HSC 2013 Maths Marathon (archive) (3 Viewers)

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

-o- said:
this thread is stressing my out so much dont know how to do half these questions...

half the questions on here are Q12-14 hardness

nakruf · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
but i cant because sy's questions are hard

would you like him to spoonfeed you the answer?

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

nakruf said:
would you like him to spoonfeed you the answer?

as a matter of fact i'd really like that -_-

Sy123 · Oct 18, 2013

Re: HSC 2013 2U Marathon

ColdLipstick said:

Bumping for more exposure

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
Yea well done

$$i) The point$ \ \ P(a, b) \ \ $lies on the function$ \ \ y=\sqrt{1-x^2}$

$$Find the equation of tangent to the function at$ \ \ P$

$ii) Hence prove that the tangent to a circle is perpendicular to the radius$

i)

$y=\sqrt{1-x^2}$

$\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}$

$M=\frac{-a}{\sqrt{1-a^2}}\ \ = \therefore \frac{-a}{b} \ \ since \ \ b=\sqrt{1-a^2}$

ii)

$$tangents gradient$ = \frac{-a}{b} \ \ $and OP is$\ \ = radius= \ \ \frac{rise}{run} = \frac{b}{a}$

$\frac{-a}{b} \star \frac{b}{a} = -1$

$\therefore $tangent to a circle is perpendicular to radius$$

nakruf · Oct 18, 2013

Re: HSC 2013 2U Marathon

$(i) find \frac {d}{dx} (xlnx)$

$$(ii) hence evaluate \int_{1}^{3} lnx dx$

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

nakruf said:
$(i) find \frac {d}{dx} (xlnx)$

$$(ii) hence evaluate \int_{1}^{3} lnx dx$

I)

$u=x\ \ \ \ v=lnx$

$u'=1 \ \ \ \ v'= 1/x$

$\frac{d}{dx}(xlnx)\ \ = u'v+v'u\ \ (1)(lnx)+1 \ \ = lnx +1$

ii)

$\int_{1}^{3} lnx dx$

$[{xln(x)-x}]_{1}^{3}$

$=(3ln3 -3)-(ln1 -1) \ \ =3ln3 - 2$
----------------------------------------------------------------
$$Find the volume generated by area of $\ \ f(x) = ln(x) $ bounded by$ \ x=2 \ \ and \ \ x=3 \ \ $ being rotated about the$\ \ y-axis$

Menomaths · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
I)

$u=x\ \ \ \ v=lnx$

$u'=1 \ \ \ \ v'= 1/x$

$\frac{d}{dx}(xlnx)\ \ = u'v+v'u\ \ (1)(lnx)+1 \ \ = lnx +1$

ii)

$\int_{1}^{3} lnx dx$

$[{xln(x)-x}]_{1}^{3}$

$=(3ln3 -3)-(ln1 -1) \ \ =3ln3 - 2$

How do you integrate the lnx?

Sy123 · Oct 18, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
How do you integrate the lnx?

The proper answer is:

$\frac{d}{dx} x\ln x = \ln x + 1$

Integrate both sides

$\therefore \ x\ln x = \int \ln x \ dx + \int 1 \ dx$

$\therefore \ \ \int \ln x \ dx = x\ln x- x + c$

Then you just sub in values

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
The proper answer is:

$\frac{d}{dx} x\ln x = \ln x + 1$

Integrate both sides

$\therefore \ x\ln x = \int \ln x \ dx + \int 1 \ dx$

$\therefore \ \ \int \ln x \ dx = x\ln x- x + c$

Then you just sub in values

can you calm down i thought the whole state knew that step its not my fault i dont like showing every step Mr sy

Sy123 · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
can you calm down i thought the whole state knew that step its not my fault i dont like showing every step Mr sy

I know you've memorised the integral of ln x and you just want to take every opportunity to show off your supreme memorisation power

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
I know you've memorised the integral of ln x and you just want to take every opportunity to show off your supreme memorisation power

xD MEMORIES GREATER THAN LOGIC ANY DAY OF THE MONTH BUDDY

Sy123 · Oct 18, 2013

Re: HSC 2013 2U Marathon

$$By making use of differentiation by first principles of$ \ \sin x$

$EVALUATE$

$\lim_{h\to 0} \frac{\sin h}{h}$

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$By making use of differentiation by first principles of$ \ \sin x$

$EVALUATE$

$\lim_{h\to 0} \frac{\sin h}{h}$

$$was first principal $\frac{ f(h+m) - f(x) }{m} ?$

$\frac{\frac{sin(h+m)}{h+m} -\frac{sin h}{h}}{m}$

is this rights so far?

1

Sy123 · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
1

Do I really need to say, provide proof

Menomaths · Oct 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
The proper answer is:

$\frac{d}{dx} x\ln x = \ln x + 1$

Integrate both sides

$\therefore \ x\ln x = \int \ln x \ dx + \int 1 \ dx$

$\therefore \ \ \int \ln x \ dx = x\ln x- x + c$

Then you just sub in values

Omg so dull I still don't get it =.=...

nakruf · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$$was first principal $\frac{ f(x+h) - f(x) }{h} ?$
forgot i had google

1

top is f(x+h)-f(x)

Sy123 · Oct 18, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Omg so dull I still don't get it =.=...

Its like saying

$\frac{d}{dx} \sin x = \cos x$

THEREFORE

$\int \cos x = \sin x + c$

we are just reversing what we've done, in this case I re-arranged the integrals

$\frac{d}{dx} x\ln x = \ln x + 1$

$x\ln x = \int (\ln x + 1) \ dx$

$x \ln x = \int \ln x \ dx + \int 1 \ dx$

Menomaths · Oct 18, 2013

Re: HSC 2013 2U Marathon

I understood up until there

Menomaths · Oct 18, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
I understood up until there

wtf... Yeah I get it. Just derped like hell...

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