VCE Specialist 3/4 Question Thread!

[Aelru]

V = 1/2*2*2*3 - 1/2*(2-h)*(2-h)*3= 6 - 3/2*(2-h)^2 [hint: use similar triangles]

Ah.. turns out I calculated how much 'empty space' was left (white bit), not realising that I haven't completed the equation by minus-ing the total possible volume by the 'empty space' to obtain the volume of water at any time.

SOOO, I got the '1/2*(2-h)*(2-h)*3', neglecting the entire volume. *slaps myself*

How do you guys know how to solve it so easily? <common question i believe

[saba.ay]

Question 14, Chapter  7 Review Maths Quest:

Two compounds A and B are put together in a solution where they react to form a compound X. To form compound X, twice as much compound A is needed as compound B. The rate at which X is formed is proportional to the unused amounts of A and B present at time t. If initially 5g of A and 8g of B are put together, set up a differential equation for the amount of X present at time t, assuming no X is present initially.

How would I write the equation for dx/dt?
Please and thank you.

[kamil9876]

so let denote the amount of present.

So the amount of used up is and the amount of used up is . So we know that (assuming conservation of mass ). Then I'm assuming for some constant is what is meant by "proportional to amounts of A and B present".

We also know and so we have enough equations to eliminate the and and hence get the differential equation.

[saba.ay]

so let denote the amount of present.

So the amount of used up is and the amount of used up is . So we know that (assuming conservation of mass ). Then I'm assuming for some constant is what is meant by "proportional to amounts of A and B present".

We also know and so we have enough equations to eliminate the and and hence get the differential equation.

Thank you so much.

[Aelru]

[SocialRhubarb]

Let's draw up a right angled triangle.
The hypotenuse can be H, the horizontal distance between the wharf and the boat can be x, and the vertical distance is given to us in the question: it is 5 metres.

Using Pythagoras' theorem:



    We take the positive square root because H is a distance and must therefore be positive.

[Aelru]

Let's draw up a right angled triangle.
The hypotenuse can be H, the horizontal distance between the wharf and the boat can be x, and the vertical distance is given to us in the question: it is 5 metres.

Using Pythagoras' theorem:



    We take the positive square root because H is a distance and must therefore be positive.

Very well explained. Thanks .
I got to the triangle, but used angles instead,totally forgetting about the use of pythagoras *facepalm*

[SocialRhubarb]

It is possible to do it with angles.

Spoiler

If we have a right angled triangle with H as the hypotenuse, x as the horizontal edge, and a vertical side length of 5, with the angle theta being in between the H and the vertical side.





, using quotient rule and chain rule because theta is a function of x.










 

Suffice to say, the Pythagoras method is simpler.

[Homer]

{z:|z-2i|=1} find the exact maximum and minimum values of Arg(z)

Not sure what to do?

[brightsky]

construct the two tangent lines to the given circle which pass through the origin. also construct line segments from the two points of tangency to the centre of the circle. now we have two right angled triangles, with side lengths in a ratio of 1:2:sqrt(5). all we need to do now is find the angle that each of the tangents makes with the positive x-axis. this can be accomplished using trig.

minimum Arg(z) = arctan(2/1) = arctan(2)
maximum Arg(z) = pi - arctan(2)

[Homer]

ans: min pi/3 max 2pi/3

[brightsky]

woops drew the wrong triangle in my head. the method's there though.

correction: the side lengths of the two triangles are in a 1:sqrt(3):2 ratio. so min Arg z = pi/2 - arcsin(opp/hyp) = pi/2 - arcsin(1/2) = pi/2 - pi/6 = pi/3. max Arg z = pi/2 + arcsin(opp/hyp) = pi/2 + pi/6 = 2pi/3.

sorry about that.

[Homer]

thats good as, thanks brightsky

[UoM]

Will we ever encounter questions regarding volumes being rotated around a line apart from the x/y axis? e.g. Rotated around x=2 or something?

[pi]

In SACs possibly. I had a question about a rotation around an asymptote (that wasn't lying on an axis)