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

Commented byabony1303 last updated on 14/Jul/20

Pls help

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 )

Commented byabony1303 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.

Commented bymr W last updated on 14/Jul/20

do you know this:  ds=(√((dx)^2 +(dy)^2 ))=(√(1+((dy/dx))^2 )) dx

Commented bymr 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

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 )

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)