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What is the arc length of #f(x)=6x^(3/2)+1 # on #x in [5,7]#?

Answer 1

Emily Ammerman

#L=4/243(568sqrt142-203sqrt406)# units.

#f(x)=6x^(3/2)+1#

#f'(x)=9sqrtx#

Arc length is given by:

#L=int_5^7sqrt(1+81x)dx#

Integrate directly:

#L=2/243[(1+81x)^(3/2)]_5^7#

Hence

#L=4/243(568sqrt142-203sqrt406)#

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Answer 2

Charles Anctil

To find the arc length of ( f(x) = 6x^{3/2} + 1 ) on the interval ([5, 7]), we use the formula for arc length:

[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} , dx ]

where ( a = 5 ) and ( b = 7 ). First, we find ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = \frac{d}{dx}(6x^{3/2} + 1) ] [ = \frac{d}{dx}(6x^{3/2}) ] [ = 9x^{1/2} ]

Now, we substitute ( \frac{dy}{dx} ) into the arc length formula:

[ L = \int_{5}^{7} \sqrt{1 + (9x^{1/2})^2} , dx ] [ = \int_{5}^{7} \sqrt{1 + 81x} , dx ]

This integral can be evaluated using various methods, such as substitution or integration by parts, to find the exact arc length of the function ( f(x) ) on the interval ([5, 7]).

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Answer from HIX Tutor

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