If I remember correctly, the brain basically remembers everything, but the recalling process is affected by the significance of the event in your life. That's why events that affect your emotionally are easier to recall.
Oh it's my bad. I disregarded ground for a minute and thought the Voltage Sources would create a potential of -10V over there. ( -12 + 2 V ). Meh.
Then as technodude says, v0 = v2 and the problem is solved. Thanks all. Sorry to confuse you with a bad diagram.
Ok I took v_2 = -2 and did the calculations but I still don't get the right answers.
The answers given in the textbook are v_o = -30V and i_o=3.5mA
Here are my calculations:
\frac{12-^-2}{4}= \frac{-2-v_2}{8} \ \longrightarrow 28 = -2-v_2 \ \longrightarrow v_2 = -30V
\frac{-2-^-30}{8} +...
Find V_o and i_o. Assume an ideal op amp.
http://img236.imageshack.us/img236/2426/633bb0.jpg
I tried applying KCL at v1 and v2 nodes but then got stuck because I had 2 equations with 3 unknowns.
\frac{12-v_1}{4} = \frac{v_1-v_2}{8} and
\frac{v_1-v_2}{8}+\frac{-10-v_2}{20}= i_o...
So can I set up the integral this way ?
\int_{-4\sqrt{2}}^{4\sqrt{2}} \ \int_{-\sqrt{16- \frac{(u+4)^2}{2}}}^{\sqrt{16- \frac{(u+4)^2}{2}}} \ \int_{\sqrt{ \left( \frac{u+v}{2} \right)^2 + \left( \frac{u-v}{2} \right)^2}}^{\frac{4-u}{2}} \,dz\,dv\,du
P.S. Plz let this be right. Editing the...
So from this I get:
\frac{16-8u+u^2}{4} = \frac{2u^2+2v^2}{4}
16-8u+u^2=2u^2+2v^2
If I go further:
16 = u^2+8u+v^2
16 + 16 = u^2+8u+16+v^2
32 = (u+4)^2 + v^2
\frac{(u+4)^2}{32} + \frac{v^2}{16} = 1
Now this is an equation of an ellipse but it is not centered along the z axis. Did I...
Ok so this is what I did.
x = (u+v)/2 & y = (u-v)/2
\frac{4-u}{2} = \sqrt{ \left( \frac{u+v}{2} \right)^2+ \left( \frac{u-v}{2} \right)^2}
Am I on the right path ?
*edit, yeah u're right. fixed now.
That's what I have been trying to do but setting
\frac{4-x-y}{2} = \sqrt{x^2+y^2} leads me to a dead end.
I get
4-x-y-x+\frac{x^2}{4} + \frac{xy}{4} -y+\frac{xy}{4}+\frac{y^2}{4} = x^2 + y^2
4-2x-2y-2xy = \frac{3}{4}x^2 + \frac{3}{4}y^2 ?
For the first one, you find the derivative first.
f(x) = 3x^2-1 \ \longrightarrow \ f'(x) = 6x then you substitude for x. That doesn't give 6. Now that you have the slope you can find the tangent line.
For the second line, I think that would be the gradient of the secant line through those...
I got stuck on the 3rd question and now I kinda don't know what to do. Can someone help me a bit ?
http://img79.imageshack.us/img79/1786/scan5eq3.png
I drew a graph and shaded the region W.
http://img127.imageshack.us/img127/8185/q3qw2.jpg
I thought it would be a good idea to integrate...
Let's say you found the area of the cross-section to be A. Well, if it is \frac{1.1 \times 10^9}{1m^2}, how much is it for \frac{x}{A}. And notice that this is not the final answer. It will give you the maximum force that that specific thickness of steel wire can resist.