Author Topic: A few Specialist Problems (Read 20789 times) Tweet Share

GerrySly · « **Reply #45 on:** July 09, 2009, 08:45:47 pm »

Quote from: TrueTears on July 09, 2009, 08:24:34 pm

http://vcenotes.com/forum/index.php/topic,9192.msg121038.html#msg121038

That would help immensely.

Thanks for that TrueTears, helped partially, got some clarifications now

The answer is $\frac{dQ}{dt}=6-\frac{Q}{10}$ and I am able to get the 6 as $Inflow=3 g/L \cdot 2 L/min=6 g/min$ but I am having trouble getting the Outflow. By my understanding of kj_'s post the outflow is $\frac{mass}{volume}$ and the volume exits and enters the container at the same amount therefore there is no $t$ variable.

This means that the answer should be $\frac{dQ}{dt}=6-\frac{Q}{20}$ right? Not sure how they got it over 10.

Damo17 · « **Reply #46 on:** July 09, 2009, 09:03:48 pm »

Quote from: GerrySly on July 09, 2009, 08:45:47 pm

Quote from: TrueTears on July 09, 2009, 08:24:34 pm
http://vcenotes.com/forum/index.php/topic,9192.msg121038.html#msg121038

That would help immensely.
Thanks for that TrueTears, helped partially, got some clarifications now

The answer is $\frac{dQ}{dt}=6-\frac{Q}{10}$ and I am able to get the 6 as $Inflow=3 g/L \cdot 2 L/min=6 g/min$ but I am having trouble getting the Outflow. By my understanding of kj_'s post the outflow is $\frac{mass}{volume}$ and the volume exits and enters the container at the same amount therefore there is no $t$ variable.

This means that the answer should be $\frac{dQ}{dt}=6-\frac{Q}{20}$ right? Not sure how they got it over 10.

$Concentration = \frac{mass}{volume}$

$Outflow= \frac{mass}{volume} \times outflow \ rate$

$Outflow= \frac{Q}{20} \times 2 = \frac{Q}{10}$

GerrySly · « **Reply #47 on:** July 14, 2009, 12:09:27 pm »

Just a clarification question, is this normal notation? The umlat on the top of the y? Seems a tad strange to me...

GerrySly · « **Reply #48 on:** July 14, 2009, 12:30:55 pm »

Another simple question haha

I was able to get the equation for displacement (x) as $x=5(9-t^2)$ , which allowed me to find the time at which the rock hits the ground, $t=3$ . Now when I subbed it into the equation for velocity, $v=-10t$ I get $-30 m/s$ . I am not sure why I am getting a negative but they are getting a positive?

kamil9876 · « **Reply #49 on:** July 14, 2009, 12:55:25 pm »

Yes, Issac Newton used this notation.

It's because the word 'speed' refers to magnitude.
Btw if you havn't already seen then there is a formula:

$v^2=u^2+2as$

Which is derived from using the method that you used. ie: sub in $t=\frac{v-u}{a}$ into $s=\frac{1}{2}at^2+ut$ and you get the formula.

Anyways, the formula i provided saves time for this problem.

GerrySly · « **Reply #50 on:** July 14, 2009, 01:31:13 pm »

Quote from: kamil9876 on July 14, 2009, 12:55:25 pm

Yes, Issac Newton used this notation.

It's because the word 'speed' refers to magnitude.
Btw if you havn't already seen then there is a formula:

$v^2=u^2+2as$

Which is derived from using the method that you used. ie: sub in $t=\frac{v-u}{a}$ into $s=\frac{1}{2}at^2+ut$ and you get the formula.

Anyways, the formula i provided saves time for this problem.

Ah ok, yeah I was going to use that formula but decided against for some stupid reason. Thanks for that

GerrySly · « **Reply #51 on:** July 14, 2009, 02:46:51 pm »

Alright just another simple question, I managed to get $\int_0^{\frac{\pi}{6}} \sin^{n}{(x)}\cos{(x)}.dx=\frac{1}{64}$ down to the following equation $\frac{1}{64}=\frac{2^{-n}}{2(n+1)}$ , and was wondering is there anyway to solve that by hand? I used nSolve on the calculator and got 3.

GerrySly · « **Reply #52 on:** July 14, 2009, 04:07:02 pm »

No idea where to even begin with this one, most differential equations I solve I am given $y=$ and then I differentiate as many times as needed and sub in to solve.

/0 · « **Reply #53 on:** July 14, 2009, 04:13:58 pm »

$\frac{dy}{dx} = 1-y$

Now we flip so you get

$\frac{dx}{dy} =\frac{1}{1-y}$

Now you can solve for x in terms of y by integrating both sides. Later you can solve for y again.

dcc · « **Reply #54 on:** July 14, 2009, 06:11:05 pm »

But does $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ ?

/0 · « **Reply #55 on:** July 14, 2009, 06:28:30 pm »

er.... does it?

dcc · « **Reply #56 on:** July 14, 2009, 06:31:03 pm »

Quote from: /0 on July 14, 2009, 06:28:30 pm

er.... does it?

I'm willing to say sometimes. Other times it would be nonsense.

kamil9876 · « **Reply #57 on:** July 14, 2009, 06:58:06 pm »

/0, your username provides one idea of a problematic scenario.

how about $y=x^2$ . what is $\frac{d}{dy}(x)|_{y=1}$ ? is it 0.5 or -0.5? But for one-to-one function this will not arise such as the logarithms and stuff that you ussually deal with, or application problems where you stick to some domain.

GerrySly · « **Reply #58 on:** July 16, 2009, 05:34:33 pm »

A bowl can be described as the solid of revolution formed by rotating the graph of $y=\frac{1}{4}x^2$ around the y-axis for $0 \le y \le 25$

Find the volume of the bowl $V=1250\pi$

The bowl is filled with water and then, at time $t=0$ , the water begins to run out of a small hole in the bottom. The rate at which the water runs out is proportional to the depth, $h$ , of the water at time $t$ . Let $V$ denote that volume of water at time $t$ .

Show that $\frac{dh}{dt}=\frac{-k}{4\pi}, k > 0$

Now this is the part that I am having trouble with. I know I must do the whole $\frac{dh}{dt}\propto$ to something, then introduce a constant of proportionality but I am not sure where to go with this.

Any help is appreciated

TrueTears · « **Reply #59 on:** July 16, 2009, 05:36:34 pm »

I thought it was $\frac{dV}{dt} \propto h$

So $\frac{dV}{dt} = -kh$ where $k > 0$

Now find $\frac{dV}{dh}$ [Let y = h, integrate and find V as a function of h] then use chain rule.

Search the forums now!

We have moved!

Author Topic: A few Specialist Problems (Read 20789 times) Tweet Share

GerrySly

Re: A few Specialist Problems

Damo17

Re: A few Specialist Problems

GerrySly

Re: A few Specialist Problems

GerrySly

Re: A few Specialist Problems

kamil9876

Re: A few Specialist Problems

GerrySly

Re: A few Specialist Problems

GerrySly

Re: A few Specialist Problems

GerrySly

Re: A few Specialist Problems

/0

Re: A few Specialist Problems

dcc

Re: A few Specialist Problems

/0

Re: A few Specialist Problems

dcc

Re: A few Specialist Problems

kamil9876

Re: A few Specialist Problems

GerrySly

Re: A few Specialist Problems

TrueTears

Re: A few Specialist Problems

Recent Posts

User:
Password: