# What is the arc length of #f(x)=sqrt(18-x^2) # on #x in [0,3]#?

##### 2 Answers

The arc length

#### Explanation:

The formula for the arc length

#L = int_a^b sqrt(1+(dy/dx)^2) dx#

First, let's find the derivative of our function, which I'll rename

#y = sqrt(18-x^2)#

#dy/dx = 1/(2sqrt(18-x^2)) * d/dx(18-x^2)#

#dy/dx = 1/(2sqrt(18-x^2)) * (-2x) = (-x)/sqrt(18-x^2)#

Plugging in what we have here...

#L = int_0^3 sqrt(1+(dy/dx)^2) dx#

#L = int_0^3 sqrt(1+((-x)/sqrt(18-x^2))^2) dx#

#L = int_0^3 sqrt(1+(x^2)/(18-x^2)) dx#

#L = int_0^3 sqrt(((18-x^2)+x^2)/(18-x^2)) dx#

#L = int_0^3 sqrt(18/(18-x^2)) dx#

Using the formula

#L = sqrt18 int_0^3 dx/sqrt(18-x^2) = [sqrt18arcsin(x/sqrt18)]_0^3#

#= sqrt18 arcsin(3/sqrt18) - sqrt18 arcsin(0/sqrt18)#

#= sqrt18 arcsin(1/sqrt2) - sqrt18 arcsin(0)#

#= sqrt18 * (pi/4)#

#L = (3pi sqrt2)/4#

*Final Answer*

Alternatively (and I didn't realize this until after I solved it brute-force with calculus), you could just use the fact that this curve is part of a circle with a radius of

Since the ending x-point of the arc is at

#y = sqrt(18-(3)^2) = sqrt(18-9) = sqrt9 = 3#

Since the circle is centered at the origin, and the x and y coordinates of the endpoint of the arc are equal (i.e. the line

Since the starting point is on the y-axis, this means that the arc covers **algebraic** formula for arc length, we can see much more simply that:

#L_"circle" = ("angle") * ("radius") = (pi/4) * (3sqrt2) = (3pi sqrt2)/4#

*Final Answer*