# What is the arc length of #f(t)=(sqrt(t-1),2-8t) # over #t in [1,3]#?

Arc length is given by:

Simplify:

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 last substitution:

Insert the limits of integration:

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To find the arc length of the curve defined by the function ( f(t) = (\sqrt{t-1}, 2-8t) ) over the interval ( t \in [1,3] ), we use the arc length formula:

[ L = \int_{a}^{b} \sqrt{[f'(t)]^2 + [g'(t)]^2} dt ]

Where ( f(t) = f(t) ) and ( g(t) = 2-8t ). Then, we find the derivatives of ( f(t) ) and ( g(t) ) with respect to ( t ).

[ f'(t) = \frac{1}{2\sqrt{t-1}} ] [ g'(t) = -8 ]

Now, plug these derivatives into the arc length formula:

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

[ L = \int_{1}^{3} \sqrt{\frac{1}{4(t-1)} + 64} dt ]

[ L = \int_{1}^{3} \sqrt{\frac{1+256(t-1)}{4(t-1)}} dt ]

[ L = \int_{1}^{3} \sqrt{\frac{256t-251}{4(t-1)}} dt ]

[ L = \int_{1}^{3} \frac{\sqrt{256t-251}}{2\sqrt{t-1}} dt ]

[ L = \frac{1}{2} \int_{1}^{3} \frac{\sqrt{256t-251}}{\sqrt{t-1}} dt ]

This integral can be solved using various techniques, such as substitution or integration by parts. After evaluating the integral, you'll find the arc length of the curve.

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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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