Halp with math1001 sample quiz!!! (1 Viewer)

[Riproot]

Wait, how do you find the modulus of (cos(pi/6) - isin(pi/6))?

Well, you should really take a negative out of the cos because cos(-theta) = cos(theta)
and a negative into sin: -sin(theta) = sin(-theta)

So it becomes cos(-pi/6) + isin(-pi/6) = cis(-pi/6)
And therefore the modulus is 1 because there is no coefficient before the cis.

 

[LookingAtYOU]

Got it.
Love your clear explanations

Accpt my FB request.

 

[Riproot]

I haven't done a Q like 6 in AGES so I'll get back to you on dat later.

 

[LookingAtYOU]

Can you halp me with Q11?
I just wish that they supplied the working out for these Q's too..
Uni is so troublesome >.<

 

[Riproot]

Q8

The top line says "z^3 = 1/i"

 

[Riproot]

Can you halp me with Q11?
I just wish that they supplied the working out for these Q's too..
Uni is so troublesome >.<

I'm not really familiar with how to do 11 because I don't really follow the lectures and kind of have no idea what is going on in uni maffs and we didn't learn that notation in extension two.

 

[LookingAtYOU]

Thanks rip!

 

[kaz1]

Can you halp me with Q11?
I just wish that they supplied the working out for these Q's too..
Uni is so troublesome >.<

most of these q's are pretty standard m8, you should be able to get them by looking at a few examples in lecture notes or something

for that Q you just differentiate in terms of x while leaving y constant and the sub the value given in the partial derivative you found

 

[LookingAtYOU]

^ can you show me the working out so i can follow kaz1?

Thanks dude!

 

[kaz1]

f(x,y)=xcosy*e^x
using product rule
f_x=cosy*e^x+xcosy*e^x
f_x(1,0)=1*e+1*1*e
f_x(1,0)=2e

 

[LookingAtYOU]

thanks kaz1

 

[powlmao]

This looks so hard :S

 

[zxreth]

This looks so hard :S

sorry if this is harsh, but it's probably cause you're only doing 2u maths. most 4u people would know this stuff

 

[Obvious]

How'd everybody find the quiz? I heard a shitload of people complaining about how difficult it was afterwards.