Menu Close

A-particle-s-trajectory-is-described-by-x-e-t-e-t-y-2t-Find-the-distance-that-the-particle-traveled-for-0-t-2-

Tinku Tara 0 Comments
Pin




Question Number 103400 by abony1303 last updated on 14/Jul/20
A particle′s trajectory is described by x=e^t +e^(−t) y=2t Find the distance that the particle traveled for 0≤t≤2
Aparticle′strajectoryisdescribedbyx=et+e−ty=2tFindthedistancethattheparticletraveledfor0⩽t⩽2
Commented by abony1303 last updated on 14/Jul/20
Pls help
Plshelp
Answered by mr W last updated on 14/Jul/20
dx=(e^t −e^(−t) )dt dy=2dt ds=(√((dx)^2 +(dy)^2 ))=(√((e^t −e^(−t) )^2 +4)) dt =(e^t +e^(−t) )dt s(t)=∫_0 ^t ds=∫_0 ^t (e^t +e^(−t) )dt=[e^t −e^(−t) ]_0 ^t =e^t −e^(−t) ⇒s(2)=e^2 −(1/e^2 )
dx=(et−e−t)dtdy=2dtds=(dx)2+(dy)2=(et−e−t)2+4dt=(et+e−t)dts(t)=∫0tds=∫0t(et+e−t)dt=[et−e−t]0t=et−e−t⇒s(2)=e2−1e2
Commented by abony1303 last updated on 14/Jul/20
thank you ser. Can you pls explain why you differentiated x and y, and I can′t understand the formula in 3rd raw.
thankyouser.Canyouplsexplainwhyyoudifferentiatedxandy,andIcan′tunderstandtheformulain3rdraw.
Commented by mr W last updated on 14/Jul/20
do you know this: ds=(√((dx)^2 +(dy)^2 ))=(√(1+((dy/dx))^2 )) dx
doyouknowthis:ds=(dx)2+(dy)2=1+(dydx)2dx
Commented by mr W last updated on 14/Jul/20
if x=f(t), y=g(t), in time dt the object covers a distance in x−direction dx=f′(t)dt and in y−direction dy=g′(t)dt. the total distance it travels in this time is ds=(√((dx)^2 +(dy)^2 ))=(√((f′(t))^2 +(g′(t))^2 ))dt
ifx=f(t),y=g(t),intimedttheobjectcoversadistanceinx−directiondx=f′(t)dtandiny−directiondy=g′(t)dt.thetotaldistanceittravelsinthistimeisds=(dx)2+(dy)2=(f′(t))2+(g′(t))2dt
Answered by OlafThorendsen last updated on 14/Jul/20
0≤t≤2 ⇔ 0≤y≤4 x = e^t +e^(−t) = 2cht = 2ch(y/2) (dx/dy) = 2×(1/2)sh(y/2) = sh(y/2) d = ∫_0 ^4 (√(1+((dx/dy))^2 ))dy d = ∫_0 ^4 (√(1+sh^2 (y/2)))dy d = ∫_0 ^4 ch(y/2)dy d = [2sh(y/2)]_0 ^4 = 2sh2 = e^2 −(1/e^2 )
0⩽t⩽2⇔0⩽y⩽4x=et+e−t=2cht=2chy2dxdy=2×12shy2=shy2d=∫041+(dxdy)2dyd=∫041+sh2y2dyd=∫04chy2dyd=[2shy2]04=2sh2=e2−1e2
Answered by Dwaipayan Shikari last updated on 14/Jul/20
(dx/dt)=e^t −e^(−t) (dy/dt)=2 Resultant =(√(△x^2 +△y^2 ))=(√((e^t −e^(−t) )^2 +4 ))=e^t +e^(−t) Time is bounded (0,2) ∫_0 ^2 e^t −e^(−t) dt=e^2 −(1/e^2 ) (Resultant tracetory)
dxdt=et−e−tdydt=2Resultant=△x2+△y2=(et−e−t)2+4=et+e−tTimeisbounded(0,2)∫02et−e−tdt=e2−1e2(Resultanttracetory)

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *