The parameter t varies from 0 to 1. Arc length of the curve on the interval 0 le

Question

The parameter t varies from 0 to 1. Arc length of the curve on the interval 0 lessthanorequalto t lessthanorequalto 1 is L = integral_a^n Squareroot [dx/dt]^2 + [dy/dt]^2 + [dz/dt]^2 dt = integral_0^1 Squareroot (Squareroot 2)^2 + (e^t)^2 + (-e^-t)^2 dt = integral_0^1 Squareroot 2 + e^2t + e^-2t dt = integral_0^1 Squareroot (e^t + e^-t)^2 dt = integral_0^2 (e^t + e^-t)dt = [e^t - e^-t]_0^1 = (e^1 - e^-1) - (e^0 - e^-0) = (e - e^-1) - (1 - 1) = e - e^-1 Hence, the required arc length of the given curve on the interval 0 lessthanorequalto t lessthanorequalto 1 is

Explanation / Answer

[0 to 1] [2+e2t+e-2t] dt

[0 to 1] [(2*1)+e2t+e-2t] dt

[0 to 1] [(2*eo)+e2t+e-2t] dt

[0 to 1] [(2*et-t)+e2t+e-2t] dt

[0 to 1] [(2ete-t)+e2t+e-2t] dt

[0 to 1] [(2ete-t)+(et)2+(e-t)2] dt

[0 to 1] [(et)2+(2ete-t)+(e-t)2] dt

expression inside suqreroot is of form a2+2ab+b2 which equals to (a+b)2

[0 to 1] [(et+e-t)2] dt

[0 to 1] (et+e-t) dt

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