help with ANOTHER ruse 4U Q (1 Viewer)

[mreditor16]

OK I just got it out and checked the solution and my method was the exact same.

First of all forget about making two assumptions, just use one.

Then use the given result to change all the k+1Cr terms to (kCr+kCr-1) terms AND since the first term is k+1C0 just change it to kC0 for uniformity sake i.e. the L.H.S becomes:

kC0/x - (kC1 + kC0)/x+1 + (kC2 + kC1)/x+2 ....

Becomes the grouping shown in the solution

Then you can apply the result of pt ii.

Quite messy but it works out

great job! thanks man

pssh who said you have "limited problem solving ability"?

 

[Fade1233]

Ah cant attach images in a new thread. So will just post my question up here:



I used SP=ePM and then worked out height using pythagoras but that doesnt work. So I don't know how they got height as that?
If u want will rep.

 

[emilios]

Ah cant attach images in a new thread. So will just post my question up here:
View attachment 30973
View attachment 30974

I used SP=ePM and then worked out height using pythagoras but that doesnt work. So I don't know how they got height as that?
If u want will rep.

i) x^2/4 + y^2/1 = 1 ; found from the intercepts of the ellipse
ii) The height of the vertical cross section varies. How exactly does it vary? Well lets call the height of the parabolic arc h. Now this is important: h WILL BE A POINT ON THE PARABOLA! So if we find the equation of the parabola, we can deduce an expression for h dependant on x right? Ok so we know that h=ax^2+bx+c , since it's a parabola. Plug in some known points. When x=0, h=2 therefore c=2. When x=2, h=0 therefore 4a+2b=-2. When x=-2, h=0 therefore 4a-2b=-2. Solving simultaneously yields b=0 and a= -1/2 .
Thus, h= -x^2/2 +2 i.e. h=1/2(4-x^2)
iii) Not sure, I think you just plug in sqrt(3) into h for the height and for the width you use 2y= sqrt(4-x^2) and plug in sqrt(3). So I think the max height of the door is like 1/2 and the max width is 1
iv.) dV = dA*dx
We need to find the area bounded by each parabolic segment. There's two ways: one is a plug and chug formula given by A= 2/3 *base*height and the other is a little more intuitive. We use Simpson's Rule because it uses a parabolic arc to estimate the area under curve. When the given curve IS a parabola, it generates an exact value. So we take two sub-intervals, each of height y (don't confuse this height with the height of the parabola!).
Applying Simpsons rule: A=y/3 *( First Value + 4*Second Value + Last Value) .On the parabolic segment, the three given points are 0, h and 0 and the value of y is taken from the equation of the ellipse (y= sqrt(4-x^2)/2). The area works out to be sqrt(4-x^2)/6 * (0 + 4h + 0 ) i.e. A = 1/3(4-x^2)^3/2. The volume is the integral of this evaluated from -2 to sqrt(3)

 

[MrBeefJerky]

Hey mreditor16 which year is this from?

 

[Kurosaki]

It's Ruse MX2 2009 trial I believe.

 

[mreditor16]

It's Ruse MX2 2009 trial I believe.

+1

 

[dunjaaa]

- Made a mistake in my first post lol

 

[Fade1233]

i) x^2/4 + y^2/1 = 1 ; found from the intercepts of the ellipse
ii) The height of the vertical cross section varies. How exactly does it vary? Well lets call the height of the parabolic arc h. Now this is important: h WILL BE A POINT ON THE PARABOLA! So if we find the equation of the parabola, we can deduce an expression for h dependant on x right? Ok so we know that h=ax^2+bx+c , since it's a parabola. Plug in some known points. When x=0, h=2 therefore c=2. When x=2, h=0 therefore 4a+2b=-2. When x=-2, h=0 therefore 4a-2b=-2. Solving simultaneously yields b=0 and a= -1/2 .
Thus, h= -x^2/2 +2 i.e. h=1/2(4-x^2)
iii) Not sure, I think you just plug in sqrt(3) into h for the height and for the width you use 2y= sqrt(4-x^2) and plug in sqrt(3). So I think the max height of the door is like 1/2 and the max width is 1
iv.) dV = dA*dx
We need to find the area bounded by each parabolic segment. There's two ways: one is a plug and chug formula given by A= 2/3 *base*height and the other is a little more intuitive. We use Simpson's Rule because it uses a parabolic arc to estimate the area under curve. When the given curve IS a parabola, it generates an exact value. So we take two sub-intervals, each of height y (don't confuse this height with the height of the parabola!).
Applying Simpsons rule: A=y/3 *( First Value + 4*Second Value + Last Value) .On the parabolic segment, the three given points are 0, h and 0 and the value of y is taken from the equation of the ellipse (y= sqrt(4-x^2)/2). The area works out to be sqrt(4-x^2)/6 * (0 + 4h + 0 ) i.e. A = 1/3(4-x^2)^3/2. The volume is the integral of this evaluated from -2 to sqrt(3)

Ahhh. Now I see. So you dont have to use Ellipse properties. Wasnt it painful typing all that? Repped. Thanks

View attachment 30976 - Made a mistake in my first post lol

Ah thanks man . hAHA I dont even know why I used SP=ePM cos I saw sqrt(3) as a foci so I am like maybe I should use SP=ePM and you can find the directrix. And then I thought you might have to use pythagoras at the centre using the found solution with x. Hahaha didnt need to go through all that. How did you computerise those? And repped.

And hahaha I forgot to mention I just needed part 2. Sorry for making you type all that guys.
Thanks again guys.