₀3
n > 6.8227...

So, all terms with n => 7 satisfy T_n > 300, and the smallest of them is:
T₇ = 0.5 * 3^7-1 = 364.5

 

[Estel]

Is log taken as log to base e?

 

[CM_Tutor]

ln always means log_e

log on the calculator means log₁₀

Using just 'log' can mean any base, because it doesn't matter - ie. in log laws like log xⁿ = nlog x, or in questions like find x if log(x + 3) = log(7 - 2x)

Alternately, using just 'log' can mean log₁₀ as is the case with a calculator.

Sometimes it might also be used to mean log_e, but this is comparatively rare.

Thus deciding what log on its own means is context-dependent. IMO, you should avoid using log on its own (ie without a base), as it is open to misinterpretation - unless a question does so first, in which case you should follow the questions lead.

 

[xeriphic]

i know e^loge 2 is 2 but how do i explain it, can anyone help

 

[Estel]

let x=e^ln2
lnx=lne^ln2
lnx=ln2.lne
x=2
e^ln2=2

 

[George W. Bush]

Originally posted by Estel
Is log taken as log to base e?

In NSW the convention is log = ln, but for most of the rest of the world it's not. You can't go wrong with specifying a base

 

[xeriphic]

lol i understand it now thanks

also can you tell me ways to differentiate a logerithm, will it be fine to only use the formula

dy/dx ln f(x) = f'(x)/f(x)

i'm confused the text book went off talking about how

dy/dx = 1/(dx/dy) not sure if it is important

quote from text book for differentiation of logarithms

Given y = lnx
then this means x = e^y
from which dx/dy = e^y <--- don't understand where dx/dy come from
but dy/dx = 1/(dx/dy)
= 1/e^y
= 1/x

do i have to follow the above working out or just apply the formula would be fine

 

[Estel]

Third line: lnx=ln2.lne
After you establish lne=1 you can simply "cancel out the ln's" (not the correct mathematical terminology and I wonder if CM or my previous maths teacher will start sending death threats).

I don't know calculus (I'm in Yr 11)
but as for your other questions
dx/dy means to differentiate in respect to y as opposed to dx/dy which is differentiating in respect to x.
You DO need to remember dy/dx=1/(dy/dx) as that is in the syllabus.

To check what working would be required I suggest you look at the mathematics syllabus section B which gives a few worked examples and discusses what can be memorised and what must be developed...
CM and some others will probably give a better reply later.



Edit: I mean dx/dy, not dy/dx, = 1/(dy/dx), as you said... damn typo.

 

[George W. Bush]

It's 2U maths - the formula should be fine. (unless it says prove of course)

from which dx/dy = e^y <--- don't understand where dx/dy come from

They differentiated with respect to y. It's like when you have y=e^x, dy/dx = e^x, except in this case they interchanged x and y.

 

[xeriphic]

thanks for the help guys, lol i understand it now, really appreciated, i guess i forgot to think of it that way bush, it really clea things up now

i checked the sylabus and it says

Differentiation of loge f(x) for simple functions f(x)

so what does that mean only the formula, by the way i can't find the part where it says you have to know dy/dx = 1/(dx/dy), is it in the calculus section rather than the exponential and logarithm