# What is the arclength of #(t-3,t^2)# on #t in [1,2]#?

The arclength is

Arclength is given by:

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

Reverse the substitution:

Insert the limits of integration:

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To find the arc length of the curve described by the parametric equations (x(t) = t - 3) and (y(t) = t^2) on the interval ([1, 2]), we use the formula for arc length:

[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt ]

Substituting the given parametric equations, we have:

[ L = \int_{1}^{2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt ]

[ L = \int_{1}^{2} \sqrt{(1)^2 + (2t)^2} dt ]

[ L = \int_{1}^{2} \sqrt{1 + 4t^2} dt ]

Now, we integrate:

[ L = \int_{1}^{2} \sqrt{1 + 4t^2} dt = \left[\frac{1}{4}t\sqrt{1 + 4t^2} + \frac{1}{8}\ln\left(2t + \sqrt{1 + 4t^2}\right)\right]_{1}^{2} ]

[ L = \left(\frac{1}{4}(2)\sqrt{1 + 4(2)^2} + \frac{1}{8}\ln\left(2(2) + \sqrt{1 + 4(2)^2}\right)\right) - \left(\frac{1}{4}(1)\sqrt{1 + 4(1)^2} + \frac{1}{8}\ln\left(2(1) + \sqrt{1 + 4(1)^2}\right)\right) ]

[ L = \left(\frac{1}{4}(2)\sqrt{17} + \frac{1}{8}\ln\left(4 + \sqrt{17}\right)\right) - \left(\frac{1}{4}\sqrt{5} + \frac{1}{8}\ln\left(2 + \sqrt{5}\right)\right) ]

[ L \approx 3.964 ]

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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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