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math1001: differential calculus (1 Viewer)

famunlau

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Originally posted by Winston
but ur just stating the cross section, you only say what the shape is when that certain plane intersects the figure right?
um....I think its best to work out the equation...it will make more sense to teh markers...
 

famunlau

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Originally posted by Carlito
hrmm still dont get 1.a.iii or 2.b
1 a iii) We have to prove lim x->2 of the function is = 4
Using Squeeze law...we know that
-1 <= cos ( 1/(x-2)) <=1
hence -(x-2) <= (x-2) cos (1/(x-2))<=x-2
when we take the lim x->2
we get
0<=(x-2)cos(1/cos(x-2)<=0
because the two sides are equal
therefore....
by squeeze law
lim x-> 2 (x-2)cos(1/(x-2)) = 0
therefore the lim x->2 of teh original f(x)
is ....lim x->2 x^2( 1+ lim x->2 (x-2)cos(1/(x-2)))
= 2^2 (1 + 0)
= 4
for 2b...too hard to explain...read the notes....just remember that
-1<=-sinx<=1
and that cos x is continuous and defferentiable for all Re x...
then just use MVT...
 

Carlito

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Yeah i got 2b now, but you actually explained 1.b.iii which i already got, but i used yours to check and i did virtually the same thing so thats good.
 

famunlau

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Originally posted by Carlito
Yeah i got 2b now, but you actually explained 1.b.iii which i already got, but i used yours to check and i did virtually the same thing so thats good.
sorry....woops.....the cross sectional thing right?..
ok at y = 0
z = sqrt ( 4 - (x-1)^2 -(-3/2)^2)
= sqrt (7/4 - (x-1)^2)
which is an equation of a semicircle with centre (1,0) on xz plane with radius of root (7/4) units
at y =3
z = sqrt ( 4- (x-1)^2 -0)
= sqrt (4- (x-1)^2)
which is an equation of a semicircle with centre (1,0) on xz plane with radius of 2 units

hope that helps
 

famunlau

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Originally posted by Winston
carl have u got 2a(iii) yet? =\
2 a) iii) u have 4 intervals
(-oo,-2], [-2,0], [0,5], [5,oo)
these intervals are from ur critical pts which should have been at x = -2,0,5 from part ii)
so now each interval is in form [a,b]
so at f(a) for the first interval
its positive...and f(b) is negative
therefore there is a root....ie change of sign...
repeat for each interval
u should get 4 roots or 4 change of signs for f(x)....
 

Winston

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Originally posted by famunlau
2 a) iii) u have 4 intervals
(-oo,-2], [-2,0], [0,5], [5,oo)
these intervals are from ur critical pts which should have been at x = -2,0,5 from part ii)
so now each interval is in form [a,b]
so at f(a) for the first interval
its positive...and f(b) is negative
therefore there is a root....ie change of sign...
repeat for each interval
u should get 4 roots or 4 change of signs for f(x)....
ohh yea i did the - infinity and - 2 one and -infinity we proved from part i already thats its positive, and -2 gave a negative answer, so a change of sign thus a root, umm how did u exactly determine those intervals like the (-inf, -2]
 

famunlau

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well they have to be the intervals...because intervals are taken between critical pts....and the end pts of a function...
 

Winston

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Originally posted by famunlau
well they have to be the intervals...because intervals are taken between critical pts....and the end pts of a function...
ahh i c lol see how behind i am :D... ok anyways im done now hahaha thanks for ur help :)... nightios!
 

Carlito

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Thanks guys for all your help, I managed to get it all done in the end.
 

Winston

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Originally posted by Carlito
Thanks guys for all your help, I managed to get it all done in the end.
yeah ditto.... thanks famunlau, and thanks grace bros :p... and thanks wyvern, :p
 

tooheyz

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Originally posted by Winston
yeah ditto.... thanks famunlau, and thanks grace bros :p... and thanks wyvern, :p
its not grace bros no more man, myers :)
 

tooheyz

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grace bros cleaning service? mate i never heard of them

but thats beside the point man... aye u guys still meeting up for donuts or what?

hahahaha on the bos at uni *sigh*
slight addiction...
 

Carlito

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Originally posted by tooheyz
grace bros cleaning service? mate i never heard of them

but thats beside the point man... aye u guys still meeting up for donuts or what?
i would of if i didnt have a gay timetable of 5 hours straight on fridays :(
 

Winston

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Originally posted by Carlito
i would of if i didnt have a gay timetable of 5 hours straight on fridays :(
try 8 - 6 on monday straight with 1 hr break :)

and 8 - 4 on tuesday with 2 hr break

and 9 - 3 on wednesday with no breaks.
 

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