Specialist Maths (Exam 1): Discussion, Questions & Potential Solutions

[TheAspiringDoc]

Jazzycab 2b I got 4’s instead of 8’s. 9c I got pi

[DinWell]

I haven't had a good look at why properly yet, but if you don't take the negative, you'll end up with a negative distance (i.e. evaluate the integral and test)

Hey, for the differential equation, I don't think that's the correct answer.

[hegeorge1908]

Yea i don’t think the DE question is right either. my coefficient for the denominator was 3/2

[DinWell]

Jazzycab 2b I got 4’s instead of 8’s. 9c I got pi

I got the same as jazzycab for those two.

[fur2018]

i think it should be 32 on top for Q8b
so 32/ (2t +16)^3/2

[fur2018]

for Q 4, why can't b be -1/3?

[mzhao]

for Q 4, why can't b be -1/3?

a, b are integers

[fur2018]

a, b are integers

Yeah u are right.

[jazzycab]

Yea i don’t think the DE question is right either. my coefficient for the denominator was 3/2

i think it should be 32 on top for Q8b
so 32/ (2t +16)^3/2

Hey, for the differential equation, I don't think that's the correct answer.

It seems I completely omitted the 3 in the numerator of the DE when attempting to solve (i.e. I solved \(\frac{dQ}{dt}=-\frac{Q}{16+2t}\))
I've checked mine and updated in my answers. mzhao's answer below is correct

[mzhao]

i think it should be 32 on top for Q8b
so 32/ (2t +16)^3/2

Yep I agree with


jazzycab's denominator exponent has to change, so the numerator must also change to satisfy the initial condition of:

[HamConspiracy]

How did everyone find the graph?

[DinWell]

How did everyone find the graph?

Personally, I found it okay. I did heaps of practice on sketching graphs like that at the start of the year so I had a method of doing it. How about you?

[jazzycab]

Jazzycab 2b I got 4’s instead of 8’s. 9c I got pi

I've just checked both with the CAS and mine are correct

[HamConspiracy]

Yeah, it put me off to be begin with since I actually thought addition pf ordinates graphs were off the course. But I just went full paranoid on making sure the scale of each partial fraction and that each part passed through the right intercepts, so it went OK.

[DinWell]

Yeah, it put me off to be begin with since I actually thought addition pf ordinates graphs were off the course. But I just went full paranoid on making sure the scale of each partial fraction and that each part passed through the right intercepts, so it went OK.

You don't need to use addition of ordinates for it. An easy way is to find the axis intersects, all the asymptotes and the regions where it's positive and negative. With all that info, there's only one shape it can really be.