Author Topic: Dekoyl's Questions (Read 24841 times) Tweet Share

dekoyl · « **Reply #30 on:** April 29, 2009, 10:40:35 pm »

Quote from: Mao on April 29, 2009, 10:33:20 pm

This one is tricky, but it goes like this:

performing linear transformations on the variable x, $\int_{-1/2}^{1/2} \cos^{-1} (2x)\; dx = \int_{0}^{1} \cos^{-1} (2(x-1/2))\; dx$

This area is the same as taking an integral in the y axis from 0 to pi

$I = \int_{0}^{\pi} \frac{1}{2}\cos (y) + \frac{1}{2}. dx$

Ah okay. So basically with these questions, we tranform the function to a domain that we can work with (ie. use the rectangle - integral method)?
If so then that's okay

Got it now - Thanks Mao

dekoyl · « **Reply #31 on:** May 03, 2009, 10:23:36 pm »

When we do something like:
Find the volume of the solid generated by revolving about the x-axis the region bounded by $y=2tan(x), x = -\frac{\pi}{4}, x = \frac{\pi}{4}$ and $y=0$
why is it:
$V = \pi \int_0^{\frac{\pi}{4}}4tan^{2}(x).dx + \pi \int_{-\frac{\pi}{4}}^{0}4tan^{2}(x).dx$
and not
$V = \pi \int_0^{\frac{\pi}{4}}4tan^{2}(x).dx - \pi \int_{-\frac{\pi}{4}}^{0}4tan^{2}(x).dx$
Note the + and - sign change before the 2nd part of the equation.
I thought that region was below the x-axis so it should be a negative sign?

Thanks - sorry if I'm not very clear.

Mao · « **Reply #32 on:** May 03, 2009, 10:33:12 pm »

but how can you have negative volume

It can also be $V = \pi \int_0^{\frac{\pi}{4}}4tan^{2}(x).dx - \pi \int_0^{-\frac{\pi}{4}}4tan^{2}(x).dx$ (notice the change in terminals)

When the function is squared, it is always positive. Hence there are no negative volume

dekoyl · « **Reply #33 on:** May 03, 2009, 10:37:59 pm »

/sigh. That was stupid.

Thanks Mao. =|

dekoyl · « **Reply #34 on:** May 04, 2009, 08:58:33 pm »

Does anyone want to have a go at this one?

$\int\frac{x}{2x^2+3x-4}.dx$
I let $u=2x^2+3x-4$ which led to $\frac{1}{4}\int\frac{1}{u}.du - \frac{3}{4}\int\frac{1}{2x^2+3x-4}.dx$ then I used partial fractions to split the latter bit up. However, the answer I got did not match CAS' answer.

Thanks.

Mao · « **Reply #35 on:** May 04, 2009, 09:08:51 pm »

Seeing log terms, they are probably different by a constant. What did you get from your hand calculations?

Remember that $\log_e (kx) = \log_e (k) + \log_e (x)$

dekoyl · « **Reply #36 on:** May 04, 2009, 09:13:09 pm »

Quote from: Mao on May 04, 2009, 09:08:51 pm

Seeing log terms, they are probably different by a constant. What did you get from your hand calculations?

Remember that $\log_e (kx) = \log_e (k) + \log_e (x)$

I rubbed it out but what I had in the logs were something similar to $x+\frac{3-2\sqrt{14}}{4}$ and $x+\frac{3+2\sqrt{14}}{4}$ but the calculator had $\sqrt{41}$ instead of $\sqrt{14}$ :S

dekoyl · « **Reply #37 on:** May 04, 2009, 09:55:41 pm »

Answer from calculator:
$\int{{{x}\over{2\,x^2+3\,x-4}}}\,dx = {{\sqrt{41}\,\log \left(2\,x^2+3\,x-4\right)+3\,\log \left(4\,x+ \sqrt{41}+3\right)-3\,\log \left(4\,x-\sqrt{41}+3\right)}\over{4\, \sqrt{41}}}$

dekoyl · « **Reply #38 on:** May 06, 2009, 08:34:07 pm »

I'm not sure what I'm doing wrong -
The region bounded by the graph of $y=\frac{1}{\sqrt{x^2+9}}$ , the x-axis, the y-axis and the line x=4 is rotated about the y-axis. Find the volume of the solid formed.

I end up with $V = \int_{\frac{1}{5}}^{\frac{1}{3}}(\frac{1}{y^2}-9)dy$ which yields $\frac{4}{5}\pi$
The answer is $4\pi$ . I'm not sure where I went wrong. =\

Thanks

kamil9876 · « **Reply #39 on:** May 06, 2009, 08:52:48 pm »

You should draw a diagram. The shape is in fact a 'hybrid shape' i.e: there is one rectangle being rotated and one curve next to it. This extra shape is a cylinder with radius 4 and height 1/5. Hence the volume of this portion is:

$\frac{4^2 \pi}{5}=\frac{16\pi}{5}$ which is exactly what you are missing.

dekoyl · « **Reply #40 on:** May 06, 2009, 09:11:37 pm »

Ah I see now. Thanks kamil!

kamil9876 · « **Reply #41 on:** May 06, 2009, 09:39:33 pm »

I can see that it did not just be looking at the terminals, the volume was between y=1/3 and y=1/5 whereas the whole shape is between 1/3 and 0.

I think you are confusing this with rotation about the x axis. In that case the cylinder would be taken into account.

dekoyl · « **Reply #42 on:** June 13, 2009, 08:19:21 pm »

Haven't touched spec. for a while so a bit rusty.. I've got a few starter questions that might help me get back on my feet:
Q:If the population of a city increases at a rate which is proportional to the current population and, if the population of a certain city was 100,000 in 1980 and 120,000 in 1990, express, in terms of t, the population t years after 1980.
What I did was:
$\frac{dP}{dt}=kp$
$\int \frac{dP}{P}=\int k.dt$
$log_e|P|+c = kt$
$P=Ae^{kt}$
$\implies A=100 000$
$120 000 = 100 000e^{10t}$
$\frac{6}{5} = e^{10k}$
$\frac{1}{10}log_e\left(\frac{6}{5}\right)=k$
$\therefore P=100 000e^{\frac{1}{10}log_e\left(\frac{6}{5}\right)t}$ <--- My answer
Answer: $100000e^{0.02t}$ <--- Book answer

/0 · « **Reply #43 on:** June 13, 2009, 09:52:57 pm »

Seems alright... maybe it's because $\frac{1}{10}\log_e(\frac{6}{5})=0.018$ which is very close to what they have and they just rounded up.

dekoyl · « **Reply #44 on:** June 14, 2009, 01:18:43 pm »

Q: A light penetrates a slab of glass, the rate of loss of intensity of he light with respect to the depth of the glass is proportional to the intensity at that depth. If the light loses 5% of its intensity penetrating 10% of the glass, what percentage is lost in penetrating the slab?

I have trouble understanding what the question wants let alone setting up an equation for it

Thanks for any help.

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