VCE Specialist 3/4 Question Thread!

[soccerboi]

Oh yes i forgot to change the terminals lol. Thanks

[soccerboi]

Can someone show me how to do this:
Find the solution to the differential eqn: =,given that y=2 when x=0

I keep getting it in the form y=log_e|(2-x)(2+x)|+c but answer says it should be in the form  y=log_e||+c

Thanks

[HenryP]

Can someone show me how to do this:
Find the solution to the differential eqn: =,given that y=2 when x=0

I keep getting it in the form y=log_e|(2-x)(2+x)|+c but answer says it should be in the form  y=log_e||+c

Thanks

Hey, I assume you split the fraction into partial fractions with the denominators being 2-x and 2+x .
I'm pretty sure the mistake you're making is that when integrating you should be getting
That should fix the problem you're having, hope this helps

[soccerboi]

Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?

[HenryP]

Probably best not to oversimplify it, but essentially yes. You can prove it to yourself by doing the substitution u= 2-x

[Phy124]

Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?

Just follow;



and you shouldn't go wrong.

[kensan]

Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?

Just follow;



and you shouldn't go wrong.

That doesn't work with x^2 does it?

[Truck]

Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?

Just follow;



and you shouldn't go wrong.

That doesn't work with x^2 does it?

The integral of 1/x^2 is -1/x, so nope.

[Phy124]

Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?

Just follow;



and you shouldn't go wrong.

That doesn't work with x^2 does it?

I was referring to and his question of what to do with the negative in front of the x.

i.e.

[HenryP]

I believe there is nothing wrong with that derivation. As to why doesn't work is because the formula should only work such the case the power of x is 1

[b^3]

The region is the overlapping region that is in the first quadrant.

So we need to look at rotating around the y-axis, that is we need to look at it in two parts. That is the region b/w y=x^2 and the y-axis for 0 to 1 and the region between y=2-x^2 and the y-axis for y=1 to 2.
Now the formula we have for rotating around the y-axis is
So we need x^2.
So
and
Now to rotate that


Now I did that in a rush, so there probably is a mistake in there somewhere, so if there is someone pick me up on it. Now to watch the Grand Prix....

Anyways hope that helps.

[Jenny_2108]

Hey, I think we can calculate one area and then double it because the upper and below parts are both equal

[Special At Specialist]

That's an interesting question. In order to avoid confusing myself, I'm going to start with a much simplified version of that problem and then work my way upwards: Suppose they only wanted you to calculate the volume of the solid formed when the area between the curve y = 2 - x^2 and the lines x = 0 and x = 1 was rotated around the y-axis.
How would I calculate that volume?

[Mao]

That's an interesting question. In order to avoid confusing myself, I'm going to start with a much simplified version of that problem and then work my way upwards: Suppose they only wanted you to calculate the volume of the solid formed when the area between the curve y = 2 - x^2 and the lines x = 0 and x = 1 was rotated around the y-axis.
How would I calculate that volume?

Break it down into two parts:

The bottom half 0<x<1, 0<y<1, is just a simple cylinder. Volume is thus

The top half can be constructed from a series of thin horizontal disks. Volume of each of these disks are , thus we must integrate over y.

We know that:
- the region we're integrating is from y=1 to y=2 (this is the curved part)
-

Thus,

Total volume is thus

[Bhootnike]

what is the general approach to finding the complex roots of a polynomial if you are given a real root eg. z=a?
if you sub in that factor in the form (z-a) into the original polynomial, is that going to help in any way??