# Find the exact length of the curve. x = et + e−t, y = 5 − 2t, 0 ≤ t ≤ 4

This question aims to find the length of the curve by applying line integral along the curve.

It is difficult to find the exact equation of the function along the curve so we need a certain formula to find the exact measurements. Line integral solves this problem as it is a type of integration that is performed on the functions present along the curve.

The line integral along the curve is also called path integral or curve integral. It can be found by finding the sum of all the points present on the curve with some differential vector along the curve.

The values of x and y are given and these are:

$x = e^t + e^{- t}$

$y = 5 – 2t$

The limits are as follows:

$0 \leq t \leq 4$

By using the formula to find the length $l$ of the curve:

$L = \int_{a}^{b} \sqrt { (\frac { dx } { dt } ) ^ 2 + (\frac { dy } { dt } ) ^ 2 } \, dt$

$\frac{dx}{dt} = e^t – e^{-t}$

$\frac{dy}{dt} = -2$

$L = \int_{0}^{4} \sqrt { ( e ^ t – e ^ {-t} ) ^ 2 + ( – 2 ) ^ 2 } \, dt$

$L = \int_{0}^{4} \sqrt { e ^ 2t – 2 + e ^ {-2t} + 4 } \, dt$

$L = \int_{0}^{4} \sqrt { ( e ^ t – e ^ {-t} ) ^ 2 } \, dt$

$L = \int_{0}^{4} \sqrt { e ^ t – e ^ {-t} } \, dt$

$L = [ e ^ t – e ^ { -t } ] ^ { 4 } _ {0} dt$

$L = e ^ 4 – e ^ { -4 } – e ^ 0 + e ^ 0$

$L = e ^ 4 – e ^ { -4 }$

## Numerical Results

The length $L$ of the curve is $e ^ 4 – e ^ { -4 }$.

## Example

Find the length of the curve if the limits are $$0 \leq t \leq 2$. $L = \int_{a}^{b} \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$ $\frac{dx}{dt} = e^t – e^{-t}$ $\frac{dy}{dt} =- 2$ $L = \int_{0}^{2} \sqrt { ( e ^ t – e ^ {-t} )^2 + (-2)^2}\, dt$ $L = \int_{0}^{2} \sqrt {e^2t – 2 + e^{-2t} + 4 }\, dt$ $L = \int_{0}^{2} \sqrt {(e^t – e^{-t} )^2 }\, dt$ $L = \int_{0}^{2} \sqrt { e ^ t – e ^ {-t} } \, dt$ By putting the limits: $L = e ^ 2 – e ^ { -2 } – e ^ 0 + e ^ 0$ $L = e ^ 2 – e ^ { -2 }$ The length$ L $of the curve is$ e ^ 2 – e ^ { -2} \$

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