# Arc Length?

Can someone help me? My answer is wrong, I got sqrt(2)e^(pi) - sqrt(2)

### 2 Answers

- SlowfingerLv 61 month ago
Your answer is correct.

Integral is √2 (e^π - 1)

Use product rule to find

dx/dt = d/dt(e^(4t) cos(4t))

= d/dt(e^(4t)) * cos(4t) + e^(4t) d/dt(cos(4t))

= 4 e^(4 t) (cos(4t) - sin(4t))

dy/dt = d/dt(e^(4t) sin(4t))

= d/dt(e^(4t)) * sin(4t) + e^(4t) d/dt(sin(4t))

= 4 e^(4t) (sin(4t) + cos(4t))

Calculate the expression under square root

(dx/dt)² + (dy/dt)² =

= 16e^(8t) [(cos 4t - sin 4t)² + (sin 4t + cos 4t)²]

expand

= 16e^(8t) [(cos²4t - 2cos4t sin 4t + sin²4t) + (sin²4t + 2cos4t sin 4t + cos²4t)]

= 16e^(8t) * 2(sin²4t + cos²4t)

= 32e^(8t)

integral from 0 to (π/4) of √(32e^(8t)) dt =

= 4 √2 integral from 0 to (π/4) of e^(4t)

= √2 [e^(4(π/4)) - e^(4*0)]

= √2 (e^π - 1)

≈ 31.312

- AmyLv 71 month ago
I agree with your answer. Perhaps you are entering it incorrectly?

dx/dt = 4e^(4t) (cos(4t) - sin(4t))

dy/dt = 4e^(4t) (cos(4t) + sin(4t))

∫ √( (4e^(4t))^2 (cos^2 - 2cos*sin + sin^2 + cos^2 + 2cos*sin + sin^2) ) dt

= ∫ √( (4e^(4t))^2 * 2 ) dt

= ∫ 4√2 e^(4t) dt

= √2 e^(4t)

L = √2 e^π - √2