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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: aronno on July 10, 2009, 10:39:56 pm

Title: SPEC. MATHS essentials
Post by: aronno on July 10, 2009, 10:39:56 pm: i need help with
chapter 11 extended response;
Q.1 (b) i, ii, iii.

Q.3 (c) I just cant crack this one!!! >:(
Title: Re: SPEC. MATHS essentials
Post by: /0 on July 10, 2009, 11:05:57 pm: b)i)
We are given that $\frac{dh}{dt}= \frac{dh}{dV} \times \frac{dV}{dt}$

When the water level is at a height $h$ , the volume of water will be

$V=\pi \int_0^h (2\sqrt{y})^2 dy=[2y^2]_0^h = 2\pi h^2$

$\Rightarrow \frac{dV}{dh} = 4\pi h$

$\Rightarrow \frac{dh}{dV} = \frac{1}{4\pi h}$

$\frac{dV}{dt} = -kh$ , as per the information.

$\therefore \frac{dh}{dt} =\frac{1}{4\pi h} \times -kh = \frac{-k}{4\pi}$

b) ii)

The initial height is $25$ . After 30 seconds it is 0.

$\Rightarrow \int_0^{30} \frac{dh}{dt} dt = -25$

$\int_0^{30} \frac{-k}{4\pi}dt = -25$

$\therefore k = \frac{10\pi}{3}$

b)iii)

$\frac{dh}{dt} = \frac{-k}{4\pi}$

$dh = \frac{-k}{4\pi}dt$

$\int dh = \int \frac{-k}{4\pi}dt$

$h = \frac{-tk}{4\pi}+C$

At $t = 0$ , $h=25$

$\Rightarrow C = 25$

$\therefore h = \frac{-kt}{4\pi}+25$

But $k=\frac{10\pi}{3}$

$\therefore h = -\frac{5}{6}t+25$

3c)
The radius of the surface at any time is given by the x-coordinate of the parabola. Since the bucket is circular,

$A = \pi x^2=\pi\left(\frac{12+y}{0.12}\right)=\frac{25(y+12)}{3}$

$\therefore \frac{dv}{dt} = -\frac{\sqrt{h}}{A} = -\frac{\sqrt{h}}{\left( \frac{25\pi(y+12)}{3}\right)}=\frac{-3\sqrt{h}}{25\pi(h+12)}$
Title: Re: SPEC. MATHS essentials
Post by: aronno on July 11, 2009, 12:24:00 am: thanks dude.
can any one help with;
4. (a)

thanks
Title: Re: SPEC. MATHS essentials
Post by: /0 on July 11, 2009, 05:00:42 pm: no problem

The volume of water when the depth is $h$ is given when $x^2+y^2=a^2$ , $-a \leq y \leq -a+h$ is revolved around the y-axis (a graph would be helpful).

$x = \pm \sqrt{a^2-y^2}$

So the volume is $V = \pi \int_{-a}^{-a+h} \left(\sqrt{a^2-y^2}\right)^2dy$
Title: Re: SPEC. MATHS essentials
Post by: TrueTears on July 11, 2009, 05:10:15 pm: Why not just solve for $x^2$ .

$x^2 = a^2 - y^2$

Saves the trouble of square rooting and then squaring it again [if the equation was harder you'd want to do it as easily and quickly as possible.]
Title: Re: SPEC. MATHS essentials
Post by: /0 on July 11, 2009, 05:18:18 pm: Sure, that works too