Answer:
d=54.93
Step-by-step explanation:
From exercise we have y=2x^{3/2} on interval [0,9]
We calculate the arc length of the curve on the given interval.
We have the formula:
d=\int\limits^a_b {\sqrt{1+f'(x)^2}} \, dx
we get
f(x)=y=2x^{3/2}
f'(x)=y'=3x^{1/2}
d=\int\limits^9_0 {\sqrt{1+(3x^{1/2})²}\, dx
d=\int\limits^9_0 {\sqrt{1+9x}}\, dx
d=[\frac{2·(9x+1)^{3/2}}{27}]_0^9
d=\frac{2·(82^{3/2}-2)}{27}
d=54.93
We use the site geogebra.org to drawn a graph.
