Mathematics Question Thread

[pokemonlv10]

Referring to my above post, this is the actual image, i posted the wrong one sorry! It goes from f'(x) to f(x) and i was trying to sketch the original function.

[RuiAce]

C and D both look appropriate for that one. That is one weird question.

[pokemonlv10]

Ah alright then. Also, what would you draw when x < -4 then? When drawing a graph from f'(x) to f(x), what would you do when its a turning point on the x axis? (for f'(x))


EDIT: Thank you for the response below!

[RuiAce]

The graph would just keep shooting downwards.

Your sketch implied that there would be a horizontal asymptote. But actually the graph can just keep going on without a problem.

[jaskirat]

Need help with questions iii and iv. checked the answers but still a little confused :/

[RuiAce]

Need help with questions iii and iv. checked the answers but still a little confused :/

Here is a breakdown of the BoS answers.

[bsdfjnlkasn]

Hey there,

Am so confused about the first part of the last (ii)...
Would really appreciate an explanation

[gilliesb18]

Hello...
Just wondering if someone could help me on the following question:
Find the equation of the locus of a point that moves so that it's distance from the line 3x + 4y +5= 0 is always 4 units.
I hardly know where to start so I would appreciate any help!!

Thanks heaps and heaps...

[angelahchan]

Hi, could someone please help explain part (ii)? I don't understand why C is 0. The answers say when t=0, v=0 therefore C=0, but doesn't part (iii) contradict with (ii) since (iii) seems to say when t=0, v=1500?

[img]http://i.imgur.com/tLOFDnR.png[/img]

[Natasha.97]

Hello...
Just wondering if someone could help me on the following question:
Find the equation of the locus of a point that moves so that its distance from the line 3x + 4y +5= 0 is always 4 units.
I hardly know where to start so I would appreciate any help!!

Thanks heaps and heaps...

Hi!

- Draw a diagram to help visualise, using the condition given in the question (If it is always 4 units away from the line, that must mean that the locus is parallel to the equation: the perpendicular distance between the equation and the locus is 4)
- Substitute the values of the equation and an arbitrary point P(x,y)  into the perpendicular distance formula (there should be two equations for the locus: one above and one below 3x+4y+5=0)

I have attached my working below, hope this helps

[sidzeman]

Hey could someone help me with the final question from the 2002 hsc?
I'm not sure how to do part 1 and 3
For part ii, i got the Di/Dx as positive on both sides of x = 0, so I'm not sure what I'm doing wrong.

[jamonwindeyer]

Hey there,

Am so confused about the first part of the last (ii)...
Would really appreciate an explanation

So we need to spot that this situation models the volume we found previously. When \(\theta=0\), the volume would have been a full hemisphere (tilted). The height of the water is actually \(AB\), where the original depth is \(OB\). You need to find the value of \(\theta\) that sets \(A\) halfway between \(O\) and \(B\). That is, \(OA=\frac{r}{2}\). You should be able to do some right angled trigonometry to find the angle:

[jamonwindeyer]

Hi, could someone please help explain part (ii)? I don't understand why C is 0. The answers say when t=0, v=0 therefore C=0, but doesn't part (iii) contradict with (ii) since (iii) seems to say when t=0, v=1500?

[img]http://i.imgur.com/tLOFDnR.png[/img]

\(C=0\) because we don't care about the initial quantity in Part (ii)! We only want how much has flowed in, so the initial amount is irrelevant to us. So, we can effectively disregard the constant!

[Mary_a]

Okay, so I thought averaging 62 in all the past papers I did was bad, but I just sat the 2016 paper and I got 42.

No matter how much work I'm doing, it's not paying off!

Where do I go from here?

[jaskirat]

need help with this question