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)