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How do you find the arc length of the curve #y=1+6x^(3/2)# over the interval [0, 1]?

Answer 1

Christopher Allers

Use the arc length formula.

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

#y'=9sqrtx#

Arc length is given by:

#L=int_0^1sqrt(1+81x)dx#

Integrate directly:

#L=2/243[(1+81x)^(3/2)]_0^1#

Hence

#L=2/243(82sqrt82-1)#

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Answer 2

Emma Aldridge

To find the arc length of the curve ( y = 1 + 6x^{3/2} ) over the interval ([0, 1]), follow these steps:

Compute the derivative of the function ( y = 1 + 6x^{3/2} ) with respect to ( x ).
Use the derivative to find an expression for ( \sqrt{1 + (f'(x))^2} ).
Set up the arc length integral: [ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} , dx ] where ( a ) and ( b ) are the lower and upper limits of the interval, respectively.
Plug in the expression for ( \sqrt{1 + (f'(x))^2} ) and integrate over the interval ([0, 1]).
Evaluate the definite integral to find the arc length.

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Answer from HIX Tutor

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.