VCE Specialist 3/4 Question Thread!

[Eugenet17]

Can i get some help on this question, heres a screenshot:
http://gyazo.com/c69f5c57fa983d6bdcc403d6b782b72d

[b^3]

Can i get some help on this question, heres a screenshot:
http://gyazo.com/c69f5c57fa983d6bdcc403d6b782b72d

This was answered a week or so ago. You need to look at it as rotating two regions around the axis, the first being the area between the curve and the axis over those values, the second being the square that is formed, taking the first away from the second to leave the volume that we're after.

Whoops, my bad, misread the question (this is why you should always do a little sketch before doing vol of revs questions!) In that case you have instead and integrate that. It's the same method with a few adjustments. That's because we're rotating the volume "below" (to the left to the y axis) x=1 and minusing the volume "below" y=tan^-1(x).

[Eugenet17]

'That's because we're rotating the volume "below" (to the left to the y axis) x=1 and minusing the volume "below" y=tan^-1(x).'

I'm not really sure what you mean here, could you clarify what you mean by rotating the volume below x=1?

[b^3]

The reason I have "below" in quotation marks is that it's "below" in the sense if you were looking with the axis being horizontal (hence the "to the left" bit in there above), (so in the image below that'd be the area in the blue rectangle that's unshaded).

So in the image below you want to rotate the blue square around the axis (which is just rotating the line ) and then subtract the volume you get from rotating the area in that blue rectangle that's not shaded. You could think of it as creating the yellow cylinder, then cutting out the vol of rev you get from the non-shaded region.


Hope that helps!

[lzxnl]

Awesome, thanks for showing the two methods.
But I have some questions about them that I was wondering if you can please clarify:

Well, I guessed based on experience. If you want to integrate f(x), can you see that the derivative of x*f(x) = f(x) + x f'(x)? So, in some cases, x f'(x) is easier to integrate. That's what I did in the first method.

Now, look carefully at both sides.
int(u dv/dx dx) = uv - int(v du/dx dx)
On the left hand side, we have u dv/dx to be integrate. On the right hand side, we have v du/dx to be integrated. Basically, on the left you have a product of u and the derivative of v under the integral sign, whereas on the right hand side you have v and the derivative of u under the integral sign. Sure, you're integrating both sides, but you're integrating different things.


Then, remember how we get to choose u and v? Well, when letting u = (ln x)^2 and dv/dx = 1, we can choose v to be anything as long as dv/dx = 1. I've just chosen v = x for simplicity as it does satisfy dv/dx = 1 and the resulting integral is simpler.

[b^3]

Just to point out, Integration by parts while useful, isn't on the spesh course. So for those of you who are looking at it and going wtf, don't stress about it.

[Eugenet17]

thanks for clearing that up b^3, helped alot

[M_BONG]

I'm having a hard time with these


Can anyone please explain the terminals when you are integrating with respect to "y"

In the screenshot, why are the terminals root 3 and 0? Isn't root 3 the POI between the two graphs? Why is that the terminals then?


By the way, the question is trying to find area bound below the curves (ie. the strip part)

[Phy124]

Lets look at it in a way that's more familiar to VCE students:



The area between the two curves and x axis is the integral of the upper curve minus the lower curve from to (this is where they intersect).

[M_BONG]

Lets look at it in a way that's more familiar to VCE students:

(Image removed from quote.)

The area between the two curves and x axis is the integral of the upper curve minus the lower curve from to (this is where they intersect).

Thanks for the reply; but what I was asking is: why are the terminals chosen the point of intersection (root3) and 0.

You didn't answer my question; I was asking why we choose the terminals as the POI, not what the terminals should be.

Is the understanding behind it beyond the scope of VCE?

If we were differentiating with respect to x (dx) we wouldn't choose the POI right? We would choose the endpoints? So why is it here we choose the POI?

Thanks!

[Thorium]

Thanks for the reply; but what I was asking is: why are the terminals chosen the point of intersection (root3) and 0.

You didn't answer my question; I was asking why we choose the terminals as the POI, not what the terminals should be.

Is the understanding behind it beyond the scope of VCE?

If we were differentiating with respect to x (dx) we wouldn't choose the POI right? We would choose the endpoints? So why is it here we choose the POI?

Thanks!

The terminals are those because the question is asking us to find the area bound by the two curves and the x axis. And we know that this us between y=0 and root3. We happen to choose root3 because that is the point that the two curves stop bounding the required region.

We can also have dx in our calculations, but it might be more time consuming. I will try to solve it using dx.

EDIT: i have attached the solution using dx. Here we have to split the region bound into two parts. And apparently this approach is quicker, not time consuming.

[Phy124]

Thanks for the reply; but what I was asking is: why are the terminals chosen the point of intersection (root3) and 0.

You didn't answer my question; I was asking why we choose the terminals as the POI, not what the terminals should be.

Is the understanding behind it beyond the scope of VCE?

If we were differentiating with respect to x (dx) we wouldn't choose the POI right? We would choose the endpoints? So why is it here we choose the POI?

Thanks!

So from the above diagram we have

If you were integrating with respect to x it would be the following instead:

[M_BONG]

(Image removed from quote.)
So from the above diagram we have

If you were integrating with respect to x it would be the following instead:

Thank you so much!

So my teacher (in the attachment) didn't rotate the graph, so it looked like (to me) he was only integrating half the graph with respect to y. But I should have known better and rotated it myself.

Clarified the problem. Thanks heaps!

[Phy124]

Thank you so much!

So my teacher (in the attachment) didn't rotate the graph, so it looked like (to me) he was only integrating half the graph with respect to y. But I should have known better and rotated it myself.

Clarified the problem. Thanks heaps!

No problem!

Just as a note rotation of the graph isn't entirely necessary. I just did it because VCE students are used to integrating with respect to x which is the standard horizontal axis so it's easier to visualise if the graph is redrawn with y being the horizontal axis when integrating with respect to y (Y)

[Sayf44]

how do we use mathematical symbols on this? I need to ask a question using them. thanks