$$\text { Find the center of mass of the wire that lies along the curve } r \text { and has density } \delta \text {. }$$
-$$r ( t ) = \left( 5 t ^ { 2 } - 1 \right) \mathbf { i } + 4 t k , - 1 \leq t \leq 1 ; \delta ( x , y , z ) = 3 \sqrt { 20 x + 36 }$$
A) $( 464,0,0 )$
B) $( 464,2,0 )$
C) $\left( \frac { 58 } { 37 } , 0,0 \right)$
D) $\left( \frac { 58 } { 37 } , 2,0 \right)$

Question

Quizplus · Accepted Answer

The center of mass of a wire in three dimensions can be found using the formulas for the coordinates of the center of mass ($x_{cm}, y_{cm}, z_{cm}$), which involve integrating the position vectors weighted by the density function over the length of the curve. Given the parametric representation of the curve $r(t) = (5t^2 - 1)\mathbf{i} + 4t\mathbf{k}$ and the density function $\delta(x, y, z) = 3\sqrt{20x + 36}$, we can calculate the center of mass by integrating over the parameter $t$ from $-1$ to $1$. The correct answer must account for the density function and the given curve, leading to the calculation of the integrals for $x_{cm}, y_{cm}, z_{cm}$. The specific values in the options suggest that detailed integration and application of the density function were performed to find the center of mass coordinates. The correct answer involves correctly applying these principles and performing the necessary calculations, which leads to the specific numerical result given in option C.

Find the center of mass of the wire that lies along the curve r and has density δ. \text { Find the center of mass of the wire that lies along the curve } r \text { and has density } \delta \text {. } Find the center of mass of the wire that lies along the curve r and has density δ.

$\text { Find the center of mass of the wire that lies along the curve } r \text { and has density } \delta \text {. }$