What is the arclength of $(t^{2} ln t, {(ln t)}^{2})$ $(t^2lnt,(lnt)^2)$ on $t \in [1, 2]$ $t in [1,2]$ ?

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What is the arclength of $(t^{2} ln t, {(ln t)}^{2})$ $(t^2lnt,(lnt)^2)$ on $t \in [1, 2]$ $t in [1,2]$ ?

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Answer 1 · 2016-11-16T18:23:44

I used Wolframalpha to evaluate the integral.

$L = 2.815$ $L = 2.815$

Explanation:

Here is a reference for Arc length of parametric equations

From the reference, the equation for arclength is:

$L = \int_{α}^{β} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ $L = int_alpha^beta sqrt((dx/dt)^2 + (dy/dt)^2)dt$

$\frac{d x}{d t} = \frac{d [t^{2} ln (t)]}{d t} = 2 t ln (t) + t$ $dx/dt = (d[t^2ln(t)])/dt = 2tln(t) + t$

$\frac{d y}{d t} = \frac{d [{(ln (t))}^{2}]}{d t} = \frac{2 ln (t)}{t}$ $dy/dt = (d[(ln(t))^2])/dt = (2ln(t))/t$

I used Wolframalpha to evaluate the integral:

$L = \int_{1}^{2} \sqrt{{(2 t ln (t) + t)}^{2} + {(\frac{2 ln (t)}{t})}^{2}} d t = 2.815$ $L = int_1^2 sqrt((2tln(t) + t)^2 + ((2ln(t))/t)^2)dt = 2.815$

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What is the arclength of $(t^{2} ln t, {(ln t)}^{2})$ $(t^2lnt,(lnt)^2)$ on $t \in [1, 2]$ $t in [1,2]$ ?

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Explanation:

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What is the arclength of (t^2lnt,(lnt)^2)(t2lnt,(lnt)2)(t^2lnt,(lnt)^2) on t in [1,2]t∈[1,2]t in [1,2]?

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Explanation:

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What is the arclength of $(t^{2} ln t, {(ln t)}^{2})$ $(t^2lnt,(lnt)^2)$ on $t \in [1, 2]$ $t in [1,2]$ ?