Question

ECUACIONES DE LONGITUD DE ARCO RESOLVER Y=3/2x^1/2 cuando x=0 hasta x=5

Answer 1

Veamos una forma paramétrica

$x(t)=t^2\;,\; y(t)=\dfrac{3}{2}t\;\; ,\; \text{donde }0\leq t\leq \sqrt5\\ \\ \\ \displaystyle L=\int_{0}^{\sqrt{5}}\sqrt{x_t^2+y_t^2}\;dt\\ \\ \\ L=\int_{0}^{\sqrt{5}}\sqrt{(2t)^2+\dfrac{9}{4}}\;dt\\ \\ \\ L=\int_{0}^{\sqrt{5}}\sqrt{4t^2+\dfrac{9}{4}}\;dt$

$\displaystyle L=2\int_{0}^{\sqrt{5}}\sqrt{t^2+\dfrac{9}{16}}\;dt\\ \\ \\ \text{Por tabla: }\\\\ \boxed{\int\sqrt{x^2+a^2}\,dx=\dfrac{x\sqrt{x^2+a^2}}{2}+\frac{a^2\ln(x+\sqrt{x^2+a^2})}{2}+C}\\ \\ \\ \text{Entonces:}\\ \\ L=2\left.\left(\dfrac{t\sqrt{t^2+9/16}}{2}+\frac{9/16\ln(t+\sqrt{t^2+9/16})}{2}\right)\right|_{0}^{\sqrt{5}}\\ \\ \\ \boxed{\boxed{L=\frac{9}{16}\ln\left(\frac{\sqrt{89}}{3}+\frac{4\sqrt{5}}{3}\right)+\frac{\sqrt{445}}{4}}}$