Mathematics Marathon (2 Viewers)

studentcheese · May 2, 2009

AnandDNA said:
what do you use to be able to type your answer like that??

It's latex - math equation editor. You can only do it in quick reply. Go to the bottom and there should be a button that says, 'launch math equation editor'.

AnandDNA · May 2, 2009

$y^2=(x+1)^4 v=\pi \int_{0}^{4}(x+1)^4 dx =\pi \left [(x+1)^5 \right ]\binom{0}{4} =\pi \left [\frac{5^5}{5}-\frac{1^5}{5} \right ] =\pi \left [\frac{3124}{5} \right ]u^3$

how can i put the equation on different lines cause its all togther there. I tried on latex to put spaces but it just comes all togther

Find the sum of all numbers between 1 and 200 that are not divisible by 6

studentcheese · May 2, 2009

The part of the curve y = 2x - x^2 enclosed by the x axis is rotated about the x axis through 180 degrees. Find the volume of the solid formed.

nerdsforever · May 2, 2009

Sum of numbers from 1 to 200: 200/2(1+200) = 100 x 201 = 20100

6, 12, 18, 24.....
sum of numbers divisble by 6: 33/2[2x6 + (33-1)6] = 3366

sum of numbers from 1 to 200 that are not divisble by 6: 20100 - 3366 = 16734

Correct me if I am wrong

(pretty sure it's wrong)

gigglinJess · May 3, 2009

studentcheese said:
The part of the curve y = 2x - x^2 enclosed by the x axis is rotated about the x axis through 180 degrees. Find the volume of the solid formed.

$y=2x-x^{2}=0$
$x(2-x)=0$
$\therefore x=0 or x=2$

$V=\frac{\pi }{2}\int_{0}^{2}(2x-x^{2})dx$
$= \frac{\pi }{2} \left [x^{2}-\frac{x^{3}}{3} \right ]_{0}^{2}$
$= \frac{\pi }{2} \left [\left (4-\frac{8}{3} \right ) - 0\right ]$
$= \frac{\pi }{2} \left (\frac{4}{3} \right )$
$= \frac{4\pi }{6} = \frac{2\pi }{3} units^{3}$

gigglinJess · May 3, 2009

nerdsforever said:
Sum of numbers from 1 to 200: 200/2(1+200) = 100 x 201 = 20100

6, 12, 18, 24.....
sum of numbers divisble by 6: 33/2[2x6 + (33-1)6] = 3366

sum of numbers from 1 to 200 that are not divisble by 6: 20100 - 3366 = 16734

Correct me if I am wrong (pretty sure it's wrong)

Hmmm i don't think its right either, but do you solve it using induction or something?

Q
Use simpons rule and 3 function values to find the approx area between the curve y=logex, the x axis, x=1 and x=5

gurmies · May 3, 2009

nerdsforever said:
Sum of numbers from 1 to 200: 200/2(1+200) = 100 x 201 = 20100

6, 12, 18, 24.....
sum of numbers divisble by 6: 33/2[2x6 + (33-1)6] = 3366

sum of numbers from 1 to 200 that are not divisble by 6: 20100 - 3366 = 16734

Correct me if I am wrong (pretty sure it's wrong)

It's correct...

gigglinJess said:
Q

Use simpons rule and 3 function values to find the approx area between the curve y=logex, the x axis, x=1 and x=5

$A \approx \frac{h}{3}[f(x_{1})+4f(x_{2})+f(x_{3})]$

$= \frac{2}{3}[\ln1+4\ln3+\ln5]$

$\approx 4.003u^{2}$

New Question:

$If \,\, y = 2^{x}, \,\, find \,\, \frac{dy}{dx} \,\, when \,\, x=2$

Eddykungfu · May 3, 2009

Here we go

gigglinJess · May 3, 2009

gurmies said:
New Question:

$If \,\, y = 2^{x}, \,\, find \,\, \frac{dy}{dx} \,\, when \,\, x=2$

$\frac{dy}{dx}=x.2^{x-1}$
when x=2
$\frac{dy}{dx}=2.2^{2-1}$
$=4$

Q. Find the equation of the tangent to the curve y = ln(4x - 1) at the point where the curve cuts the axis

jet · May 3, 2009

gigglinJess said:
$\frac{dy}{dx}=x.2^{x-1}$
when x=2
$\frac{dy}{dx}=2.2^{2-1}$
$=4$

actually, $\frac{d}{dx} \left ( 2^x \right ) = \, $ln$ \, 2 \cdot 2^x$

Drongoski · May 3, 2009

Eddykungfu's Question

$\100dpi \int_{0}^{4}x\cdot e^{x^{2}}\ dx\ \ =\ \ \frac{1}{2}\int_{0}^{4}e^{x^{2}}\cdot 2xdx\\ \\ \ \ =\ \ \frac{1}{2}\left [e^{x^{2}} \right ]^{4}_{0}\ \ =\ \ =\ \ \frac{1}{2}\left [e^{4^{2}}-e^{0} \right ]\ \ =\ \ \frac{1}{2}(e^{16}-1)\\ \\$

gurmies · May 3, 2009

gigglinJess said:
$\frac{dy}{dx}=x.2^{x-1}$
when x=2
$\frac{dy}{dx}=2.2^{2-1}$
$=4$

Notice that x is an exponent in this question. This is why I posted it, to perhaps trick/teach a few people xD. You can't treat exponents like the way you have. Consider using change of base rules or something similar...

gigglinJess said:
Q. Find the equation of the tangent to the curve y = ln(4x - 1) at the point where the curve cuts the axis

$y = \ln(4x-1)$

$\frac{dy}{dx}=\frac{4}{4x-1}$

$When \,\, curve \,\, crosses \,\, x-axis, \,\, y = 0$

$\ln(4x-1)=0$

$4x - 1 = 1$

$x = \frac{1}{2}$

$At \,\, (\frac{1}{2},0), M_{T}=4$

$\therefore Eqn \,\, of \,\, tangent \,\, is \,\, y=4(x-\frac{1}{2})$

$y = 4x - 2$

jet · May 3, 2009

I'll do gurmies' old one then,
$\frac{dy}{dx} = ln \, 2 \cdot 2^x \\ $When $ x = 2, \frac{dy}{dx} = 4ln2 = ln16$
Took this from last year's HSC:
$Solve $ ln \,x - \frac{3}{ln \, x} = 2, $ justifying whether all of your solutions are valid$

azureus88 · May 3, 2009

[maths]\ln x-\frac{3}{\ln x}=2\\(\ln x)^2-2\ln x-3=0\\(\ln x-3)(\ln x+1)=0\\\ln x=-1,3\\x=\frac{1}{e},e^3[/maths]

They're both valid?? Or is there something i missed.

Dumbledore · May 4, 2009

study-freak · May 4, 2009

gurmies · May 4, 2009

Not sure how to best explain this, but if it's anything that deviates from linear, another power of x will come out. If linear, only numerical values will come out when integrating.

Drongoski · May 4, 2009

gurmies said:
Not sure how to best explain this, but if it's anything that deviates from linear, another power of x will come out. If linear, only numerical values will come out when integrating.

gurmies

I can explain it. I've to rush off to tutor. If not done when i get back about 10 pm will try to post explanation.

gurmies · May 4, 2009

Awesome, perhaps you can articulate it better than I did (not difficult)

jet · May 4, 2009

In the meantime...
$Find the integral $ \int^{4}_{0}4 - x \, dx \\ $Hence, find the volume of the solid of revolution formed by rotating the graph of $ y = 4 - x^2 $above the x-axis, about the y- axis$

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