Exam 1 Discussion

[smartypants234]

How did everyone go?


Did anyone else get sqrt(3)pi/2 - (3pi/8) +(3ln(0.5))/4

[BLACKCATT]

Was the angle with positive axis worth 2 marks or 1?

[chuck981996]

Hey guys, how did you all go?

These are my numerical answers... I checked them with my teacher:

1a)

  b)

  c)


2c)


3a)

  b) or


4)


5b)

  c)


6b)


7a)

  c)


8ai)

 aii)

   c)

[Aman476]

How did everyone go?


Did anyone else get sqrt(3)pi/2 - (3pi/8) +(3ln(0.5))/4

I got that

[Aman476]

Hey guys, how did you all go?

These are my numerical answers... I checked them with my teacher:

1a) 1/srt(2)i - 1/sqrt(6)j - 1/sqrt(3)k
b)pi/4
c)m=6+5sqrt(2)

2c)sqrt(5)

3a)z^2 + 1
b) z = 2 +- sqrt(2)i or z = +-i

4) 4/9

5b) 8S(u^2)du, 0<u<sqrt(3)/2
c) sqrt(3)

6b) pi(1 + ln(5/3))

7a) [0, inf)
c) pi(sqrt(3)/6 - 1/8) - 1/2ln(2)

8ai) T1 = 5gsec(theta)
aii) T2 = 5gtan(theta)
c) pi/3

I got that to but i stupidly forgot to flip the gradient in that question so i said -9/4 i wann kms haha


Actually 7c i got the same as smartypants234 ....

[Aman476]

Only thing i couldn't do was prove that sec(theta) was greater than tan(theta) i did it in words, but not mathematically, i just said sec(theta) is an element of [1,inifinity) and tan(theta) was an element of [0, infinity) so since sec starts above 0 and increases faster it is greater than tan (i didn't even prove that it increases faster si just said it haah) had no clue for that, well i've lost two marks for forgetting to flip to gradient of normal and that prove question hahaha fk! (and that was way easier than last year, so an a+ will probs be 39+) damnnnnn

[forchina]

For 5)b isn't it negative 8S(u^2)du ? As u=cos(6x), du/dx = -sin(6x). SO putting a negative would cancel it out in the integral

Also what did everyone else get for the last question. It was either pi/2 or tan^-1(2)? except im not sure

[Aman476]

For 5)b isn't it negative 8S(u^2)du ? As u=cos(6x), du/dx = -sin(6x). SO putting a negative would cancel in in the integral would cancel it out

I put the negative in and ended up with sqrt(3) so idk?

[BLACKCATT]

Was the angle with positive axis worth 2 marks or 1?

Anyone?

[jt33]

For 5)b isn't it negative 8S(u^2)du ? As u=cos(6x), du/dx = -sin(6x). SO putting a negative would cancel it out in the integral

Also what did everyone else get for the last question. It was either pi/2 or tan^-1(2)? except im not sure

it was pi/2, T1>T2

[Aman476]

For 5)b isn't it negative 8S(u^2)du ? As u=cos(6x), du/dx = -sin(6x). SO putting a negative would cancel it out in the integral

Also what did everyone else get for the last question. It was either pi/2 or tan^-1(2)? except im not sure

I got those two angles but chose pi/3, since as the angle increases sec(theta) increases (above 10g) and as the angle decreases tan(theta) decreases, so sinces tan-1(2) is a bigger angle i said pi/3.......or you could have just used what they told you in the previous question sec(theta)>tan(theta).....i only realised that after haha

did anyone prove it?

[jt33]

it was pi/2, T1>T2

sorry, i meant pi/3

[ivysan]

hi anyone can scan and upload the paper? thanks

[Aman476]

it was pi/2, T1>T2

if theta = pi/2, sec(theta) (1/cos(theta)) = infinity?? praying you meant pi/3

Awesome just so your post - nearly had a heart attack haha

[Robert123]

How did everyone go?


Did anyone else get sqrt(3)pi/2 - (3pi/8) +(3ln(0.5))/4

I didn't have the 3 in front of the log as it cancelld out for me when I removed the 3 from in front of the arctan.
Also, not sure if they would want it further to -ln(2) instead of ln(0.5).

For 5)b isn't it negative 8S(u^2)du ? As u=cos(6x), du/dx = -sin(6x). SO putting a negative would cancel it out in the integral

Also what did everyone else get for the last question. It was either pi/2 or tan^-1(2)? except im not sure

For the negative in front of the integral, you could of removed it by switching the terminals around