Author Topic: A few Specialist Problems (Read 20028 times) Tweet Share

GerrySly · « **Reply #15 on:** March 23, 2009, 07:53:47 pm »

Quote from: Mao on March 23, 2009, 07:42:31 pm

Quote from: GerrySly on March 23, 2009, 04:02:40 pm
Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

dividing through by x to get
$\implies (x + 1) + \frac{2}{x}$

as $x\to -\infty$ , $\frac{2}{x}\to 0^-$ , so on the left, the graph becomes $(x+1) + 0^- = x+1$ . Do the same thing on the right to obtain $(x+1)+0^+ = x+1$ . Hence $y=x+1$ is a slant asymptote.

Ah ok, yeah that is one of my problems heh, asymptotes that are equations, thanks for that Mao. It seems you are the only one who can help me lol

GerrySly · « **Reply #16 on:** April 19, 2009, 03:57:14 pm »

An inverted cone with depth 50 cm and radius 25cm is intially full. Water drains out at 0.5 litres per minute. The depth of water in the cone is h cm at time t minutes.(Find an expression for $\frac{dh}{dt}$ )

kamil9876 · « **Reply #17 on:** April 19, 2009, 04:14:10 pm »

$V=\frac{1}{3}h\pi r^2$

$\frac{dV}{dt}=-500$

The slope of the line on the side is a constant $\frac{50}{25}=\frac{h}{r}$
Use that expression to find r in terms of h. Then you can sub that into the equation for V to get V in terms of h. This yields an easy situation.

GerrySly · « **Reply #18 on:** April 19, 2009, 05:13:32 pm »

Quote from: kamil9876 on April 19, 2009, 04:14:10 pm

$V=\frac{1}{3}h\pi r^2$

$\frac{dV}{dt}=-500$

The slope of the line on the side is a constant $\frac{50}{25}=\frac{h}{r}$
Use that expression to find r in terms of h. Then you can sub that into the equation for V to get V in terms of h. This yields an easy situation.

$\frac{dV}{dt}=-500$ , why is that -500 and not -1/2?

I attempted the equation and ended with the answer $-\frac{1}{72\pi h^2}$ , I think I am not fully understanding something.

2 weeks of no maths at all has really hit me hard now

dcc · « **Reply #19 on:** April 19, 2009, 05:17:02 pm »

Quote from: GerrySly on April 19, 2009, 05:13:32 pm

$\frac{dV}{dt}=-500$ , why is that -500 and not -1/2?

1 litre = 1000 cm^3. Since all the other measurements are in cm, it makes sense to get volume in cm as well.

kamil9876 · « **Reply #20 on:** April 19, 2009, 05:45:28 pm »

$r=0.5h$
$V=\frac{1}{12}\pi h^3$
$\frac{dV}{dh}=\frac{1}{4}\pi h^2$
$\frac{dh}{dV}=\frac{4}{\pi h^2}$

and now chain rule it to get -2000 on the numerator for dh/dt.

GerrySly · « **Reply #21 on:** April 19, 2009, 08:43:21 pm »

Thanks for that guys, appreciate the help, starting to get back into the swing of things. Got another though that I can't seem to do

I know it has something to do with using the relationship between $A m^2$ and $x$ but I am not sure how to do that as x is constantly changing so it can't be the radius (which I could then use $A=6x$ , to solve)

Mao · « **Reply #22 on:** April 19, 2009, 09:22:54 pm »

the circular cross section can be modelled by a circle of radius 2 with a center of (2,0): $(x-2)^2 + y^2 = 4$
[think of this as the tank is laying on its side with water flowing out to the left (-x direction)]

Then, at any given x value between 0 and 4, the width is 2y

$|y| = \sqrt{4-(x-2)^2}$

$\implies 2y = 2\sqrt{4-(x-2)^2}$

The surface area A can then be expressed as $A = (2y)\times 6 = 12\sqrt{4-(x-2)^2} = 12\sqrt{4x-x^2}$ [this area is the area of water that is exposed to atmosphere, ambiguous wording...]

hence, $\frac{dx}{dt} = \frac{-0.025\sqrt{x}}{12\sqrt{4x-x^2}} = -\frac{1}{480\sqrt{4-x}}$

(if I remember correctly, that is)

GerrySly · « **Reply #23 on:** May 03, 2009, 04:14:01 pm »

Having some troubles with those two questions. I thought I had an idea of how to do 10 a) but I have trouble working it out without 3 bits of information. I thought that I just subbed it into the equation $v^2=u^2+2as$ , and then solved but with two unknowns I am not sure where to go from there

$\begin{align*}v^2&=(1.2)^2 + 2(a)(3.2)\\&=(6.4)a+1.44\end{align*}$

Number 11 I know I have to split it into two parts and solve, i.e. forwards and backwards, but no idea where to go from there.

Help appreciated

Flaming_Arrow · « **Reply #24 on:** May 03, 2009, 04:37:28 pm »

10.

a = a
u = 1.2
d = 3.2
v = 0

$0 = 1.44 + 2 * a * 3.2$

$0 = 1.44 + 6.4a$

$-1.44= 6.4a$

$a= - 0.225$

GerrySly · « **Reply #25 on:** May 03, 2009, 06:13:52 pm »

Quote from: Flaming_Arrow on May 03, 2009, 04:37:28 pm

10.

a = a
u = 1.2
d = 3.2
v = 0

$0 = 1.44 + 2 * a * 3.2$

$0 = 1.44 + 6.4a$

$-1.44= 6.4a$

$a= - 0.225$

Oh wow, probably should have figured that out heh, thanks

$\begin{align*} t&=\frac{v-u}{a}\\ &=\frac{0-1.2}{-0.225}\\ &=\frac{-\frac{12}{10}}{-\frac{9}{40}}\\ &=\frac{16}{3} s \end{align*}$

Flaming_Arrow · « **Reply #26 on:** May 03, 2009, 06:30:38 pm »

11.

d = ut
d = 12 * 4 = 48m

u = -4 a= ? t= 20 d=48

d= -80 + 0.5at^2

48 = -80 + 0.5 * a * 400

a = 0.64 ms^-2

i don't get what b is asking

GerrySly · « **Reply #27 on:** May 03, 2009, 07:04:06 pm »

Thanks for that, I thought B was asking for the time in which the final velocity of the first section = 0. So when the particle slows to a stop and begins going backwards

The way I thought to do that was the following

$\begin{align*} v&=\int -0.64 dt\\ &=-(0.64)t+c v(12)&=4\\ 4&=-(0.64)(12)+c\\ c&=11.68\\ \therefore v(t)&=-(0.64)t+(11.68) \end{align*}$

Then solve that for when $v(t)=0$ , but I keep getting $\frac{73}{4}$ when the answer is $\frac{55}{4}$

Mao · « **Reply #28 on:** May 03, 2009, 10:47:34 pm »

11. Setting $t_0$ at 12 seconds, the displacement x is +48m, velocity is +4. to get back to origin, it must have a displacement of -48.

$s = ut + \frac{1}{2}at^2$

$\implies -48 = 80 + 200a \implies a = -\frac{16}{25}$

The time the particle has a negative velocity:

$v = u + at \implies 0 = 4 - \frac{16}{25}\cdot t \implies t = 6.25$ (this is the point where the particle momentarily stops)

Hence it spends the rest of the time, 13.75 seconds, going backwards.

GerrySly · « **Reply #29 on:** May 31, 2009, 06:32:23 pm »

Show that

$\begin{align*} \sin{(\theta)}+\cos{(\theta)}i&=cis \left ( \frac{\pi}{2} - \theta \right )\\ \cos{(\theta)}-\sin{(\theta)}i&=cis \left ( -\theta \right )\\ \sin{(\theta)}-\cos{(\theta)}i&=cis \left ( \theta - \frac{\pi}{2} \right )\\ \end{align*}$

Just showing me how to do the first one should be helpful, thanks

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Author Topic: A few Specialist Problems (Read 20028 times) Tweet Share

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