What is the arclength of #f(x)=x+xsqrt(x+3)# on #x in [-3,0]#?

Question

Cameron Alexander · Accepted Answer

#s~~6.54696#

The formula for arc length of a curve defined by #f(x)# is:

#s=int_a^bsqrt(1+f'(x)^2)#

First let's find the derivative #f'(x)#:

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

#f'(x)=1+(x+3)^(1/2)+1/2x(x+3)^(-1/2)# (by the product rule)

#f'(x)=1+sqrt(x+3)+x/(2sqrt(x+3))#

Now let's evaluate #f'(x)^2#

#f'(x)^2=[1+sqrt(x+3)+x/(2sqrt(x+3))]^2#

#=1+sqrt(x+3)+x/(2sqrt(x+3))+sqrt(x+3)+(x+3)+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x/(2sqrt(x+3))+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x^2/(4(x+3))#

#=1+2sqrt(x+3)+cancel(2)x/(cancel(2)sqrt(x+3))+(x+3)+cancel(2)(x/cancel(2))+x^2/(4(x+3))#

#=x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+4#

Now you would need to evaluate this (very nasty) integral:

#int_-3^0sqrt(x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+5)#

I punched it into Wolfram|Alpha and got the following result:

So the answer is

#s~~6.54696#

Elijah Abram · Answer

undefined To find the arc length of the function $ f(x) = x + x\sqrt{x+3} $ on the interval $[-3, 0]$, we need to use the arc length formula:

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

where $ a $ and $ b $ are the lower and upper limits of the interval, and $ \frac{dy}{dx} $ is the derivative of $ f(x) $ with respect to $ x $.

First, find $ \frac{dy}{dx} $:

$$ f'(x) = 1 + \frac{d}{dx} (x\sqrt{x+3}) $$

Using the product rule:

$$ f'(x) = 1 + \sqrt{x+3} + x \left(\frac{1}{2\sqrt{x+3}}\right) $$

$$ f'(x) = 1 + \sqrt{x+3} + \frac{x}{2\sqrt{x+3}} $$

Now, plug $ f'(x) $ into the arc length formula and integrate over the interval $[-3, 0]$:

$$ L = \int_{-3}^{0} \sqrt{1 + \left(1 + \sqrt{x+3} + \frac{x}{2\sqrt{x+3}}\right)^2} \, dx $$

This integral can be solved numerically.