$$\text { Find the required quantity given the wire that lies along the curve } r \text { and has density } \delta .$$
-Moment of inertia $\mathrm { I } _ { \mathrm { y } }$ about the $\mathrm { y }$-axis, where $\mathbf { r } ( \mathrm { t } ) = ( 4 - 4 \mathrm { t } ) \mathrm { i } + 3 \mathrm { t } \mathrm { j } , 0 \leq \mathrm { t } \leq 1 ; \delta ( \mathrm { x } , \mathrm { y } , \mathrm { z } ) = 4$
A) $\mathrm { I } _ { \mathrm { y } } = - \frac { 1600 } { 3 }$
B) $\mathrm { I } _ { \mathrm { y } } = - \frac { 320 } { 3 }$
C) $\mathrm { I } _ { \mathrm { y } } = \frac { 320 } { 3 }$
D) $\mathrm { I } _ { \mathrm { y } } = \frac { 1600 } { 3 }$

Question

Quizplus · Accepted Answer

The moment of inertia $I_y$ about the y-axis for a wire with density $\delta$ is given by the integral $$I_y = \int (x^2 \delta) ds,$$ where $ds$ is the element of arc length along the curve. Given $\mathbf{r}(t) = (4-4t)\mathbf{i} + 3t\mathbf{j}$ and $\delta(x, y, z) = 4$, we first find $ds$. The derivative of $\mathbf{r}(t)$ with respect to $t$ is $\mathbf{r}'(t) = -4\mathbf{i} + 3\mathbf{j}$, so the magnitude of $\mathbf{r}'(t)$ is $\sqrt{(-4)^2 + 3^2} = 5$. Thus, $ds = 5dt$. The x-coordinate squared is $(4-4t)^2$, so the integral becomes $$I_y = \int_0^1 4(4-4t)^2 \cdot 5 dt = 80\int_0^1 (1-t)^2 dt.$$ Solving this integral gives $$80\left[\frac{(1-t)^3}{3}\right]_0^1 = 80\left[\frac{1}{3}\right] = \frac{80}{3} \times 4 = \frac{320}{3}.$$

Find the required quantity given the wire that lies along the curve r and has density δ.\text { Find the required quantity given the wire that lies along the curve } r \text { and has density } \delta . Find the required quantity given the wire that lies along the curve r and has density δ.

$\text { Find the required quantity given the wire that lies along the curve } r \text { and has density } \delta .$