Find, in terms of p, the approximate change in curve tan(2x+1) as x increases from 1 to 1+p, where p is small. Note that tan(2x+1) has no stationary point as its dydx cannot be 0
remember the radian thing? apply that. remember the tan p = p if p is very small? apply that.
so use your angle addition rule for tan (x +p) = (tan+tan)/(1 + tantan) thingie.
and sub in your tan p = p.
tot is just differentiate tan(2x+1)?
Originally posted by hisoka:tot is just differentiate tan(2x+1)?
u get 2 (sec 2x+1)^2
Then what>?
for x, when 1 ==> 1 + p,
tan (2(1) +1) ==> tan (2(1+p) +1)
tan (3) ==> tan ( 3 + 2p)
==> (tan3 + tan 2p) / [1 + tan(3)tan(2p)]
let tan p = p since p is small.
==> (tan3 + 2p) / [1 + tan(3)2p]
change = final - initial
= (tan3 + 2p) / [1 + tan(3)2p] - tan (3)
go clean it up more.
for x, when 1 ==> 1 + p, this is because since p is so small it doesnt matter?
let tan p = p since p is small. -> i never knew this XD
this is for radian approximation...
1 ==> 1+p
is my way of saying as 1 approaches 1+p,
than the next line explains what happens to tan(2x + 1) when this approximation happens.
if this is out of your syllabus, than just skip this thingie bah.
Originally posted by skythewood:this is for radian approximation...
1 ==> 1+p
is my way of saying as 1 approaches 1+p,
than the next line explains what happens to tan(2x + 1) when this approximation happens.
if this is out of your syllabus, than just skip this thingie bah.
the tys answers required you to differentiate first.
Ill give you the solution cos i cant understand what the sol. means here it is:
x=1
8x= p
dydx = 2 (sec (2x+1))^2
= 3 (sec 3) ^2
8y = 2 (sec 3)^2 * p
= 2.04p
care to xplain? thanks
that looks really... exact.
approximation?
approx the tan curve to be straight line with gradient 2 (sec (2x+1))^2 = 2.04
using simple forumla Y = mX + C.
when x = 1, Y = something...
when x = 1+ p, Y = something + 2.04p