Amaths help please
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Ermm...can anyone help me on the the topic relative velocity in AM.
I need help knowing how to draw the diagram
l0l...
like they always say the wind is blowing 050 bearing from smthsmth
can somebody explain how to draw pls?....sry i m unable to give any question,donno what to give...l0l... -
050 bearing means, from the vertical line(north) you measure 50 degrees and draw a line
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u noe vectors?
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o.o...
ya,i know vectors..
since no question very hard to help,i put some question..
Q: A town A is 320km due west of a town B. On a day when there is no wind an aircraft takes 50min to fly from A to B. On a second day, when the wind is blowing from a direction 350, the aircraft also takes 50min for the journey from A to B.Find
a)the course set by the polot on this second day,
b)the speed,in km/h,of the wind..
thx for the help,best got diagram if possible..hahaha,i need to learn the diagram to solve the question. -
some variables are missing. what is the speed of the wind?
cos if no wind and got wind also takes the same time, then the air speed and resultant velocity should be the same. -
y not?
anyway no answers? like maybe 070 and 133.4 -
l0l...dunno,the question from TYS,oso nv stat wind speed....
another question then..
Q:A motor boat travels in a straight line across a river which flows at 3m/s between straight parallel banks 200m apart. The motor boat, which has a top speed of 6m/s in still water, travels directly from a point A on one bank to a point B,150m downstream of A, on the opposite bank. Assuming that the motoring boat is travelling at top speed, find, to the neareset second, the time it takes to travel from A to B -
Sorry hijack thread for a while ah
For differentoation of 1/cosX
do we re-express it as cosX^-1 and do normal differentiation or we have to use quotient rule? -
can same time, but.......Originally posted by angelfairy:y cant be same time?
anyway no answers provided? like maybe 070 degrees and 133.4 kmh-1? -
1/cosx is just secx, differentiate is just secxtanxOriginally posted by lpx88:Sorry hijack thread for a while ah
For differentoation of 1/cosX
do we re-express it as cosX^-1 and do normal differentiation or we have to use quotient rule? -
d/dx(sec x)Originally posted by lpx88:Sorry hijack thread for a while ah
For differentoation of 1/cosX
do we re-express it as cosX^-1 and do normal differentiation or we have to use quotient rule?
use quotient rule. -
Qn at first is secx....but tys nvr say d/dx secx gives us what..so i converted it to 1/cosX?
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then do as tamago said, btw for the diagram i got the same one as yours except my internal angle is 80 not 100...Originally posted by lpx88:Qn at first is secx....but tys nvr say d/dx secx gives us what..so i converted it to 1/cosX? -
o.0...u got the answer.
a)070
b)133km/h
how u get?...l0l.....
can explain how u get the diagram,sry,i really suck at maths. -
Originally posted by Evo4eva:l0l...dunno,the question from TYS,oso nv stat wind speed....
another question then..
Q:A motor boat travels in a straight line across a river which flows at 3m/s between straight parallel banks 200m apart. The motor boat, which has a top speed of 6m/s in still water, travels directly from a point A on one bank to a point B,150m downstream of A, on the opposite bank. Assuming that the motoring boat is travelling at top speed, find, to the neareset second, the time it takes to travel from A to B
With Pythagoras theorem find distance the boat will travel
Then, find angle X. U can find this by using sohcahtoa(sin = oppo/hyp..bla bla), then use 90 degrees, minus off this angle u just found. Answer u get will be angle X
Use sine rule and get angle B
180 degrees - angle X - angle B= Angle Z
With angle Z, u can use sine rule or cosine rule to get final velocity
Dist/ Velocity = time -
changed cos direction of travel shld be east.
v² = (384km/h)² + (384km/h)² - 2(384km/h)(384km/h) cos 20º
v = 133.7km/h -
X2....Then jus use sine rule can get answer liao..Originally posted by ^tamago^:changed cos direction of travel shld be east.
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sin a = 200/250 = 0.8Originally posted by lpx88:With Pythagoras theorem find distance the boat will travel
Then, find angle X. U can find this by using sohcahtoa(sin = oppo/hyp..bla bla), then use 90 degrees, minus off this angle u just found. Answer u get will be angle X
Use sine rule and get angle B
180 degrees - angle X - angle B= Angle Z
With angle Z, u can use sine rule or cosine rule to get final velocity
Dist/ Velocity = time
6/sin a = 3/sin b
sin b = 0.5×0.8=0.4
Z = 180º - asin(0.8 ) - asin(0.4) = 103.3º
v² = (3m/s)² + (6m/s)² - 2(3m/s)(6m/s) cos 103.3º
v = 7.30m/s
t = (250m)/(7.30m/s) = 34s (nearest sec) -
thx thx for all ur help.. -
Last time, I ask my school teachers about this. They give me a straight "no".Originally posted by lpx88:Sorry hijack thread for a while ah
For differentoation of 1/cosX
do we re-express it as cosX^-1 and do normal differentiation or we have to use quotient rule?
Never write cos^-1 x in your ans. Rewrite it as sec x. or write it as (cos x)^-1
They give me a reason that cos^-1 x will be misunderstood by them as arc cosine x which represents the inverse function of cosine as this symbol is commonly used by the calculator. Many students are misunderstood due to the confusion of these symbols sin^-1, cos^-1, tan^-1.
If you want to express 1/cos x in a simpler way, they say just simply put sec x or (cos x )^-1 -
So using the chain rule...Originally posted by tanjun:Last time, I ask my school teachers about this. They give me a straight "no".
Never write cos^-1 x in your ans. Rewrite it as sec x. or write it as (cos x)^-1
They give me a reason that cos^-1 x will be misunderstood by them as arc cosine x which represents the inverse function of cosine as this symbol is commonly used by the calculator. Many students are misunderstood due to the confusion of these symbols sin^-1, cos^-1, tan^-1.
If you want to express 1/cos x in a simpler way, they say just simply put sec x or (cos x )^-1
(cos x) ^-1 = - (cos X) ^-2 . - (sin X)
= sin X / ( cos X ) ^2
Is it?Anyway to simplify further? -
Originally posted by lpx88:
to simplify: sin x/cos²x = (sin x/cos x)*(1/cos x) = sec x tan x
So using the chain rule...
(cos x) ^-1 = - (cos X) ^-2 . - (sin X)
= sin X / ( cos X ) ^2
Is it?Anyway to simplify further?
therefore, d/dx (sec x) = sec x tan x
just to reiteriate...
if the number is a positive integer, then it means power and the two terms can be used interchangeably, i.e.
cos²x = (cos x)²
cos³x = (cos x)³
if it's -1 it means inverse and not power. i.e. cos¨¹ 60º = acos 60º = 0.5, cos¨¹ 90º = acos 90º = 0
cos¨²x is not meaningful and if u mean the inverse squared, it is best to use (cos x)¨². -
Originally posted by ^tamago^:
Ah i see...thanks
[b]to simplify: sin x/cos²x = (sin x/cos x)*(1/cos x) = sec x tan x
therefore, d/dx (sec x) = sec x tan x
just to reiteriate...
if the number is a positive integer, then it means power and the two terms can be used interchangeably, i.e.
cos²x = (cos x)²
cos³x = (cos x)³
if it's -1 it means inverse and not power. i.e. cos¨¹ 60º = acos 60º = 0.5, cos¨¹ 90º = acos 90º = 0
cos¨²x is not meaningful and if u mean the inverse squared, it is best to use (cos x)¨².
[/b]
But i abt blur bout this part..
if it's -1 it means inverse and not power. i.e. cos¨¹ 60º = acos 60º = 0.5, cos¨¹ 90º = acos 90º = 0
y suddenly got 'a' come in one? -
acos means inverse cos. but dun use at o or a lvls. it's only if someone else writes acos then u noe wat it means.Originally posted by lpx88:Ah i see...thanks
But i abt blur bout this part..
if it's -1 it means inverse and not power. i.e. cos¨¹ 60º = acos 60º = 0.5, cos¨¹ 90º = acos 90º = 0
y suddenly got 'a' come in one?
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..........scare me siaOriginally posted by ^tamago^:acos means inverse cos. but dun use at o or a lvls. it's only if someone else writes acos then u noe wat it means.