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

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Let L be the parametric arclength, then the formula is:

Substitute the above into the formula:

Integration performed by WolframAlpha

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To find the arc length of the curve defined by the parametric equations (t^2ln(t), ln(t^2)) for t in the interval [1, 2], we use the formula for arc length:

L = ∫[a,b] √[ (dx/dt)^2 + (dy/dt)^2 ] dt

Substituting the parametric equations into the formula, we have:

L = ∫[1,2] √[ (d/dt(t^2ln(t)))^2 + (d/dt(ln(t^2)))^2 ] dt

After taking the derivatives and simplifying, we get:

L = ∫[1,2] √[ (2tln(t) + t)^2 + (2/t)^2 ] dt

Now, we integrate this expression over the interval [1,2] to find the arc length. This integral might not have a closed-form solution, so numerical methods or software may be used to approximate the value.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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