**Applied Maths Before/After**

no scope codgod

xJEx wrote: »

OMG yay! But wait , somebody said last night you have to do 10 in radian mode, did you?? I cant remember what mode my calculator was in!

And i got e = 1/16 for the last bit but then i got an increase in 1.25 cm

e is the amount of speed conserved though, not the height it rises to. So in the last part if it rose to 1/16th the height, then its speed would have been 1/4 of the original speed so e would have been 1/4, not 1/16.

xJEx

no scope codgod wrote: »

e = kd (d is thickness)
1/3 = k(2.5)
2.5/3 = k
k = 5/6

1/4 = kd
1/4 = 5d/6
6/4 = 5d
6 = 20d
d = 6/20 = 3/10 = 3.33

Is what I did

e isnt proportional to d though, e DECRESES by a factor thats proportion to d

xJEx

no scope codgod wrote: »

e is the amount of speed conserved though, not the height it rises to. So in the last part if it rose to 1/16th the height, then its speed would have been 1/4 of the original speed so e would have been 1/4, not 1/16.

sorry yeah i got e to be 1/4

xJEx

Jackobyte wrote: »

My differential didn't require the use of any trigonometry, only logs, so radians didn't matter. I formed a different equation than you though.

Oh grand! I didnt know why you had to put it in radians i only used logs too! So we're right then? our answer?

Jackobyte

Kingkumar wrote: »

should it not be K(q-V) where q is the volume of tank when its full as the rate is proportional to how much water is left in the tank?

Yeah, this was like mine. dv/dt = k(1-v) where v=water gone as a fraction of the whole container.

no scope codgod wrote: »

e is the amount of speed conserved though, not the height it rises to. So in the last part if it rose to 1/16th the height, then its speed would have been 1/4 of the original speed so e would have been 1/4, not 1/16.

****, I think you could be You are right.
I did:
mgh=1/2 m v^2
Then:
1/2 m v^2 e = mgh(1/16)
mgh(e)=mgh(1/16)
Therefore: e=1/16. However, that e should have been included in the v, making it (ev)^2; thus making e^2=1/16.

Shit...

xJEx

Jackobyte wrote: »

Yeah, this was like mine. dv/dt = k(1-v) where v=water gone as a fraction of the whole container.

i had that but i wasnt sure how to solve it so i changed it to -dVdt = kV
and i started with v = 1, as in none gone, and then v=1/2, as in half gone

so like i said the rate of decrease of the volume is proportional to the fraction of the volume remaining. seeing as i had the minus sign isnt that correct? it must be if we got the same answer

xJEx

can anoyne tell me what they got for end of 1, 8 or the tensions in 4?

Jackobyte

xJEx wrote: »

can anoyne tell me what they got for end of 1, 8 or the tensions in 4?

Q1a: 7 seconds?
Q8(last part): 5r/3

xJEx

Jackobyte wrote: »

Q1a: 7 seconds?
Q8(last part): 5r/3

That same question was on 1992 actually, the 8 so yeah thats right
and yeah thats what I got in 1 too

xJEx

anyobyd get 63.32N and then 71.24 N for first bitta 4?

John Meighan

xJEx wrote: »

anyobyd get 63.32N and then 71.24 N for first bitta 4?

Yea, that's what I got, I know nobody does hydrostatics but did anyone get 400 for a?

xJEx

John Meighan wrote: »

Yea, that's what I got, I know nobody does hydrostatics but did anyone get 400 for a?

yay i got all of 4 correct then sorry didnt do it.
did you do 3 cause i got confused on it and im dying to know what the two angle answers were!

John Meighan

xJEx wrote: »

yay i got all of 4 correct then sorry didnt do it.
did you do 3 cause i got confused on it and im dying to know what the two angle answers were!

I think 30 degrees for first one and 45 degrees for second one that answer could be a bit dodgy though.

xJEx

John Meighan wrote: »

I think 30 degrees for first one and 45 degrees for second one that answer could be a bit dodgy though.

yeah i got them, but i had like 2sin(theta)cos(theta) - sin^2(theta) in the range so that had to be a max and i wasnt sure how to find the max theta so i just kinda guessed and got 5 dont know if ill get the marks for method even if my answer is right :P do they have to give u full marks if you get the answer?

John Meighan

xJEx wrote: »

yeah i got them, but i had like 2sin(theta)cos(theta) - sin^2(theta) in the range so that had to be a max and i wasnt sure how to find the max theta so i just kinda guessed and got 5 dont know if ill get the marks for method even if my answer is right :P do they have to give u full marks if you get the answer?

I changed that using formulas to sin2(theta) +.5 -Cos2(theta) and said sin max at 90 degrees and cos min at 90 degrees.

Jackobyte

xJEx wrote: »

yeah i got them, but i had like 2sin(theta)cos(theta) - sin^2(theta) in the range so that had to be a max and i wasnt sure how to find the max theta so i just kinda guessed and got 5 dont know if ill get the marks for method even if my answer is right :P do they have to give u full marks if you get the answer?

I let that equal y, then differentiated and let equal 0 for max/min. Think I got 30 degrees.

xJEx

John Meighan wrote: »

I changed that using formulas to sin2(theta) +.5 -Cos2(theta) and said sin max at 90 degrees and cos min at 90 degrees.

yeah well i knew 90 gives a max sin and a min cos so i just kind of went from there, 45 degrees gave the whole bracket bit a value of 0.5
dammit jack may be right.. his gives 0.6 or something

MarieCurie22

what angle did you guys get for q2 a?

MarieCurie22

my answers are so f*cked up in comparison....was any body like "DISC!!!!????? WHAT!!?!?!?!??!! AGAIN!!!!?!"

MarieCurie22

DID YE GET 1 AND 8 **sorry cap locks** seconds for q1?

Undeadfred

Is there anywhere online that has the answers?

Monsieur Folie

MarieCurie22 wrote: »

DID YE GET 1 AND 8 **sorry cap locks** seconds for q1?

I think that's what I got, interval of 7 seconds. Though I am the least trustworthy person in this thread when it comes to Applied Maths so idk. :P

Bionicle

I've been doing some of the questions and can confirm that for q5(b), the answers are e=1/2, e=1/3, thickness is 3.75 cm and that the answers for q10 (a) is 1.5, (b) 60 m/s and 36m. I'll get crcking on more questions if you guys want. Any in particular?

Undeadfred

Bionicle wrote: »

I've been doing some of the questions and can confirm that for q5(b), the answers are e=1/2, e=1/3, thickness is 3.75 cm and that the answers for q10 (a) is 1.5, (b) 60 m/s and 36m. I'll get crcking on more questions if you guys want. Any in particular?

Are you a teacher?

Bionicle

Undeadfred wrote: »

Are you a teacher?

Maths college student.

appliedmaths

I have put up (draft) solutions to qu1,2,3,4,5,10 on CBS Kilkenny Moodle.
Hope they are correct! But pls feel free...!

ahmdoda

what did people get for question 10 part b ii?

xJEx

Bionicle wrote: »

I've been doing some of the questions and can confirm that for q5(b), the answers are e=1/2, e=1/3, thickness is 3.75 cm and that the answers for q10 (a) is 1.5, (b) 60 m/s and 36m. I'll get crcking on more questions if you guys want. Any in particular?

**** YEAH. got all them. in your face everyone who thought it was 3.3 for the thickness! Majority does not always rule :P

xJEx

Bionicle wrote: »

I've been doing some of the questions and can confirm that for q5(b), the answers are e=1/2, e=1/3, thickness is 3.75 cm and that the answers for q10 (a) is 1.5, (b) 60 m/s and 36m. I'll get crcking on more questions if you guys want. Any in particular?

Please do 3, 4 and the last bit of 10

oh and the value of e for 5a please

xJEx

appliedmaths wrote: »

I have put up (draft) solutions to qu1,2,3,4,5,10 on CBS Kilkenny Moodle.
Hope they are correct! But pls feel free...!

I got the same answer for ten but when i set up my differential equation i had a minus in front of the dV/dt, is that wrong? Also, i left it as 79.32 mins, that ok do you think?