question the velocity vector of a particle moving in the xy - plane has components given by dx/dt=-5 ln(8t +…

question the velocity vector of a particle moving in the xy - plane has components given by dx/dt=-5 ln(8t + 3) and dy/dt=-7 ln(8t² + 1). find the distance traveled by the particle along the path from t = 6 to t = 7? use a calculator and round to three decimal places if necessary.

question the velocity vector of a particle moving in the xy - plane has components given by dx/dt=-5 ln(8t + 3) and dy/dt=-7 ln(8t² + 1). find the distance traveled by the particle along the path from t = 6 to t = 7? use a calculator and round to three decimal places if necessary.

Answer

Explanation:

Step1: Recall arc - length formula

The distance $s$ traveled by a particle with velocity components $x'(t)$ and $y'(t)$ from $t = a$ to $t = b$ is given by $s=\int_{a}^{b}\sqrt{(x'(t))^{2}+(y'(t))^{2}}dt$. Here, $x'(t)=\frac{dx}{dt}=- 5\ln(8t + 3)$ and $y'(t)=\frac{dy}{dt}=-7\ln(8t^{2}+1)$, $a = 6$, $b = 7$. So, $s=\int_{6}^{7}\sqrt{(-5\ln(8t + 3))^{2}+(-7\ln(8t^{2}+1))^{2}}dt=\int_{6}^{7}\sqrt{25\ln^{2}(8t + 3)+49\ln^{2}(8t^{2}+1)}dt$.

Step2: Use a calculator

Use a graphing calculator or a numerical - integration tool (such as the integral function on a TI - 84 Plus: $\int_{a}^{b}f(x)dx$). Enter the function $y=\sqrt{25\ln^{2}(8x + 3)+49\ln^{2}(8x^{2}+1)}$, $a = 6$, and $b = 7$.

Answer:

$s\approx13.197$