Question

Find the length of the curve y=x³/12+1/x, 1 less than or equal to x less than or equal to 4.
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Answer 1

Given,

y=x³/12+1/x

We have to find the length of the curve.

Solution :

we know that,

If we want to find the arc length(l) of the graph of y=f(x) from x=a to x=b, then it can be found by

$\sf \: L = \displaystyle \int _a^b \rm \: \sqrt{1 + {( \frac{dy}{dx} })^{2} } \: \: dx$

Now,

at first we have to find the 1st derivatives of y

$\rm \: \frac{dy}{dx} = \frac{d}{dx} ( \frac{ {x}^{3} }{ 12} + \frac{1}{x} ) \\ \\ = 3 \times \frac{ {x}^{2} }{12} - \frac{1}{ {x}^{2} } \\ \\ = \frac{ {x}^{2} }{4} - \frac{1}{ {x}^{2} }$

Now,

let the length of the curve be = L

$\sf \: L = \displaystyle \int _1^4 \rm \: \sqrt{1 + {( \frac{dy}{dx} })^{2} } \: \: dx \\ \\ \rm = \displaystyle \int _1^4 \rm \: \sqrt{1 + ( { \frac{ {x}^{2} }{4} - \frac{1}{ {x}^{2} } })^{2} } \: \: dx \\ \\ = \displaystyle \int _1^4 \rm \: \sqrt{1 + \frac{ {x}^{4} }{16} - \frac{1}{2} + \frac{1}{ {x}^{4} } } \: \: dx \\ \\ \rm = \displaystyle \int _1^4 \rm \: \sqrt{ {( \frac{ {x}^{4} }{16} + \frac{1}{2} + \frac{1}{ {x}^{4} } })^{2} } \: \: dx \\ \\ \rm = \displaystyle \int _1^4 \rm \: \sqrt{ {( \frac{ {x}^{2} }{4} + \frac{1}{ {x}^{2} } })^{2} } \: \: dx \\ \\ \rm = \displaystyle \int _1^4 \rm \: ( \frac{ {x}^{2} }{4} + \frac{1}{ {x}^{2} } )\: \: dx \\ \\ \rm = [ \frac{ {x}^{3} }{12} - \frac{1}{x} ]_1^4 \\ \\ = ( \frac{ {4}^{3} }{12} - \frac{1}{4} ) - ( \frac{ {1}^{3} }{12} - \frac{1}{1} ) \\ \\ = \frac{64}{12} - \frac{1}{4} - \frac{1}{12} + 1 \\ \\ = 4$

So the length of the curve is = 4 unit