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[Estel]

Syllabus Reference 15.1
Inverse functions
(dx/dy).(dy/dx)=1
i.e. (dx/dy)=1/(dy/dx)

 

[xeriphic]

i see, it goes in more depth when in trig functions, i guess i'll have to start now, thanks

 

[CM_Tutor]

Xeriphic, dy/dx = 1/(dx/dy) is a direct consequence of the chain rule. BTW, don't stress about it too much, it doesn't come up all that often.

Estel, whilst I wouldn't call it "cancel out the ln's", it is true to say that if ln A = ln B then A = B - and I don't send death threats.

 

[xeriphic]

that's a relief, though still not sure about how chain rule is used in this situation

 

[redslert]

i think this thread is getting way too long....makes it hard for people who want to help out more difficult!
cause no way i'm reading so many pages to find out where people are up to....

 

[charmed_cuties]

i'm having troubles in series application especially all the long equations i end up with and wen plugging it o nthe calculator
i had a series application question and i got it all wrong i lost 4 marks just from that

 

[Estel]

Well practice series application questions then lol.

 

[xeriphic]

you mean like time repayment and super annuation, i don't really like that too, then again if you take care in reading the question, such as the rate of interest and length like quarterly and annually, the process to all the questions are pretty much the same as you only have to manipulate the figures to begin with, agreeing with Estel, indeed practise will be perfect

 

[Seraph]

Ok if a $2000 loan is offerred At 18 % p.a interest , charged monthly, over 4 years. if no repayments need to be paid for the first 3months , find the amount of each repayment.

Ive taken this as 45 terms , yet im still not gettting the correct answer...
(oh btw this is a time payment question heh)

 

[Xayma]

T₁=1.015($2000)
T₃=1.015³($2000)
T₄=1.015⁴($2000)-M
T₅=1.015⁵($2000)-(1+1.015)M
.
.
.
T₄₈=1.015⁴⁸($2000)-(1+1.015+...+1.015⁴⁴)M

but the loan is paid back after 48 months.

(1+1.015+...+1.015⁴⁴)M=1.015⁴⁸($2000)
M=1.015⁴⁸*2000*.015/(1.015⁴⁵-1)
=$64.25

Now just to check something 45M=$2891.25 hmm interest sounds a little low but it looks right.

 

[Seraph]

i see waht youve done , taking out the ^45 ^46 and ^47 month pretty much ..

but arent those the last three months??? or doesnt it matter ?

Also i dont see why jsut taking 45 months wouldnt have worked hmmm

 

[~dEjA vOuX~]

I'm stuck on this question...can someone please help me ?

differentiate cosx sin^4x

 

[Xayma]

f(x)=cos x sin⁴ x

f'(x)=sin⁴x.-sinx+cosx.4sin³x.cosx
=4cos²x sin³x -sin⁵x
=sin³x(4cos²x-sin²x)

 

[Calculon]

d/dx cosxsin⁴x
d/dx cosx(1-cos²x)²
d/dx cosx(1-2cos²x+cos⁴x)
d/dx cosx - 2cos³x + cos⁵x
-sinx +6sinxcos²x -5sinxcos⁴x

this uses the function rule and the identity sin²x+cos²x = 1

 

[~dEjA vOuX~]

Hahaha I'm so stupid for getting stuck on such an easy question... Thanks

 

[Xayma]

Now to prove mine and calculons are both correct:

sin³x(4cos²x-sin²x)
=(1-cos²x)sinx(4cos²x-1+cos²x)
=(1-cos²x)sinx(5cos²x-1)
=(sinx-sinxcos²x)(5cos²x-1)
=-sinx+sinxcos²x+5sinxcos²x-5sinxcos⁴x
=-sinx+6sinxcos²x-5sinxcos⁴x

 

[Calculon]

Originally posted by 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.

Apparently for the HSC (this is taken from a 2u lecture I saw at usyd last year) ln=log=log_e

 

[Sarah168]

This looks like a really easy question (and it probably is) but Im just stumped and I asked some 3u people at school and they couldnt even do it i think im just missing something really obvious but yeah here it is:

Integrate 2x + 5/ x + 4

The answer at the back is 2x - 3 ln (x+4) + c

 

[CrashOveride]

You split it up via partial fractions.

[2(x+4) - 3] / [x + 4]

Which is 2(x+4)/(x+4) - 3 / (x+4)

Integrate that, and u get 2x - 3ln(x+4) + c

The 3u people should consider dropping =p

 

[Sarah168]

damn, I KNEW there was some splitting up involved cos the previous q's in the exercise involved split fractions, I just couldnt figure it out...my basic algebra and fractions knowledge is abysmal

hahaha, the 3u ppl that i asked are actually top 3u notch as well...strange