2 Unit Revising Marathon (1 Viewer)

MrBrightside · Aug 10, 2011

Differentiate: (2x^2 + 1) / (3x^2 - 4)

hence, evaluate: The integral with limits b = 1, a = 0 x / (3x^2 - 4)^2 dx

SpiralFlex · Aug 10, 2011

$=\frac{4x(3x^2-4)-6x(2x^2+1)}{(3x^2-4)^2}$

$=-\frac{22x}{(3x^2-4)^2}$

$\int_{0}^{1}-\frac{22x}{(3x^2-4)^2}=\frac{2x^2 + 1}{3x^2 - 4}$

$-\frac{1}{22}\int_{0}^{1}-\frac{22x}{(3x^2-4)^2}=-\frac{1}{22}*\frac{2x^2 + 1}{3x^2 - 4}$ [With limits 1 and 0.]

$=\frac{1}{8}$

MrBrightside · Aug 10, 2011

SpiralFlex said:
$=\frac{4x(3x^2-4)-6x(2x^2+1)}{(3x^2-4)^2}$

$=-\frac{22x}{(3x^2-4)^2}$

$\int_{0}^{1}-\frac{22x}{(3x^2-4)^2}=\frac{2x^2 + 1}{3x^2 - 4}$

$-\frac{1}{22}\int_{0}^{1}-\frac{22x}{(3x^2-4)^2}=-\frac{1}{22}*\frac{2x^2 + 1}{3x^2 - 4}$ [With limits 1 and 0.]

$=\frac{1}{8}$

Correct

JennaJameson · Aug 10, 2011

Re: 2 Unit Revising Marathon HSC '10

JennaJameson said:
i found dis pr3tty intredasting

A haz mass 2m, B haz mass m, whole thing is rotated at constant angular velocity w, other deets r in teh diagram br8s

find teh tension in OA and OB

gogogogogo faggots

oi cunts

MrBrightside · Aug 11, 2011

Re: 2 Unit Revising Marathon HSC '10

Differentiate:

d/dx 17 / (3t + 1)^2

Showing all working out necessary.

nightweaver066 · Aug 11, 2011

Re: 2 Unit Revising Marathon HSC '10

MrBrightside said:
Differentiate:

d/dx 17 / (3t + 1)^2

Showing all working out necessary.

0

Or assuming you mean $\frac{d}{dt}( \frac{17}{(3t+1)^2})$ ,
$=-2 \times 17 \times (3t+1)^{-3} \times 3$
$= -\frac{102}{(3t+1)^3}$

Question:

MrBrightside · Aug 11, 2011

Re: 2 Unit Revising Marathon HSC '10

nightweaver066 said:
0

Or assuming you mean $\frac{d}{dt}( \frac{17}{(3t+1)^2})$ ,
$=-2 \times 17 \times (3t+1)^{-3} \times 3$
$= -\frac{102}{(3t+1)^3}$

Correct

MrBrightside · Aug 11, 2011

Re: 2 Unit Revising Marathon HSC '11

A couple takes out a mortgage of $150 000 over 22 years. The interest is 0.8% per month reducible. Calculate the monthly repayments.

Show FULL working

MrBrightside · Aug 11, 2011

Re: 2 Unit Revising Marathon HSC '11

P = 150 000
r = 0.8% per month = 0.008 per month
n = 22*12 = 264 months

A1 = 150 000(1.008)^1 - M

A2 = A1(1.008)^1 - M
= [150 000(1.008)^1 - M](1.008)^1 - M
=150 000(1.008)^2 - M(1.008)^1 - M
=150 000(1.008)^2 - M(1.008 +1)

A3 = A2(1.008)^1 - M
= [150 000(1.008)^2 - M(1.008 +1)](1.008)^1 - M
= 150 000(1.008)^3 - M(1.008^2 + 1.008 + 1)
= 150 000(1.008)^3 - M(1 + 1.008 + 1.008^2)

.
.
.
A264 = 150 000(1.008)^264 - M(1 + 1.008 + ... + 1.008^263)

But at A264 the loan is paid off, therefore A264 = 0

Therefore 0 = 150 000(1.008)^264 - M(1 + 1.008 + ... + 1.008^263)

M(1 + 1.008 + ... + 1.008^263) = 150 000(1.008)^264
GP: a = 1 r = 1.008 n = 264

Use GP sum formula

M( 1(1.008^264 - 1) / 1.008 - 1 ) = 150 000(1.008)^264

(now we can cross multiply to find M)

Therefore

M = [150 000(1.008^264)](1.008 - 1) / 1(1.008^264 - 1)

= $1366.77

Therefore each monthly repayment is $1366.77

MrBrightside · Aug 11, 2011

Re: 2 Unit Revising Marathon HSC '10

nightweaver066 said:
Question:

Trapezoidal rule

y = 1 + ln(x)

draw up a table of values
__y0__y1___y2_____y3____y4
x | 1 | __2_| __3__|__ 4__| __5_
y | 1 | 1.693| 2.099|2.386 | 2.609

(4 subintervals)

h = (5 - 1) / 4

therefore

Area Approximately = 1 / 2 * [1 + 2.609 + 2(1.693 + 2.099 + 2.386)]
Approximately = 7.983

Simpson's Rule

Use the same table of values as above

Area Approximately = 1/3 * [1 +2.609 + 4(1.693 + 2.386) + 2(2.099)]
Area Approximately = 8.041

MrBrightside · Aug 12, 2011

Re: 2 Unit Revising Marathon HSC '10

I'm not sure if this question is worded incorrectly. It never mentions how often the interest is charged. Like per. annum or per month. I tried this question with per month because it says "paid off in equal monthly instalments"

But I ended up getting $24 150 per monthly repayment which is incorrect.

MrBrightside · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

Find the indefinite integral (primitive function) of 3^(2x-1)

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

MrBrightside said:
Find the indefinite integral (primitive function) of 3^(2x-1)

$\int 3^2^x^-^1 \: \mathrm{d} x$

$=3^2^x^-^1*\frac{1}{2}*\ln 3+C$

$=\frac{3^2^x^-^1}{2\ln 3}+C$

MrBrightside · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

SpiralFlex said:
$\int 3^2^x^-^1 \: \mathrm{d} x$

$=3^2^x^-^1*\frac{1}{2}*\ln 3+C$

$=\frac{3^2^x^-^1}{2\ln 3}+C$

Is that function of function rule? If you don't mind explaining, could you go into further detail as to where the ln(3) comes from. and 1 / 2, I know integral of e^ax + b = 1/a * e^(ax + b) + C, but this is a log question :/

I'm a little lost

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

MrBrightside said:
Is that function of function rule? If you don't mind explaining, could you go into further detail as to where the ln(3) comes from. and 1 / 2, I know integral of e^ax + b = 1/a * e^(ax + b) + C, but this is a log question :/

I'm a little lost

If you differentiate $\frac{3^2^x^-^1}{2\ln 3}$ ,

Removing the constant to make it easier,

$\frac{1}{2\ln 3} 3^2^x^-^1$

Let's begin to differentiate,

$\frac{1}{2\ln 3} * \frac{\mathrm{d} }{\mathrm{d} x}\: 3^2^x^-^1$

$=\frac{1}{2\ln 3} * 3^2^x^-^1 * \ln 3 * 2$ [From the differential formula of $a^f^(^x^)=a^f^(^x^) *\ln a * f'(x)$ Where $a \neq e$ ]

$=3^2^x^-^1$

MrBrightside · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

SpiralFlex said:
If you differentiate $\frac{3^2^x^-^1}{2\ln 3}$ ,

Removing the constant to make it easier,

$\frac{1}{2\ln 3} 3^2^x^-^1$

Let's begin to differentiate,

$\frac{1}{2\ln 3} * \frac{\mathrm{d} }{\mathrm{d} x}\: 3^2^x^-^1$

$=\frac{1}{2\ln 3} * 3^2^x^-^1 * \ln 3 * 2$ [From the differential formula of $a^f^(^x^)=a^f^(^x^) *\ln a * f'(x)$ Where $a \neq e$ ]

$=3^2^x^-^1$

I'm sorry I don't catch you.

I understand that your trying to show me the way by reversing the process, but is the f'(x) / f(x) not used here?

Drongoski · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

no.

rolpsy · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

MrBrightside said:
I'm sorry I don't catch you. I understand that your trying to show me the way by reversing the process, but is the f'(x) / f(x) not used here?

$\begin{align*} a^x &= e^{\log_{e}(a^{x})} \\ &=e^{x.\log_{e}(a)} \\ \frac{\mathrm{d} }{\mathrm{d} x}(a^x) &= \frac{\mathrm{d} }{\mathrm{d} x}(x.\log_{e}(a)) \times e^{x.\log_{e}(a)} \\ &=e^{x.\log_{e}(a)} \ \log_{e}(a) \\ &=e^{\log_{e}(a^x)} \ \log_{e}(a) \\ \therefore \frac{\mathrm{d} }{\mathrm{d} x}(a^x)&=a^x \ \log_{e}(a) \end{align*}$

edit: or for f(x)

$\begin{align*} a^{f(x)} &= e^{\log_{e}(a^{f(x)})} \\ &=e^{f(x).\log_{e}(a)} \\ \frac{\mathrm{d} }{\mathrm{d} x}(a^{f(x)}) &= \frac{\mathrm{d} }{\mathrm{d} x}(f(x).\log_{e}(a)) \times e^{f(x).\log_{e}(a)} \\ &=e^{f(x).\log_{e}(a)}\times\log_{e}(a)\times f'(x) \\ &=e^{\log_{e}(a^{f(x)})} \times \log_{e}(a) \times f'(x)\\ \therefore \frac{\mathrm{d} }{\mathrm{d} x}(a^{f(x)})&=a^{f(x)} \times \log_{e}(a)\times f'(x) \end{align*}$

SpiralFlex said:
Where $a \neq e$

a can be e because the ln(e) will be 1, leaving the $f'(x)e^{f(x)}$ we all know and love

Drongoski · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

That's the way to do it!

rolpsy · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

new question:

find $\frac{\mathrm{d} }{\mathrm{d} x}\left ( x^x \right )$

that'll test if you understand

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