# What is the arclength of #(t-e^(2t),lnt)# on #t in [1,12]#?

The arclength is around

First, let's get a sense of how long the arclength might actually be. It will go from

to

We can tell that this distance will be very big, so that means the arclength is also very big.

To calculate the actual arclength, we'll need to get an integral in the form of

(This link does a great job of explaining arc lengths of parametric curves.)

First, split the parametric function into a function of

Now, differentiate each one and get

Now, plug these into the aforementioned integral with the appropriate bounds. You'll see that the

At this point, you should probably plug this integral into a calculator because the function probably doesn't have an antiderivative, and even if it does it would be a pain to calculate.

A calculator should spit out something around

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To find the arc length of the curve ( (t - e^{2t}, \ln t) ) on the interval ( t \in [1, 12] ), you can use the arc length formula for parametric curves:

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

Given the parametric equations ( x(t) = t - e^{2t} ) and ( y(t) = \ln t ), we need to find ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ) and then evaluate the integral over the given interval.

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

Now, plug these derivatives into the arc length formula:

[ L = \int_{1}^{12} \sqrt{(1 - 2e^{2t})^2 + \left(\frac{1}{t}\right)^2} , dt ]

You can then integrate this expression over the interval ( t \in [1, 12] ) to find the arc length.

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