HSC 2013 Maths Marathon (archive) (7 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...

HSC 2013 Maths Marathon (archive) (7 Viewers)

Member

New Member

Member

This too shall pass

Member

New Member

Member

Exaı̸̸̸̸̸̸̸̸lted Member

This too shall pass

Member

This too shall pass

Member

This too shall pass

Member

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

New Member

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

Exaı̸̸̸̸̸̸̸̸lted Member

Users Who Are Viewing This Thread (Users: 0, Guests: 7)