Post by: luken93 on February 24, 2011, 07:06:46 pm
Anyways, first question:
The simultaneous linear equations
(1)...
(2)...
Have No Solution for.....
Post by: francis8nho on February 24, 2011, 07:18:27 pm
For no solution det=0
(m-2)(m-3)-2*3=0
Simplify to get m= 0 and m = 5
Post by: luken93 on February 24, 2011, 07:26:26 pm
A. R \ {0,5}
B. R \ {0}
C. R\ {6}
D. m = 5
E. m = 0
I would've said A, but apparently not!
Post by: francis8nho on February 24, 2011, 07:40:09 pm
I just tried again using CAS for m=5, x=2-y
for m=0, yes there is no solutions
Post by: sajib_mostofa on February 24, 2011, 07:47:07 pm
Post by: francis8nho on February 24, 2011, 07:52:32 pm
Post by: sajib_mostofa on February 24, 2011, 08:06:18 pm
Post by: luken93 on February 24, 2011, 08:21:21 pm
You get no solutions when both lines have different equations but the same gradient. So sub in the value for m that fits that description and you should get the answer, which is m=0 in this case.So what was your actual method there? I rearranged them to make a linear y = equation, and then let both the gradients equal eachother which gave m = 0 and 5?
Post by: sajib_mostofa on February 24, 2011, 08:31:05 pm
By rearranging the equations to make y the subject, you can see that both equations have the same gradients but different equations, so they are parrallel lines. Hence they give no solution. However, if we sub in
When simplified, these two equations are essentially indicate the same line, hence there is infinite no. of solutions. Therefore
Post by: luken93 on February 24, 2011, 08:35:08 pm
Post by: sajib_mostofa on February 24, 2011, 08:39:03 pm
Aaahh, thanks for that
Whenever you get these type of questions where you have to determine unique, infinite, no solutions etc, the important thing is to always verify your answers or its very easy to make a mistake. Unless you only get the one answer of course.
Post by: luken93 on April 14, 2011, 03:56:36 pm
Looking for a nice solution to this, as any of mine get very messy :P
Let
Express in modulus - Argument form:
Post by: brightsky on April 14, 2011, 04:36:55 pm
For argument, the complex number is in the 4th quadrant.
So tan(x) = (1/cos(t))/(1/sin(t)) = sin(t)/cos(t) = tan(t), pi/2 < t < pi
x = arctan(tan(t)) = t, pi/2 < t < pi
But since x is in the 4th quadrant, then we have 0 < x < -pi/2.
Post by: luken93 on April 14, 2011, 04:57:16 pm
Post by: Andiio on April 14, 2011, 09:25:51 pm
Post by: luken93 on April 14, 2011, 09:59:33 pm
Is modulus-argument form just polar form? O_OYeah, |z|cis(Arg)
Have a go if you want, the answer is retarded.
Post by: evaever on April 14, 2011, 10:35:43 pm
Modulus is trivial.
For argument, the complex number is in the 4th quadrant.
So tan(x) = (1/cos(t))/(1/sin(t)) = sin(t)/cos(t) = tan(t), pi/2 < t < pi
x = arctan(tan(t)) = t, pi/2 < t < pi
But since x is in the 4th quadrant, then we have 0 < x < -pi/2.
is 0 < neg?
is the answer -2cosec(2x) cis(x-pi)?
Post by: luken93 on April 14, 2011, 10:45:31 pm
Post by: evaever on April 14, 2011, 10:53:45 pm
Post by: luken93 on April 14, 2011, 11:22:46 pm
------
That was my working to get to the answer above (apparently the correct answer), is yours any quicker/simpler?
Post by: evaever on April 15, 2011, 12:20:09 am
The answer works out to be:
------
That was my working to get to the answer above (apparently the correct answer), is yours any quicker/simpler?
Post by: luken93 on April 15, 2011, 09:05:49 am
Also, if you start with addressing the domain first, then wouldn't you change the 1/sin, because the Re(z) is -ve in the 2nd quadrant. Even though, it'd get cancelled out when squaring it.
Post by: brightsky on April 15, 2011, 10:28:49 am
But since
So
So
Post by: evaever on April 15, 2011, 10:49:10 am
So how'd you get to the (theta - pi) then? I understand what you mean, but that was the only way I could get to it...you have confused arg of z with
Also, if you start with addressing the domain first, then wouldn't you change the 1/sin, because the Re(z) is -ve in the 2nd quadrant. Even though, it'd get cancelled out when squaring it.
Post by: luken93 on April 15, 2011, 12:44:41 pm
Ok, that's making more sense, but I feel I'm missing something fundamental here...
But since, then
Soso
needs a negative in front of it, which means:
Soand
For the step I bolded above, I understand that cosec(x) is negative for pi/2 -> pi, but why do you NEED to put a negative in front of the cis because of it? Am I thinking right that the negative will make the point be in the forth quadrant rather than the second, hence putting a negative cis will return it back to the 2nd quadrant? I only made this assumption through working through some problems, I haven't actually seen a written form that explains how/why this happens?
Post by: brightsky on April 15, 2011, 02:22:42 pm
So z = -2csc(2t) * -cis(t) = 2csc(2t)cis(t)
Using this method, you don't need to worry about quadrants. It's all methodical.
Post by: luken93 on May 03, 2011, 06:52:20 pm
So we had our SAC the other day, and this is still bugging me. It was an Analysis SAC on electrical circuits, and we were given the voltage to be:
V(t) = 4sin(wt)
and the Current given was along the lines of:
I(t) = 3.5sin(wt - pi)
From there, we were given that Power = Voltage x Current
Finally, it was a show that question which asked us to find the power using the 2 functions above. However, I feel that I musty have read it wrong, as the "show that" equalled something along the lines of;
P(t) = 7sin(2wt - pi) - 7
Now please, someone give me the satisfaction that I've read it incorrectly, because as far as I know you can't multiply the two together as is, but furthermore a vertical translation?
Post by: Mao on May 04, 2011, 02:38:32 am
Post by: luken93 on May 04, 2011, 07:25:37 am
Post by: Mao on May 04, 2011, 09:25:11 am
Or even better, notice that sin(wt-pi)=-sin(wt), so P=-14sin2(wt), use double angle formula of cos to get it to the required form.
Post by: luken93 on May 04, 2011, 11:36:47 am
Post by: luken93 on May 20, 2011, 07:14:07 pm
My method was as follows:
OR
But both seem very tiresome methods :P
Any ideas?
Post by: syn14 on May 20, 2011, 07:27:56 pm
Let u = e^x
du/dx = e^x
Integrand becomes u/(u+1) as you divide du/dx in substitution to get e^x in the numerator which becomes u and the denominator becomes u+1 from direct substitution of u. Integrate u/(u+1) which is u–ln|u+1| which is e^x – ln |e^x+1|.
Post by: yawho on May 20, 2011, 07:53:15 pm
du/dx = e^x
int[2,e+1] (u+1)/u du = int[2,e+1] (1+1/u) du = [u+lnu] 2,e+1 = e+1+ln(e+1) - 2-ln2 =e-1+ln((e+1)/2)
Post by: luken93 on June 23, 2011, 04:56:06 pm
I was doing the 2010 VCAA Methods Exam, and another one of these "Find the minimum number of days/hours needed to get the Pr(X) > 0.9"
I understand the theory, that's not the problem. My problem is, how do you do it on the calc?
If I know that Pr(X = 0) + Pr(X = 1) <0.1, and the Pr(Success) = 0.2, how do I use solve on the calc for binomial distribution?
I did solve(0.8^n + nCr(n,1) x (0.2) x (0.8)^(n-1) < 0.1, n) on my Ti89, however this doesn't yield anything.
Also tried solve(BinomCdf(n, 0.2, 0, 1) < 0.1, n) but this doesn't give anything either.
Am I inputting it in the right way? I haven't been taught this in class, hence my question.
Post by: Mao on June 25, 2011, 01:31:19 am
Try entering that expression into the y-editor, and check out the 'table' functionality (around F5 iirc). This generates a list of values for various n, which you can manually check to find when it first drops below 0.1.
Post by: luken93 on July 14, 2011, 12:07:03 pm
weight of the body to be 2.5 kg wt. What would be the reading if the lift was:
a) at rest?
b) accelerating upwards at 2 m/s^2?
I kinda managed to get a), but I'm lost as to what to use for b)...
If someone would be as kind to do a diagram as well that'd be great, my physics is pretty rusty haha
Post by: moekamo on July 14, 2011, 12:26:14 pm
Now in part b, in the attached pic, by adding the forces and equating to the net force, we get N-mg=ma=2m, so N=m(g+2) = 32.804N = 3.35 kg wt. This is what the scale would read.
also i don't know why the question is using kg wt, ive never seen it in VCAA exams, the SI unit is newtons and you should probably stick with that.
Post by: luken93 on July 14, 2011, 12:39:25 pm
As for the units, this is from essentials so who knows, but the majority are in this kg wt form...
And just to check, mass is always going to be constant, it's just the acceleration that changes the apparent weight, where weight = mg yes?
And kg wt * 9.8 = N?
Post by: xZero on July 14, 2011, 12:48:57 pm
Post by: luken93 on July 14, 2011, 01:14:16 pm
have you seen the net force formulae f=ma? If you convert this formulae to units it gives N = ms^-2 * kg wt, since 9.8 is acceleration due to gravity, kg wt * 9.8 = NHaha yeah I know F=ma, just checking I was doing everything right :P
Post by: abeybaby on July 15, 2011, 12:42:37 am
Is modulus-argument form just polar form? O_O
Yes it is:
r.cis(theta) is polar form, where r is the modulus and theta is the argument, hence, modulus-argument form
Post by: luken93 on August 04, 2011, 06:56:25 pm
Cheers
Post by: tony3272 on August 04, 2011, 07:09:57 pm
Post by: moekamo on August 04, 2011, 07:13:12 pm
There is a reaction force from the bottom block acting on the upper block(R1), so by newtons 3rd law a force of the same magnitude acts on the lower block going down, also R1
Finally we have the reaction force from the ground acting on the lower block acting upwards, which is R2.
Therefore diagram C is correct with R1, mg, Mg acting down and R2 acting upwards.
Post by: luken93 on August 04, 2011, 07:33:46 pm
I, my teacher, and moekamo said C. Incorrect.
Checkpoints said E. Incorrect.
tony, you are correct. Please explain!
Post by: b^3 on August 04, 2011, 07:39:29 pm
Does that make any sense?
Post by: tony3272 on August 04, 2011, 07:45:37 pm
That's basically how i saw it, it that makes any sense.
Post by: luken93 on August 04, 2011, 09:24:34 pm
Post by: tony3272 on August 04, 2011, 09:32:59 pm