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What is the arc length of #f(x)=10+x^(3/2)/2# on #x in [0,2]#?

Answer 1

Samantha Adkison

#L=1/27(17sqrt34-32)# units.

#f(x)=10+1/2x^(3/2)#

#f'(x)=3/4sqrtx#

Arc length is given by:

#L=int_0^2sqrt(1+9/16x)dx#

Integrate directly:

#L=32/27[(1+9/16x)^(3/2)]_0^2#

Insert the limits of integration:

#L=32/27((17/8)^(3/2)-1)#

Simplify:

#L=1/27(17sqrt34-32)#

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Answer 2

Adam Adkins

To find the arc length of the function ( f(x) = 10 + \frac{x^\frac{3}{2}}{2} ) on the interval ( x \in [0,2] ), we use the formula for arc length:

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

First, find ( \frac{dy}{dx} ):

[ f'(x) = \frac{3}{4}x^\frac{1}{2} ]

Now, substitute into the arc length formula:

[ L = \int_{0}^{2} \sqrt{1 + \left(\frac{3}{4}x^\frac{1}{2}\right)^2} dx ]

[ = \int_{0}^{2} \sqrt{1 + \frac{9}{16}x} dx ]

Now, integrate:

[ = \int_{0}^{2} \sqrt{1 + \frac{9}{16}x} dx ]

[ = \frac{32}{27} \left(10 \sqrt{\frac{25}{36}} + 3 \log{\left(2 \sqrt{\frac{16}{25}} + 5\right)}\right) ]

[ \approx 7.0785 ]

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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.