VCE Specialist 3/4 Question Thread!

[veryJasonn]

thank you! so much!

i have another one:
please refer to the attachment

So first of all, disregard the fact that the dy/dx and the equation is all on side - I think that's to throw you off ^.^

Okay, so the question is basically asking you to find the gradient. If it cuts the x-axis at "a" then the point would be (a,0). Then simply apply rise of run from the point (x,y), so it would be y/(x-a). Then you move it to the left (since that's basically what there solutions have done) and then it would be dy/dx - y/(x-a) = 0

So C (if I'm not wrong) 

[devilsadvocate]

So first of all, disregard the fact that the dy/dx and the equation is all on side - I think that's to throw you off ^.^

Okay, so the question is basically asking you to find the gradient. If it cuts the x-axis at "a" then the point would be (a,0). Then simply apply rise of run from the point (x,y), so it would be y/(x-a). Then you move it to the left (since that's basically what there solutions have done) and then it would be dy/dx - y/(x-a) = 0

So C (if I'm not wrong) 

Sorry to butt in haha but it says normal... so -dx/dy = y/(x-a) and I'm guessing it's B?

[speedy]

Wait what? Are you always supposed to mark POI with crosses? I always just write the coordinates down and don't lose marks in SACs?

Same.

[devilsadvocate]

I thought of that as well! But isn't point P referring to the curve? So dy/dx = the gradient of the tangent?

Yeah dy/dx is in terms of the curve, and since the line that connects P(x,y) and (a,0) is the normal to it, the correct way to set up the differential equation would be normal = gradient between P and (a,0); P(x,y) on its own doesn't hold much significance, just a "point" on the curve

[M_BONG]

Yeah dy/dx is in terms of the curve, and since the line that connects P(x,y) and (a,0) is the normal to it, the correct way to set up the differential equation would be normal = gradient between P and (a,0); P(x,y) on its own doesn't hold much significance, just a "point" on the curve

Yup, that clarifies it! Cheers!

[magneto]

thank you everyone!

I have another one i thought i was correct but the ans said E?
where did their P go?

[devilsadvocate]

thank you everyone!

I have another one i thought i was correct but the ans said E?
where did their P go?

The question implies that P doesn't necessarily have to be present; and if you have an object on its own on a rough plane with no force pulling it up, you can have friction acting upwards to stop it from moving down, but you can't have friction acting downwards to stop it from moving up, as there's no force pulling it up. (Remember that gravity always pulls the object down the plane) Hence E would be the correct option

[magneto]

but if P is present as indicated by C shouldn't the F be in the other direction?

[devilsadvocate]

but if P is present as indicated by C shouldn't the F be in the other direction?

In that case P and F would both be acting against the mgsin(theta) force to stop it from moving down; that scenario is only possible when the object is on the point of moving down (comes in handy when you need to find min/max values of theta/coeff. of friction)

[magneto]

thank you!

another one? im so sorry...

why not A? but E again?

[devilsadvocate]

thank you!

another one? im so sorry...

why not A? but E again?

(my god that's the most disgusting question I've seen all year)

I checked the examiner's report and it says the answer is D... And I guess that makes sense.
There are three forces acting on the larger mass in total:
- Mg (weight force)
- R2 (normal reaction between ground and larger mass)
- R1 (normal reaction between larger and smaller mass)

Because R1 is acting upwards on the smaller mass, it's acting towards the ground on the larger mass (action/reaction) hence D would be the correct option...? (not sure tbh, someone feel free to correct me)

[Mieow]

How am I supposed to find the area of the shaded region? It's question 2 from 2013

[Soon]

How am I supposed to find the area of the shaded region? It's question 2 from 2013

I find it handy to put in my bound reference that 
or you could alternatively find the area of a quarter of that circle and then subtract the triangle from it

[Zues]

this question is annoying me, its about reciprocal graphs

usually if i've been given a function f(x) the turning point of this parabola say it is (2,1) then the turning point of g(x) its reciprocol should be (1/2,1) ?

why is it with this that it is the same

the function is f(x) = 1/ (x^2+4x+3)

[M_BONG]

this question is annoying me, its about reciprocal graphs

usually if i've been given a function f(x) the turning point of this parabola say it is (2,1) then the turning point of g(x) its reciprocol should be (1/2,1) ?

why is it with this that it is the same

the function is f(x) = 1/ (x^2+4x+3)

When you reciprocate something, you flip it right? As such, the output value - the y value - not the input (x-value) should be reciprocated.

So you actually flip the y-coordinate not the x-coordinate. The x-coordinate stays the same.

Tell me if you're still confused.