Mathematics Question Thread

[Shadowxo]

I just have a clarification question here:

(For iii.)

Because I get two answers, is it safe to just ignore t=75 because of the fact that it takes him t=50 to empty the contents thus I can assume that t=75 is inadmissible even if they didn't supply a domain in the Q. ?(Image removed from quote.)

Correct, the implied domain is 0≤t≤50 as this is the time he's drinking the water and the volume is decreasing
Also, in questions like this you usually take the earliest answer as that's when it first occurs

[georgiia]

Correct, the implied domain is 0≤t≤50 as this is the time he's drinking the water and the volume is decreasing
Also, in questions like this you usually take the earliest answer as that's when it first occurs

Thanks!

[RuiAce]

Anything still unattended to? Just quickly checking I wasn't sure what wasn't addressed but I'll be able to handle a few things soon - just not sure if I missed any

[Shadowxo]

Anything still unattended to? Just quickly checking I wasn't sure what wasn't addressed but I'll be able to handle a few things soon - just not sure if I missed any

Think everything's been answered

[RuiAce]

Think everything's been answered

Awesome. Even the locus one? - that one appeared in my head just now

[Shadowxo]

Hey just wanted to confirm a few things, because my teacher hasnt been very clear about this topic. How do you know when to use logs in an equation ? (besides the questions where it is obviously asking about logs?)

Also just in addition to what georgiia said,
You use logs when what you want to find is in the power.
a^b=c
log_ac=b
You can use this for any power and any base, doesn't have to be e (but for this a>0 and c>0)

Awesome. Even the locus one? - that one appeared in my head just now

Everything but that question I mean forgot about it too but I'm not familiar with solving those kinds of problems, I'll leave that to you

[RuiAce]

I was able to do that part, but theres a 2nd part to the Q

I ended up finding something (in what little time I ended up having) but it involves a 3U circle geometry theorem. Before I jump into it, what was the source of this question?

[winstondarmawan]

I ended up finding something (in what little time I ended up having) but it involves a 3U circle geometry theorem. Before I jump into it, what was the source of this question?

Fitzpatrick Textbook, not sure if 2U or 3U, because my friend sent me that photo, sorry.
I just assumed it was 2U because it was a locus Q.

[Youssk]

Hello, could I please have some help with this question.

The acceleration of a moving body is given by a=√2t+1 ms^-2. If the body starts from rest, find its velocity after 4 seconds.

[MisterNeo]

Hello, could I please have some help with this question.

The acceleration of a moving body is given by a=√2t+1 ms^-2. If the body starts from rest, find its velocity after 4 seconds.

Hi! For this question, you would integrate the acceleration to find the velocity function with a +C.
It says that the body starts from rest, so velocity is 0 when t=0. (You have to solve for C)
Then sub in t=4 into the full velocity function to find the velocity after 4 seconds!

[RuiAce]

Hello! Would appreciate help with Q3:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20270047_1538514876198918_1314381077_n.jpg?oh=3ce907b341fda76bc69f71d6d59fe730&oe=5975F432
TIA

[georgiia]

Can someone please walk me through drawing the primitive function? I haven't done these for ages and Im ok drawing the gradient function but i can remember what turns into what for primitive


Thanks!

[Shadowxo]

Can someone please walk me through drawing the primitive function? I haven't done these for ages and Im ok drawing the gradient function but i can remember what turns into what for primitive


Thanks!

Hi
So, while I haven't done primitive functions, I have done anti-differentiation / integration. So sorry if my terminology is wrong or I don't address everything etc
Where y=0 on the graph / original function, the gradient of the primitive function is 0 (basically opposite of finding the gradient function), which means either a turning point or a point of inflection. As on the graph there is no turning point at y=0 x=2, (just goes straight through 0, one side is negative one side is positive), this means x=2 on the primitive function must be a turning point, and it's a local minimum as the original function goes from negative to positive.
Where there's a turning point on the original function, there is a point of inflection on the primitive function. In this case the original function becomes more positive, then more negative, then more positive again, meaning the gradient on the primitive function will do the same.
So for this primitive function, it will start off high and be decreasing, have a couple points of inflections approx x=-1 and x=1 (not a stationary point of inflection though, and it will continue decreasing this whole time as the original graph has y as negative until x=2). Then, at x=2 there will be a turning point and it will go upwards.

Hope this helps, let me know if you're still having trouble with it or if you'd like a diagram 

[georgiia]

Hi
So, while I haven't done primitive functions, I have done anti-differentiation / integration. So sorry if my terminology is wrong or I don't address everything etc
Where y=0 on the graph / original function, the gradient of the primitive function is 0 (basically opposite of finding the gradient function), which means either a turning point or a point of inflection. As on the graph there is no turning point at y=0 x=2, (just goes straight through 0, one side is negative one side is positive), this means x=2 on the primitive function must be a turning point, and it's a local minimum as the original function goes from negative to positive.
Where there's a turning point on the original function, there is a point of inflection on the primitive function. In this case the original function becomes more positive, then more negative, then more positive again, meaning the gradient on the primitive function will do the same.
So for this primitive function, it will start off high and be decreasing, have a couple points of inflections approx x=-1 and x=1 (not a stationary point of inflection though, and it will continue decreasing this whole time as the original graph has y as negative until x=2). Then, at x=2 there will be a turning point and it will go upwards.

Hope this helps, let me know if you're still having trouble with it or if you'd like a diagram 

Thank You!! I just needed clarification as to what turns into what when you integrate and you've refreshed my memory now, Thanks!!

[Shadowxo]

Thank You!! I just needed clarification as to what turns into what when you integrate and you've refreshed my memory now, Thanks!!

No problem