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

Question

Luke Alvarez · Accepted Answer

#approx 2.96# units

#s = int_(0)^(2) sqrt(1+(y')^2) \ dx# # = int_(0)^(2) sqrt(1+x^2) \ dx# #= 1/2[ (x sqrt(x^2+1) +sinh^(-1)(x)) ]_(0)^(2)# #= (sqrt(5)+1/2 sinh^(-1)(2) )# units #approx 2.96# units Computer used for integration and numerical solution [The basic integral, # int sqrt(1+x^2) \ dx#... ... can be approached, using the identity #cosh^2 z - sinh^2 z = 1#... ....so by the sub #x = sinh z, dx = cosh z \ dz# ... and then maybe a hyperbolic double angle formula]

Isabella Ackerman · Answer

undefined To find the arc length of the curve $ y = \frac{x^2}{2} $ over the interval $[0, 1]$, you can use the formula for arc length:

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

Here, $ a = 0 $ and $ b = 1 $. First, find $ \frac{dy}{dx} $, then plug it into the formula and integrate over the given interval.

1. Find $ \frac{dy}{dx} $:
$$ \frac{dy}{dx} = x $$

2. Substitute $ \frac{dy}{dx} = x $ into the arc length formula:
$$ L = \int_{0}^{1} \sqrt{1 + x^2} \, dx $$

3. Integrate $ \sqrt{1 + x^2} $ from 0 to 1 to find the arc length.

Emily Ammerman · Answer

undefined To find the arc length of the curve $y = \frac{x^2}{2}$ over the interval $[0, 1]$, we use the formula for arc length of a curve given by:

$$ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx $$

Where $f(x)$ is the given function, and $a$ and $b$ are the lower and upper limits of integration, respectively.

For the given function $y = \frac{x^2}{2}$, we first find its derivative:

$$ f'(x) = x $$

Now, we substitute this derivative into the formula:

$$ L = \int_{0}^{1} \sqrt{1 + x^2} \, dx $$

This integral represents the arc length of the curve $y = \frac{x^2}{2}$ over the interval $[0, 1]$. We evaluate this integral to find the arc length.