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# Calc question?

Is the correct answer 3/2(e^2-1) ? the answer isn't supposed to be a decimal approximation

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- Ian HLv 71 year agoFavorite Answer
Arc length vector r(t) = <f, g, h> is given by

L = {t; from 0 to 1}∫√[(f’)^2 + (g’)^2 +(h’)^2]dt

Use c for cos, s for sine

(f’)^2 = (2c – s)^2*e^4t = e^4t[s^2 – 4cs + 4c^2]

(g’)^2 = (c + 2s)^2*e^4t = e^4t[4s^2 + 4cs + c^2]

(h’)^2 = e^4t *4

Total under square root sign will be

e^4t[5s^2 +5c^2 + 4] = 9e^4t

L = {t; from 0 to 1}∫3e^2t dt

L = |(3/2)e^2t| from 0 to 1

L = (3/2)(e^2 – 1)

Which is what you meant to write

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