Find the length of the curve. r(t)=212ti+etj+e−tk,0≤t≤1

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yagombyeR · Accepted Answer

We have to find the length of the given curve. To do this, first of all, let us calculate the derivative of each of the components for the given vector function, we get

r^{'} (t) = \sqrt{2} i + e^{t} j + (- e^{- t}) k

Now, let us find the magnitude of the derivative function that is,

| r^{'} (t) | = \sqrt{{(\sqrt{2})}^{2} + {(e^{t})}^{2} + {(- e^{- t})}^{2}}

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

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

= e^{t} + e^{- t}

This implies that the length of the curve is,

L = \int_{0}^{1} | r^{'} (t) | \cdot d t

= \int_{0}^{1} (e^{t} + e^{- t}) \cdot d t

= {[e^{t} - e^{- t}]}_{0}^{1}

= (e^{1} - e^{- 1}) - (e^{0} - e^{0})

= e - e^{- 1} - (1 - 1)

= e - \frac{1}{e}

\approx 2.35

Result: The length of the given curve is approximately 2.35 units.

Find the length of the curve. r(t)=2^1/2ti+e^tj+e^{-t}k, 0<=t<=1

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