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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: cohen on June 06, 2011, 08:10:34 pm

Title: Cohen's Specialist questions
Post by: cohen on June 06, 2011, 08:10:34 pm: I'm having some trouble figuring this question out question (Quite a few people in my class are actually). So hopefully you guys will be able to help me :D

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Concrete blocks for a road are carved out of cylindrical blocks, that are 4m high and 12m in diameter. e) Show that the remaining volume that is left after carving the road barriers is 106.4 m^3 Cylinders, that were previously described in part e, are made for the road barriers that are now hollow instead of concrete. While they are being placed at the new location, a downpour occurs and the cylinders are filled with water. The volume of each cylinder is increasing at a constant rate of 2m^3 per hour. f) Find the rate, in meters per hour, at which the height is increasing when there is a height of 3m of water in the cylinder.
I have managed to do part e, which was actually relatively easy, but I have no idea how to do part f :\
Title: Re: Cohen's Specialist questions
Post by: xZero on June 06, 2011, 08:50:39 pm: so v = pi*r^2*h, find the ratio between r and h and express r in terms of h

from the info in part e) we know that h = 4 and r = 6. the ratio should be h/4 = r/6, r = 3h/2

v = pi*(3h/2)^2*h. find dv/dh and use related rates of change. dh/dt = dh/dv * dv/dt, sub h = 3 in and BAM, you have your answer
Title: Re: Cohen's Specialist questions
Post by: cohen on June 06, 2011, 09:08:41 pm: Cheers, i ended up getting the same answer as my friend =]
Title: Re: Cohen's Specialist questions
Post by: Mao on June 07, 2011, 08:33:52 pm: Not sure why it was locked.

Unlocked.
Title: Re: Cohen's Specialist questions
Post by: cohen on June 07, 2011, 08:46:09 pm: I actually had done that myself :P, I should of posted that i had locked it, so sorry.
Thanks for your consideration though Mao.
Title: Re: Cohen's Specialist questions
Post by: cohen on July 04, 2011, 11:38:30 pm: Does anybody have any idea of how to do the analysis question 2 part a? I have no idea how to even start it D:
http://i51.tinypic.com/v8d0uc.jpg

-Thanks if anyone is able to help me :)
Title: Re: Cohen's Specialist questions
Post by: luken93 on July 04, 2011, 11:53:28 pm: Quote from: cohen on July 04, 2011, 11:38:30 pm
Does anybody have any idea of how to do the analysis question 2 part a? I have no idea how to even start it D:
http://i51.tinypic.com/v8d0uc.jpg

-Thanks if anyone is able to help me :)
x = 0, f(0) = 3pi
x = 4, f(x) = 3pi/2
x = 6, f(x) = pi

Solve by simultaneous..
Title: Re: Cohen's Specialist questions
Post by: cohen on July 05, 2011, 12:22:46 am: I should of been more specific, sorry :P

I had gotten that far, I just couldn't figure out how to solve the simultaneous equations, and the ti-nspire couldn't solve it either. :\
Title: Re: Cohen's Specialist questions
Post by: xZero on July 05, 2011, 01:47:04 am: umm this is gonna be hard to explain but ill give it my best.

First of all i didnt use simultaneous equations, i inversed the equation of the graph to $c\ Cos(\frac{x}{a})+b=y$ where the domain is $[0,3\pi]$ and the range is $[0,8]$ .

Normally the domain and the range of y=Cos(x) is $[0,\pi]$ and $[0,1]$ , from the domain change we can find a and the range will find b and c. Double the domain so this becomes a full cycle of a cos graph -> $\frac{2\pi}{n}=6\pi$ , where n=1/a, rearrange and you will get a=3.

Range remains the same so y=[0,8]. This tells us that the mid point is at y=4, hence b=4. It oscillates at a magnitude of 4 so c=4

Now we have a,b and c sub that back into the arccos and we get $f(x)=3Cos^{-1}(\frac{x-4}{4})$

Of course if you're confident with your arccos graphs you dont have to inverse it back to cosine but I'm not too familiar with the properties of arccos graphs
Title: Re: Cohen's Specialist questions
Post by: cohen on September 12, 2011, 06:27:58 pm: Got another question, it's part of my sac and the guys I've talked to don't have any idea of how to do it either.
g(x) is a function such that $g(x) = a - \frac{b}{x^2 + c}$
The function passes through the points $(0, 0)$ and $(2, \frac{2}{3})$ and is a maximum at the point x = $\frac{\sqrt{6}}{3}$

Thanks if anyones able to help me ^.^

edit: Just realised what to do for the maximum part, never mind :P (That's the only part that had originally confused me...)
Title: Re: Cohen's Specialist questions
Post by: b^3 on September 12, 2011, 06:31:05 pm: I'm guessing you wan't the values of a, b and c. Since it's part of a SAC, I won't do it for you but guide you. Substitute the two points (0,0) and (2,2/3) in to form two equations. A maximum point will have zero gradient, so find the derivative and let it equal zero, subbing the x-value in. That will gice you a thrid equation, then solve simultaneously for a,b,c.
Title: Re: Cohen's Specialist questions
Post by: Mao on September 12, 2011, 10:01:27 pm: Quote from: cohen on September 12, 2011, 06:27:58 pm
Got another question, it's part of my sac and the guys I've talked to don't have any idea of how to do it either.
g(x) is a function such that $g(x) = a - \frac{b}{x^2 + c}$
The function passes through the points $(0, 0)$ and $(2, \frac{2}{3})$ and is a maximum at the point x = $\frac{\sqrt{6}}{3}$

Thanks if anyones able to help me ^.^

edit: Just realised what to do for the maximum part, never mind :P (That's the only part that had originally confused me...)

Wait, this doesn't make any sense to me. The derivative will always show the only horizontal point is $x=0$ .

that family of functions cannot have a maximum at $\frac{\sqrt{6}}{3}$ . Are you sure you have the formula correct?