HSC 2014 MX2 Marathon (archive) (1 Viewer)

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hit patel

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Re: HSC 2014 4U Marathon

Ah. Nice question. I tried to draw the arcsin curve of sec x but couldnt but still didnt realise that.
Volume question please if possible. otherwise integration still good.
 

dunjaaa

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Re: HSC 2014 4U Marathon

what i meant was -1 <= sec(a) < sec(b) <= 1 (within the domain) which is undefined unless a=b=0 i tried editing but it still doesnt show, some bug in the system
 

Davo_01

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Re: HSC 2014 4U Marathon

Find the volume produced by rotating the area bounded by the ellipse: about the y-axis, using cylindrical shells.
 

dunjaaa

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Re: HSC 2014 4U Marathon

Can u explain what I'm doing wrong, I divide by x and bring the x inside the sqrt. By simplifying, you get lim(n->-infinity)[sqrt(1+4/x+7/x^2)+1] and you get 2. Now i go to check with my calculator by subbing in a large value, it tells me its worth -2 zzz
 

Davo_01

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Re: HSC 2014 4U Marathon

Do we let x+2 = sqrt (3) tan theta? Does that work?
Maybe, I have never tried it, but I have two different methods, one of them being rationalizing the numerator
 

Davo_01

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Re: HSC 2014 4U Marathon

Can u explain what I'm doing wrong, I divide by x and bring the x inside the sqrt. By simplifying, you get lim(n->-infinity)[sqrt(1+4/x+7/x^2)+1] and you get 2. Now i go to check with my calculator by subbing in a large value, it tells me its worth -2 zzz
This is why i chose the question :). What you need to think about is when you bring the x inside the square root bracket, are you still dividing by x?

Like does the x still remain as x?
 
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dunjaaa

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Re: HSC 2014 4U Marathon

dw I got it, typesetting solution :D
 

hit patel

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Re: HSC 2014 4U Marathon

hmm my answer was what was ur integral
Way off for both.
Integral was 8pi integral_2_0 sqrt (4-(x-2)^2). I obtained this by shifting curve and its equation or not!!!! damn i forgot to shift equation.
after shifting it would become 8pi integral _2_0 sqrt (4-(x^2)). That gets me 16 Pi^2 , i think
 
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dunjaaa

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Re: HSC 2014 4U Marathon

So what I should of done:
Multiple top and bottom by sqrt(x^2+4x+7)-x,
You get lim(n->-infinity) 4x+7/(sqrt(x^2+4x+7)-x),
Divide everything by x,
lim(n->-infinity) [4+7/x]/[-sqrt(1+4/x+7/x^2)-1] (for x negative)
=-2
 

Davo_01

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Re: HSC 2014 4U Marathon

Way off for both.
Integral was 8pi integral_2_0 sqrt (4-(x-2)^2). I obtained this by shifting curve and its equation or not!!!! damn i forgot to shift equation.
after shifting it would become 8pi integral _2_0 sqrt (4-(x^2))
you forgot the x



edit: my bad my answer is
 

hit patel

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Re: HSC 2014 4U Marathon

So what I should of done:
Multiple top and bottom by sqrt(x^2+4x+7)-x,
You get lim(n->-infinity) 4x+7/(sqrt(x^2+4x+7)-x),
Divide everything by x,
lim(n->-infinity) [4+7/x]/[-sqrt(1+4/x+7/x^2)-1] (for x negative)
=-2
Dunja can you please try my way? using the x+2 = sqrt(3) tan(theta). I am trying but I am not getting -2. I changed the limits. so that it will approach pi/2
 
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