What is the arclength of $f (t) = (t + \sqrt{ln t}, t - \sqrt{ln t}))$ $f(t) = (t+sqrt(lnt),t-sqrtlnt))$ on $t \in [1, e]$ $t in [1,e]$ ?

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What is the arclength of $f (t) = (t + \sqrt{ln t}, t - \sqrt{ln t}))$ $f(t) = (t+sqrt(lnt),t-sqrtlnt))$ on $t \in [1, e]$ $t in [1,e]$ ?

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Answer 1 · 2018-01-02T06:27:09

The arc length between $t = a$ $t=a$ and $t = b$ $t=b$ on the parametric curve $y = f (t), x = g (t)$ $y=f(t),x=g(t)$ is given by
$int_a^bsqrt((dy/dt)^2+(dx/dt)^2)\ dt$

So, from the question, we have $t$ from $1$ to $e$ and $y=t-sqrt(ln(t)),x=t+sqrt(ln(t))$ . Then, $dy/dt=1-1/(2tsqrt(ln(t)))$ and $dx/dt=1+1/(2tsqrt(ln(t)))$ .

Thus, we need to find
$int_1^esqrt((1-1/(2tsqrt(ln(t))))^2+(1+1/(2tsqrt(ln(t))))^2)\ dx$
$=int_1^esqrt(2+1/(2t^2ln(t)))\ dx$

Using numerical approximation methods, we find that the integral is approximately equal to $3$ .

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What is the arclength of $f (t) = (t + \sqrt{ln t}, t - \sqrt{ln t}))$ $f(t) = (t+sqrt(lnt),t-sqrtlnt))$ on $t \in [1, e]$ $t in [1,e]$ ?

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What is the arclength of $f (t) = (t + \sqrt{ln t}, t - \sqrt{ln t}))$ $f(t) = (t+sqrt(lnt),t-sqrtlnt))$ on $t \in [1, e]$ $t in [1,e]$ ?