VCE Specialist 3/4 Question Thread!

[eagles]

Hello!

I have a differentiation question that I would really appreciate help with.
Why is that we don't use the chain rule for q.3.ii. so that the derivative is x(ln(x))² is (ln(x))² + (2(ln(x)))/x ?
Find below a screenshot of chapter 8 review question 3 (ER) in the essentials textbook.

Thanks

[keltingmeith]

You DO do that, but remember the product rule. For this case, we'll let u=x and v=(ln(x))^2. From this, we get:

[eagles]

Oh dear, thanks for pointing that out

[alchemy]

Ah that makes much more sense (but damn these polar graphs LOL) T_T
Thanks lzxnl and EulerFan101

Polar graphs aren't even on the Spesh course...why bother?

[eagles]

Hello!

I'm having trouble understanding the third line of these solutions.
Can I please have an explanation for this?

Thank you.

[Zealous]

I'm having trouble understanding the third line of these solutions.
Can I please have an explanation for this?

Difference of two squares:



so therefore:

[IndefatigableLover]

Polar graphs aren't even on the Spesh course...why bother?

Just to understand the fundamentals and to see where everything is coming from I guess rather than learning about "What's going to be on the exam etc." It's just some light work (nothing serious really)

[keltingmeith]

Just to understand the fundamentals and to see where everything is coming from I guess rather than learning about "What's going to be on the exam etc." It's just some light work (nothing serious really)

gurl, dun worry. u do your extra maths. <3 maths, werk it.

[magneto]

i have a surface area of a revolution question:
i know its not apart of the study design but my teacher has decided to teach us for a SAC

if the region enclosed by the eclipse (x^2/a^2) + (y^2/b^2) = 1 is revolved around the x axis, find the resulting area of te resulting football shape solid
NOTE there are two cases: when a>b and b>a

[johnsmith123]

2 questions please:

 Show that dN/dt = (k-c*N)*N can be expressed as N=k/(B*e^(-k*t)-c

And

Patient is given a drug at 200mg /kg of tissue: drug leaves the body at a rate proportional to x ( the amount of drug)
Find dx/dt (where x is amount of drug) basically set up differential equation

Then: if after 30 mins there is 50% of drug remaining solve differential equation

And finally: how long does it take for 10 percent of drug to remain

Thanks

[Zealous]

"A bullet is fired vertically upwards, ignoring any air resistance, at a speed of 98ms^-1 and three seconds later a second bullet is fired vertically upwards from the same point with the same speed".
Find how long it takes after the first bullet is shot that the two bullets meet.
Answer: 11.50 seconds.
Can anyone show me how to get there?

First Bullet:


Second Bullet (this bullet is fired three seconds after, so we use t-3):


We want the bullets to be at the same point, which means that their x value is the same, so we can equate them:

[lzxnl]

Ok not sure what I have done wrong. Question also seems simple enough...

"A bullet is fired vertically upwards, ignoring any air resistance, at a speed of 98ms^-1 and three seconds later a second bullet is fired vertically upwards from the same point with the same speed".

Find how long it takes after the first bullet is shot that the two bullets meet.

Answer: 11.50 seconds.

Can anyone show me how to get there?

Let up be positive. Let t be the time after the first bullet is shot.
Position of the first bullet: ut + 1/2 at^2 = 98t - 4.9t^2
Position of the second bullet: u(t-3) + 1/2 a(t-3)^2 = 98(t-3) - 4.9(t-3)^2
Meet => positions are the same
98(t-3) - 4.9(t-3)^2 = 98t - 4.9t^2
20t - 60 - (t^2 - 6t + 9) = 20t - t^2
-60 + 6t - 9 = 0
6t = 69
t = 11.5 s

[lzxnl]

Thanks for the reply, I realise my mistake, I put (t+3) instead of (t-3). Can anyone explain why it's minus three? Bullet two is fired three seconds after, so i thought it was intuitive that its minus 3?

Think about it. Three seconds after the first bullet is fired, which is t = 3, we have the second bullet fired
Let the time after the firing of the second bullet be u. Then, we know t and u are always a constant apart (time runs at the same rate)
Also, we know that when t = 3, u = 0
Hence, u = t-3

[Thorium]

Thanks for the reply, I realise my mistake, I put (t+3) instead of (t-3). Can anyone explain why it's minus three? Bullet two is fired three seconds after, so i thought it was intuitive that its minus 3?

We can also put it this way:

If you plot the graph of path of the first bullet (x=ut+1/2at^2), it is gonna be a parabola starting at origin (t=0). And the second bullet should have the same shape of path, but it starts at t=3, which means the graph you plotted has been translated 3 units in positive x-axis. So its equation looks like x=u(t-3)+1/2a(t-3)^2.
You might know that the "-3" means that the second bullet doesnt start at the origin, it starts at t=3.

Hope that makes sense:)

[Bestie]

how would i do part c)

please and thanks