Question

Find the arc length function for the curve y = 2x^3/2 with starting point P0(36, 432).

Answer 1

Answer:

$s\approx17531$

Step-by-step explanation:

When considering a curve defined by a function f(x) and its respective derivative f'(x) that are continuous in an interval [a, b], the length s of the arc delimited by a and b is given by the equation:

$s=\int\limits^a_b {\sqrt{1+(f'(x))^2} } \, dx$

First, let's find the derivate of y(x):

$y'(x)=3\sqrt{x}$

Replacing the derivate in the previous equation:

$s=\int\limits^a_b {\sqrt{1+((3\sqrt{x}) ^2} } \, dx=\int\limits^a_b {\sqrt{1+9x} } \, dx$

Substitute u=1+9x and du=9x, so:

$s= \frac{1}{9} \int\limits^a_b u^{1/2}  \, du$

The antiderivative of $u^{1/2}$ is:

$s=\frac{2u^{3/2} }{27}$

u=1+9x, so:

$s=\frac{2(1+9x)^{3/2} }{27}$

Evaluating the antiderivative at the limits:

$s=\frac{2(1+9x)^{3/2} }{27} \left \{ {{b=432} \atop {a=31}} \right. =\frac{2(3889)^{3/2} }{27}-\frac{2(325)^{3/2} }{27} =17530.82988\approx17531$