VCE Specialist 3/4 Question Thread!

[soccerboi]

How do you antidiff xtan^-1x? Would something like this be in Exam 1 or 2?
Thanks

[paulsterio]

How do you antidiff xtan^-1x? Would something like this be in Exam 1 or 2?
Thanks

I can do it with integration by parts, but it's not on the course. So you won't have to do something like this for VCE.

[soccerboi]

How do you antidiff xtan^-1x? Would something like this be in Exam 1 or 2?
Thanks

I can do it with integration by parts, but it's not on the course. So you won't have to do something like this for VCE.

ok thanks, so i should just chuck it in my CAS?

What if it was diff xtan^-1x?

[monkeywantsabanana]

How do you antidiff xtan^-1x? Would something like this be in Exam 1 or 2?
Thanks

I can do it with integration by parts, but it's not on the course. So you won't have to do something like this for VCE.

ok thanks, so i should just chuck it in my CAS?

What if it was diff xtan^-1x?

You can do that by using product rule

[soccerboi]

oh yeh! thanks

[soccerboi]

How do i show that g'(x)>0 for >0 if g'(x)=(2tan^-1x)/(1+x²)? Can someone describe what to do or show the steps? Thanks:)

[brightsky]

look at the numerator and denominator separately. recall the graph of arctan(x). for x>0, arctan(x) > 0 from the graph. it is obvious that 1+x^2>0 for x>0 (in fact for all real x). since the numerator and denominator are both positive for positive x, then the whoe function is positive for x>0.

[soccerboi]

A bowl is modeled by rotating a curve y=x² for 0<x<1 around the y axis.

part b) If liquid is poured into a bowl at rate of R units of volume per second, find the rate of increases of the depth of liquid in the bowl when the depth is 1/4.

I dono how to start it.

[brightsky]

dy/dt = dy/dV * dV/dt
that should get you started.

[soccerboi]

When asked to factorise say P(z)=z³-5z²+8z-6 and i know z-3 is a factor, how do you find the quadratic bit of it without doing long division. I remember there was a way by just inspection, something like looking at the last term of the original p(z) but i forgot it. Can someone explain it to me?

Thanks

[Phy124]

When asked to factorise say P(z)=z³-5z²+8z-6 and i know z-3 is a factor, how do you find the quadratic bit of it without doing long division. I remember there was a way by just inspection, something like looking at the last term of the original p(z) but i forgot it. Can someone explain it to me?

Thanks

Not sure, but you might be thinking about how the three terms in brackets need to multiply together to make the last term.

(say you had (x-3)(x-2)(x+1) you know the last term of the expanded polynomial will be 6 as -3 x -2 x 1 = 6)

Meaning the remaining two must multiple to make 2 (as we already have -3 and 2 x -3 = -6).

So something like and (as ) could work, and does.

edit: Oh I misinterpreted your question, my apologies.

You have;



You know is a factor



We know that the coefficient of the term is , which is calculated by therefore



We also know that the last term in the expanded polynomial is , which will be - , thus



Lastly the coefficient of the is , which is calculated by the addition of and , therefore and hence;



Giving our desired roots of

[soccerboi]

I'm having trouble getting the middle term though, any ideas on how to do it?

[anthony1n]

so P(Z) = z³-5z²+8z-6
So you know (z-3) is a factor, the first term is z³ which means you have to mutiply z by z² to get the first term.
Similarly you can do that to get that last term of 2
which leaves you with
(z-3)(z² + ____ +2)
Ok so to get the middle term pick up either of the middle terms lets say (8z) in this example. So only two of the possible terms can be multiplied to give 8z, the first being z x 2 = 2z
The other term being the (-3) so: 2z - 3 = 8z => -3 = 6z , z=-2 So the middle term is -2z.
Alternatively since 2z + x = 8z , x=6z
What can (-3) be multiplied to, to give an answer of 6z, -3 * -2z
Sorry its kinda long and was also hard to explain online but this is what I do to solve the middle term

[Phy124]

A flexible beam is supported at its ends, which are the same horizontal level and at a distance L apart. The deflection y of the beam, measured downwards from the horizontal, satisfies the differential equation: 16 d^2y/dx^2 = L-3x, 0≤x≤L
The deflection of the beam and its inclination to the horizontal are both zero at x=0
Use calculus to find an expression for y in terms if x.

Is deflection d^2y/dx^2 and inclination dy/dx?

Thanks

Yep, is the deflection and is the gradient of the beam.

[|ll|lll|]

A flexible beam is supported at its ends, which are the same horizontal level and at a distance L apart. The deflection y of the beam, measured downwards from the horizontal, satisfies the differential equation: 16 d^2y/dx^2 = L-3x, 0≤x≤L
The deflection of the beam and its inclination to the horizontal are both zero at x=0
Use calculus to find an expression for y in terms if x.

Is deflection d^2y/dx^2 and inclination dy/dx?

Thanks

Yep, is the deflection and is the gradient of the beam.

Thanks, a little confused here... how do I solve for a specific solution with the DE and find k?