Thread: length of curve

from t=0 to t=pi/2
x=e^-t cost
y=e^-t sint

dx/dt = -e^-t cost - e^-t sint
dy/dt = -e^-t sint + e^-t cost

(dx/dt)^2 = ?
(dy/dt)^2 = ?

L = (int 0 to pi/2) [sqrt(-2e^-2t)]dt

how do you get that? when i square dx/dt and dy/dt using the (a+b)(a+b) = a^2 +2ab+b^2 method i get a wrong answer, but when i just slap on a ^2 on each term and simplify i get it right. does that squaring rule not apply here for some reason?

Originally Posted by jeph

from t=0 to t=pi/2
x=e^-t cost
y=e^-t sint

dx/dt = -e^-t cost - e^-t sint
dy/dt = -e^-t sint + e^-t cost

(dx/dt)^2 = ?
(dy/dt)^2 = ?

L = (int 0 to pi/2) [sqrt(-2e^-2t)]dt

how do you get that? when i square dx/dt and dy/dt using the (a+b)(a+b) = a^2 +2ab+b^2 method i get a wrong answer, but when i just slap on a ^2 on each term and simplify i get it right. does that squaring rule not apply here for some reason?

You want
sqrt{(dx/dt)^2 + (dy/dt)^2}

So
(dx/dt)^2 = [-e^{-t}cos(t) - e^{-t}sin(t)]^2

= [e^{-t}cos(t) + e^{-t}sin(t)]^2

= e^{-2t}cos^2(t) + 2e^{-2t}sin(t)cos(t) + e^{-2t}sin^2(t)

= [e^{-2t}cos^2(t) + e^{-2t}sin^2(t)] + 2e^{-2t}sin(t)cos(t)

= e^{-2t}[cos^2(t) + sin^2(t)] + 2e^{-2t}sin(t)cos(t)

= e^{-2t} + 2e^{-2t}sin(t)cos(t)

Similarly:
(dy/dt)^2 = [-e^{-t}sin(t) + e^{-t}cos(t)]^2

= e^{-2t} - 2e^{-2t}sin(t)cos(t)

So
sqrt{(dx/dt)^2 + (dy/dt)^2}

= sqrt{e^{-2t} + 2e^{-2t}sin(t)cos(t) + e^{-2t} - 2e^{-2t}sin(t)cos(t)}

= sqrt{2e^{-2t}}

as you said you needed.

-Dan

wow...i wish i remembered what my thoughts were while working on this problem on the exam. i dunno if i just got lucky or i really knew what i was doing at the time. the problem works just as well if you just slap on the ^2 on each of the terms.