[amp_mcq option1=”0.27″ option2=”0.67″ option3=”1″ option4=”1.22″ correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{B) }0.67}$.<\/p>\n
The formula for the arc length of a curve $y=f(x)$ over the interval $[a,b]$ is:<\/p>\n
$$L = \\int_a^b \\sqrt{1+\\left(\\frac{dy}{dx}\\right)^2} dx$$<\/p>\n
In this case, we have $y=\\frac{2}{3}x^{3\/2}$ and $a=0$ and $b=1$. Substituting these values into the formula, we get:<\/p>\n
$$L = \\int_0^1 \\sqrt{1+\\left(\\frac{d}{dx}\\left(\\frac{2}{3}x^{3\/2}\\right)\\right)^2} dx = \\int_0^1 \\sqrt{1+\\left(\\frac{2}{3}\\cdot\\frac{3}{2}x^{^{-1\/2}}\\right)^2} dx = \\int_0^1 \\sqrt{1+\\frac{4}{9}x^{-1}} dx$$<\/p>\n
We can now evaluate this integral using the substitution $u=\\frac{1}{x}$. This gives us:<\/p>\n
$$L = \\int_0^1 \\sqrt{1+\\frac{4}{9}x^{-1}} dx = \\int_1^\\infty \\sqrt{1+\\frac{4}{9}u^2} du = \\left.\\frac{2}{3}u^{3\/2}\\sqrt{1+\\frac{4}{9}u^2}\\right|_1^\\infty = \\frac{2}{3}\\sqrt{1+\\frac{4}{9}} – \\frac{2}{3} = 0.67$$<\/p>\n
Therefore, the length of the curve $y=\\frac{2}{3}x^{3\/2}$ between $x=0$ and $x=1$ is $\\boxed{\\text{B) }0.67}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0.27″ option2=”0.67″ option3=”1″ option4=”1.22″ correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20193","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n