2 Unit Revising Marathon (1 Viewer)

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

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

edit: or for f(x)

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

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

No. I was referring to anything but $e$ . I was attempting to avoid it so I would not confuse MrBrightside.

nightweaver066 · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

rolpsy said:
new question:

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

that'll test if you understand

Harrys?

$x^x = e^{lnx^x}$
$=e^{xlnx}$
$\frac{d}{dx}(e^{xlnx}) = (1 + lnx)e^{xlnx}$
$= x^x(1 + lnx)$

MrBrightside · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

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

edit: or for f(x)

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

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

umm you do know the original question was integrate right?

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

nightweaver066 said:
Harrys?

$x^x = e^{lnx^x}$
$=e^{xlnx}$
$\frac{d}{dx}(e^{xlnx}) = (1 + lnx)e^{xlnx}$
$= x^x(1 + lnx)$

It's a common question.

We should not confused MrBrightside any longer.

MrBrightside said:
umm you do know the original question was integrate right?

We were trying to help you understand it.

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

Here's a good question.

An object is dropped from a point $P$ which is $20\ln 2$ metres above the horizontal ground below. The object's motion as it falls is governed by the differential equation

$\frac{dx}{dv}=\frac{40v}{400-v^2}$

Where v m/s is the velocity of the object after it has fallen $x$ metres from $P.$

i. Integrate both sides of the differential equation with respect to $v$ to show that the displacement is given by

$x= 20\ln \frac{400}{400-v^2}$

ii. Hence show that $v^2=400(1-e^{-\frac{1}{20}x})$

iii. Hence find the speed at which the object strikes the ground.

iv. The object approaches its limiting velocity as the distance travelled gets larger and has no restriction. What percentage of the limiting velocity has the object reached when it strikes the ground? Give your answer correct to the nearest percentage.

rolpsy · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

MrBrightside said:
umm you do know the original question was integrate right?

$\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x}(a^{f(x)})=a^{f(x)} \times \log_{e}(a)\times f'(x) \\ \therefore \int a^{f(x)} \times \log_{e}(a)\times f'(x) \ dx &= a^{f(x)} \\ \textrm{because} \log_{e}(a) \textrm{ is constant:} \\ \log_{e}(a) \int f'(x) \times a^{f(x)} \ dx &= a^{f(x)} \\ \therefore \int f'(x) \ a^{f(x)} \ dx &= \frac{a^{f(x)}}{\log_{e}(a)} \end{align*}$

nightweaver066 · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

SpiralFlex said:
Here's a good question.

An object is dropped from a point $P$ which is $20\ln 2$ metres above the horizontal ground below. The object's motion as it falls is governed by the differential equation

$\frac{dx}{dv}=\frac{40v}{400-v^2}$

Where v m/s is the velocity of the object after it has fallen $x$ metres from $P.$

i. Integrate both sides of the differential equation with respect to $v$ to show that the displacement is given by

$x= 20\ln \frac{400}{400-v^2}$

ii. Hence show that $v^2=400(1-e^{-\frac{1}{20}x})$

iii. Hence find the speed at which the object strikes the ground.

iv. The object approaches its limiting velocity as the distance travelled gets larger and has no restriction. What percentage of the limiting velocity has the object reached when it strikes the ground? Give your answer correct to the nearest percentage.

i.
$\int\frac{dx}{dv}dv = \int\frac{40v}{400-v^2}dv$
$x = -20\ln(400-v^2) + c$

When the object hits the ground, v = 0, x = 0
$0 = -20\ln(400) + c$
$c = 20\ln(400)$

$\therefore x = -20\ln(400-v^2) + 20\ln(400)$
$x =20\ln(\frac{400}{400-v^2})$

ii.
$\frac{x}{20} = \ln(\frac{400}{400-v^2})$
$e^{\frac{x}{20}} = \frac{400}{400-v^2}$
$400-v^2 = \frac{400}{e^{\frac{x}{20}}}$
$v^2 = 400 - \frac{400}{e^{\frac{x}{20}}}$
$v^2 = 400(1 - e^{-\frac{x}{20}})$

iii.
When the object strikes the ground, $x = 20\ln2$
$v^2 = 400(1-e^{-\ln2})$
$v^2 = 400(1-\frac{1}{2})$
$v^2 = 200$
$v = 10\sqrt{2}ms^{-1}$

Unsure about that one...

iv. Limiting velocity = max velocity?
$\textup{Since } \frac{dx}{dv} = \frac{40v}{400-v^2}$
$\therefore \frac{dv}{dx} = \frac{400-v^2}{40v}$
$\textup{For maximum/minimum velocity, }\frac{dv}{dx}=0$
$400 - v^2 = 0$
$v = 20,\textup{as } v > 0$
$[tex]\frac{d^2v}{dx^2} = \frac{-80v - 40(400-v^2)}{(40v)^2}$
$\textup{At x = 20, } \frac{d^2v}{dx^2} < 0 \therefore \textup{x = 20 is the limiting velocity}$

$\therefore \frac{10\sqrt{2}}{20} = 71\percent \textup{percentage sign}$

Is that correct? :S

mangoz · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

SpiralFlex said:
Here's a good question.

An object is dropped from a point $P$ which is $20\ln 2$ metres above the horizontal ground below. The object's motion as it falls is governed by the differential equation

$\frac{dx}{dv}=\frac{40v}{400-v^2}$

Where v m/s is the velocity of the object after it has fallen $x$ metres from $P.$

i. Integrate both sides of the differential equation with respect to $v$ to show that the displacement is given by

$x= 20\ln \frac{400}{400-v^2}$

ii. Hence show that $v^2=400(1-e^{-\frac{1}{20}x})$

iii. Hence find the speed at which the object strikes the ground.

iv. The object approaches its limiting velocity as the distance travelled gets larger and has no restriction. What percentage of the limiting velocity has the object reached when it strikes the ground? Give your answer correct to the nearest percentage.

From trying to work out i) in my head, i don't see where the 400 can appear? in the displacement equation o.o on top

Edit: forgetting C. ): but meh my friend forgot it as well and he does 4u and got 97 for his 2u accelerated so MEH.

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

mangoz said:
From trying to work out i) in my head, i don't see where the 400 can appear? in the displacement equation o.o on top

Use paper.

nightweaver066 said:
i. $\int\frac{dx}{dv}dv = \int\frac{40v}{400-v^2}dv$
$x = -20\ln(400-v^2) + c$

When the object hits the ground, v = 0, x = 0
$0 = -20\ln(400) + c$
$c = 20\ln(400)$

$\therefore x = -20\ln(400-v^2) + 20\ln(400)$
$x =20\ln(\frac{400}{400-v^2})$

Correct. A few more to go.

mangoz · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

Lol i forgot c hahahahahah.

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

mangoz said:
Lol i forgot c hahahahahah.

You must go into retirement now.

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

nightweaver066 said:
i. $\int\frac{dx}{dv}dv = \int\frac{40v}{400-v^2}dv$
$x = -20\ln(400-v^2) + c$

When the object hits the ground, v = 0, x = 0
$0 = -20\ln(400) + c$
$c = 20\ln(400)$

$\therefore x = -20\ln(400-v^2) + 20\ln(400)$
$x =20\ln(\frac{400}{400-v^2})$

ii. $\frac{x}{20} = \ln(\frac{400}{400-v^2})$
$e^{\frac{x}{20}} = \frac{400}{400-v^2}$
$400-v^2 = \frac{400}{e^{\frac{x}{20}}}$
$v^2 = 400 - \frac{400}{e^{\frac{x}{20}}}$
$v^2 = 400(1 - e^{-\frac{x}{20}})$

iii. When the object strikes the ground, $x = 20\ln2$
$v^2 = 400(1-e^{-\frac{2ln2}{20}})$
$v^2 = 400(1-\frac{e^{10}}{e^{ln2}})$
$v^2 = 400(1-\frac{e^{10}}{2})$
$v^2 = 400-200e^{10}$
$v = 10\sqrt{4-2e^{10}}ms^{-1}$

Unsure about that one...

You were almost there. Some silly errors.

$v^2=(400-e^-^{\ln 2}})$

$v^2=400(1-e^{\frac{1}{2}})$

$v^2=400(1-\frac{1}{2})$

$v=10\sqrt{2}$ m/s.

Have you attempted the last question?

mangoz · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

SpiralFlex said:
You must go into retirement now.

Yeah probably after my HSC.

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

mangoz said:
Yeah probably after my HSC.

Do the last question in your head.

mangoz · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

SpiralFlex said:
You were almost there. Some silly errors.

$v^2=(400-e^-^{\ln 2}})$

$v^2=400(1-e^{\frac{1}{2}})$

$v^2=400(1-\frac{1}{2})$

$v=10\sqrt{2}$ m/s.

Have you attempted the last question?

70ish percent

nightweaver066 · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

Sorry for taking so long, not used to latex even though i've been trying to use it.

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

mangoz said:
70ish percent

My calculator says 71%.

mangoz · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

SpiralFlex said:
My calculator says 71%.

70.7 -_- i just wanted to write ish.

mangoz · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

nightweaver066 said:
i.
$\int\frac{dx}{dv}dv = \int\frac{40v}{400-v^2}dv$
$x = -20\ln(400-v^2) + c$

When the object hits the ground, v = 0, x = 0
$0 = -20\ln(400) + c$
$c = 20\ln(400)$

$\therefore x = -20\ln(400-v^2) + 20\ln(400)$
$x =20\ln(\frac{400}{400-v^2})$

ii.
$\frac{x}{20} = \ln(\frac{400}{400-v^2})$
$e^{\frac{x}{20}} = \frac{400}{400-v^2}$
$400-v^2 = \frac{400}{e^{\frac{x}{20}}}$
$v^2 = 400 - \frac{400}{e^{\frac{x}{20}}}$
$v^2 = 400(1 - e^{-\frac{x}{20}})$

iii.
When the object strikes the ground, $x = 20\ln2$
$v^2 = 400(1-e^{-\ln2})$
$v^2 = 400(1-\frac{1}{2})$
$v^2 = 200$
$v = 10\sqrt{2}ms^{-1}$

Unsure about that one...

iv. Limiting velocity = max velocity?
$\textup{Since } \frac{dx}{dv} = \frac{40v}{400-v^2}$
$\therefore \frac{dv}{dx} = \frac{400-v^2}{40v}$
$\textup{For maximum/minimum velocity, }\frac{dv}{dx}=0$
$400 - v^2 = 0$
$v = 20,\textup{as } v > 0$
$[tex]\frac{d^2v}{dx^2} = \frac{-80v - 40(400-v^2)}{(40v)^2}$
$\textup{At x = 20, } \frac{d^2v}{dx^2} < 0 \therefore \textup{x = 20 is the limiting velocity}$

$\therefore \frac{10\sqrt{2}}{20} = 71\percent \textup{percentage sign}$

Is that correct? :S

Yup but instead of "weaving" (get it i used ur name) through all that, u coulda just done x -> infinity in ur ii. equation.

SpiralFlex · Aug 13, 2011

Re: 2 Unit Revising Marathon HSC '10

mangoz said:
Yup but instead of "weaving" (get it i used ur name) through all that, u coulda just done x -> infinity in ur ii. equation.

Nice pun. Yes, it would be faster.

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