Find the general curvature and the point of max curvature of the curve r(t) = (t

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Note: I assume you really meant r(t) = e^(-2t)i + e^(2t)j + 2t sqrt(2)k (Otherwise, computing the arc length exactly is nearly impossible to do!) ------------------------- (a) r'(t) = -2 e^(-2t)i + 2 e^(2t)j + 2 sqrt(2)k At t = 0, the direction vector is r'(0) = -2i + 2j + 2 sqrt(2)k. Also, r(0) = 1i + 1j + 0k. ==> x = 1 - 2t, y = 1 + 2t, z = 0 + 2 sqrt(2) t. -------------- (b) ||r'(t)|| = sqrt [4e^(-4t) + 4e^(4t) + 8] = 2 sqrt [e^(-4t) + 2 + e^(4t)] = 2 sqrt [e^(-2t) + e^(2t)]^2 = 2e^(-2t) + 2e^(2t). So, the arc length equals ?(t = 0 to ln 5) [2e^(-2t) + 2e^(2t)] dt = -e^(-2t) + e^(2t) {for t = 0 to ln 5} = (25 - 1/25) - (-1 + 1) = 624/25. ----------------------- (c) From part b, ||r'(t)|| = 2e^(-2t) + 2e^(2t). Thus, T = r'(t) / ||r'(t)|| = [-e^(-2t)i + e^(2t)j + sqrt(2)k] / [e^(-2t) + e^(2t)] ==> T = -[1 + e^(4t)]^(-1) i + [e^(-4t) + 1]^(-1) j + sqrt(2) [e^(-2t) + e^(2t)]^(-1) k T' = 4e^(4t)[1 + e^(4t)]^(-2) i + 4e^(-4t)[e^(-4t) + 1]^(-2) j - sqrt(2) (-2e^(-2t) + 2e^(2t)) [e^(-2t) + e^(2t)]^(-2) k At (1, 1, 0), we have t = 0. ||r'(0)|| = 4 T'(0) = 1i + 1j - 0k ==> ||T'(0)|| = sqrt(2).

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