luken93's Methods/Spesh Question thread...

[luken93]

Yeah it turned out to be assessing double angle formulas, was fairly easy after that. Thanks anyway Mao!

[luken93]

Is there any "quick" way of doing this question? It's not that hard, but my teacher is sure that there is an easier way haha



My method was as follows:



OR



But both seem very tiresome methods

Any ideas?

[syn14]

Sorry if this is hard to follow.

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|.

[yawho]

Let u = e^x+1

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)

[luken93]

Hey again guys;

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.^(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.

[Mao]

Your problem is that the solve function requires everything to be a floating point number. The solve function would be substituting n=1.578290394 and crazy shit like that into your equation, which would choke because it requires n to be an integer. (nCr is only defined for integer values of n).

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.

[luken93]

In a lift that is accelerating downwards at 1 m/s^2, a spring balance shows the apparent
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

[moekamo]

so for a i got the mass to be 2.78 kg and the normal force if at rest to be 2.78 kg wt or 27.28 N.

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.

[luken93]

Aha so you do use the answer from part a), that makes more sense.

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?

[xZero]

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 = N

[luken93]

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 = N

Haha yeah I know F=ma, just checking I was doing everything right

[abeybaby]

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

[luken93]

Can someone explain this question to me? I thought I had it covered, but I asked my physics/spesh teacher and he got the wrong answer...

Cheers

[tony3272]

Is the answer D?

[moekamo]

ok so obviously we have gravity on the larger mass, Mg acting down. we also have the force due to gravity from the upper mass acting down, which is mg.

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.