# What is the arclength of #f(t) = (t^3-t+55,t^2-1)# on #t in [2,3]#?

Arclength is given by:

Expand the squares:

Complete the square in the square root:

Rearrange:

Apply partial fraction decomposition:

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To find the arclength of the curve defined by ( f(t) = (t^3 - t + 55, t^2 - 1) ) on the interval ( t ) in ([2,3]), 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 ]

where (a = 2) and (b = 3).

Differentiating (x(t) = t^3 - t + 55) and (y(t) = t^2 - 1) with respect to (t), we get:

[ \frac{dx}{dt} = 3t^2 - 1 ] [ \frac{dy}{dt} = 2t ]

Substituting these derivatives into the formula, we get:

[ L = \int_{2}^{3} \sqrt{(3t^2 - 1)^2 + (2t)^2} dt ]

Simplify and compute the integral:

[ L = \int_{2}^{3} \sqrt{9t^4 - 6t^2 + 1 + 4t^2} dt ] [ L = \int_{2}^{3} \sqrt{9t^4 - 2t^2 + 1} dt ]

[ L = \int_{2}^{3} \sqrt{(3t^2 - 1)^2} dt ] [ L = \int_{2}^{3} (3t^2 - 1) dt ] [ L = \left[\frac{t^3}{3} - t\right]_{2}^{3} ] [ L = \left[\frac{3^3}{3} - 3\right] - \left[\frac{2^3}{3} - 2\right] ] [ L = \left[9 - 3\right] - \left[\frac{8}{3} - 2\right] ] [ L = 6 - \left(\frac{8}{3} - 2\right) ] [ L = 6 - \frac{2}{3} ] [ L = \frac{16}{3} ]

So, the arclength of ( f(t) ) on ( t ) in ([2,3]) is ( \frac{16}{3} ).

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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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