VCE Specialist 3/4 Question Thread!

[rife168]

I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be instead of ?
That would make sense to me but if not, how do you work out that asymptote?

[pi]

I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be instead of ?
That would make sense to me but if not, how do you work out that asymptote?

Oh shit, you're completely right. is the right one, no idea how i copied it wrong from my notes book.

Sorry if I've just wasted your last couple of hours

[rife168]

I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be instead of ?
That would make sense to me but if not, how do you work out that asymptote?

Oh shit, you're completely right. is the right one, no idea how i copied it wrong from my notes book.

Sorry if I've just wasted your last couple of hours

Haha, it's alright, I now know what a cissoid is so I wouldn't call it a waste...

As for the question...


The volume of the cissoid rotated about x=2a is equivalent to the volume of the cissoid, translated by -2a units in the x direction rotated about the y axis

the new equation is thus

so


I'm a bit stuck now and I'm about to go to bed (2 SACs tomorrow)... do you reckon you could either give me a big hint that will give it away or actually show exactly how to get the answer otherwise I swear I won't be able to sleep haha...

[pi]

It's a very tough question and I think it's beyond the yr12 course, so don't stress on it.

As for a hint, think of it in halves (top and bottom half). Now think of the top half being made of infinite number of cylinders, such that their "diameter" is and hence, their individual "height" is . Using the formula for a cylinder, sub those in and see what happens when you take the sum of those cylinders when approaches 0. ie.

[luke_rulz94]

hey guys,

can someone show me how to do this :




A tank has a volume v cm3 at a depth of h m given by v =  pix4h^3/3 + h. The tank is initially empty but water is pumped in at the rate of 0.03 cubic metres per second and flows out from the bottom of the tank at 0.04x root of h cubic metres per second.

1   Find the rate, in metres per second, at which the depth is increasing when the depth is 0.5 metres, giving your answer correct to two significant figures.

2   Express the time taken for the depth to rise to 0.5 metres as a definite integral and evaluate it correct to the nearest second.

3   At what depth does the water ultimately stabilise in the tank?

[#1procrastinator]

When you are asked to implicitly differentiate something like (2/x)+(1/y)=4 why is it incorrect to get rid o the denominators first ie 2y+x=4xy i know you loset some information regardin thw domaim and range bu t shouldnt it still be similar ouldnt you write down the dom and ran?

[Bhootnike]

hey guys,

can someone show me how to do this :




A tank has a volume v cm3 at a depth of h m given by v =  pix4h^3/3 + h. The tank is initially empty but water is pumped in at the rate of 0.03 cubic metres per second and flows out from the bottom of the tank at 0.04x root of h cubic metres per second.

1   Find the rate, in metres per second, at which the depth is increasing when the depth is 0.5 metres, giving your answer correct to two significant figures.

Is the answer to part a) = 0.0072 m/s ?

[soccerboi]

A cricket ball is thrown upwards from the top of a 10m high building at a speed of 2m/s. Find the exact value of the speed at which the ball strikes the ground.

Solution used u=2 , a= -9.8 , s= -10 and got v=10root2

I dont understand why s=-10? I though that whatever you choose as the upwards direction will be opposite sign to the acceleration. so if a is negative should s be positive?
Thanks

[brightsky]

assuming rectilinear motion, we can define upwards direction as positive and downwards direction as negative (from the point at which you throw the ball). so the top of the building is 0. using these definitions then the bottom of the building is -10. displacement = final position - initial position = -10 - 0 = -10.

[TrueTears]

A cricket ball is thrown upwards from the top of a 10m high building at a speed of 2m/s. Find the exact value of the speed at which the ball strikes the ground.

Solution used u=2 , a= -9.8 , s= -10 and got v=10root2

I dont understand why s=-10? I though that whatever you choose as the upwards direction will be opposite sign to the acceleration. so if a is negative should s be positive?
Thanks

Denote upwards to be positive and downwards to be negative.

u = 2, a = -9.8 and x = -10.

the displacement is negative because the ground is "lower" than the building (assuming we defined the top of the building as our origin), it is clearly a negative displacement.

[soccerboi]

Thanks guys

[Bhootnike]

Hey
got some questions i couldnt get out  -- ONLY 6a and 7a
because i can do the rest once i can do part a)
i thought i was doing it right but yeah, nah.... :p
for q6, a)  i got 12(x-5)^2 m/s^2
which is wrong

then for q7a) i used
and got something like

please suggest some other way to go about it?!

[Lasercookie]

Am I missing something but what are the actual questions Bhootnike?

[Bhootnike]

oops
forgot to attach
/attached in op!

[Lasercookie]

The derivative of velocity is acceleration

EDIT: b^3 just pointed out the rookie mistake I did. Ignore this post.

(we can't use a = dv/dt since we don't have an expression in terms of time).

6a.
and when

is right? Is it possible that the answer was just an expanded out form of it?

7a.

, when (letting x > 0 being to the right of the origin)









?