ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: HarveyD on November 07, 2011, 02:02:14 am
-
just wondering for this
why is area enclosed the portion of the bigger circle
would the intersection between the two be the smaller one?
-
I don't really understand your question. Also, what is L? The question is cut off.
-
L is y = 1/root3 x
how do you find the shaded area?
-
The shaded area is like a 'piece' of a 'donut' right?
Find the area of the large circle, then subtract the area of the smaller circle. Then find the proportion of the angle of your 'piece'. As in, do you have half the donut? 1/4? 1/9? Etc. Then multiply the area you found before by this proportion.
edit: so I guess, if my assumption was right, then the area of the bigger circle minus the smaller is:
4pi - pi = 3pi
And the proportion of the whole thing is 1/12 (since the angle between the two boundaries is pi/6, and one revolution is 2pi. Then the area required is 3pi*1/12 = pi/4. Is that right?
-
Yes it is pi/4
I hope this explains it to you. WARNING: I suck at paint and I am not the greatest person to explain this kind of stuff, but I shall try!
-
thanks for the help :)
got another question
with the question below
couldnt C be true as well?
If F = 7 , a = 0
so constant velocity?
-
The keyword is "moves" with constant velocity. If F=7, it has constant velocity so a=0 however it is at rest.
-
what's the answer?
-
can someone please do this question ?
3 kg : F - T -3g x 1/7 = 3a
2kg : T - 2g x 1/7 =2 a
??? what next?
-
can someone please do this question ?
3 kg : F - T -3g x 1/7 = 3a
2kg : T - 2g x 1/7 =2 a
??? what next?
Something looks wrong with the equation of motion for the 3kg object, you don't normally have friction and weight force in the same direction. I think the weight force should be in the j direction and you also need a normal reaction force if there is friction. Do you have the full question?
-
no. but i got it. i worked it out! its a?
-
yep
-
no. but i got it. i worked it out! its a?
Nvm, I assumed F was friction. Still not too sure what you're asking but at least you've got the answer. :)
-
just wondering
how would you guys do this question?
Is it wrong to let
t^2 + 1 = 7t - 5
then
2t = t + 6
then solve from there
-
anyone know what the deal is with the abs. value signs in the logs?
wtf is that shit?
sometimes they just make what's in there positive and other times they leave it and i can't work it out
-
just wondering
how would you guys do this question?
Is it wrong to let
t^2 + 1 = 7t - 5
then
2t = t + 6
then solve from there
You're right, you have to find a value(s) of t that satisfy both equations. Remember to sub t back in though cos the question is asking for the point of collision! :)
anyone know what the deal is with the abs. value signs in the logs?
wtf is that shit?
sometimes they just make what's in there positive and other times they leave it and i can't work it out
Leave them there unless you can actually justify when you're removing them. Trial exam papers are usually just lazy in their solutions!
-
hmm when I do that I get the other t value (which doesnt collide)
as t = 1
the answers below has t = 8 :S
should i use that method if a similar question comes up tomorrow?
-
will marks be taken off if i use them where theyre not needed? ive always just used brackets lol
-
is this an exam 1 question?
-
Oh yup if you want to find where the paths will meet (but not necessarily collide), you can do what they did or you can equate the Cartesian equations.
-
ahh alright
thanks heaps :)
good luck tomorrow!
and yeah, this is an exam 1 question
-
i thought it was c, because i got a=0 if F=7. please correct me if i'm wrong? :\