Chapter 3.8, Problem 17E

(a)

To determine

To find: The linear density at $x=1\text{m}$.

Answer to Problem 17E

The linear density at $x=1\text{m}$ is, $\rho \left(1\right)=6\text{kg/m}$.

Explanation of Solution

Given:

The mass of the part of the metal rod which lies between its left end and a point x m away from the right end is $3x^2\text{kg}$.

Calculation:

The mass of the metal rod is, $m\left(x\right)=3x^2$.

The linear density of the rod is the derivative of the mass and hence $\rho \left(x\right)=6x$.

The linear density at $x=1\text{m}$ is $\rho \left(1\right)=6\text{kg/m}$.

(b)

To determine

To find: The linear density at $x=2\text{m}$.

Answer to Problem 17E

The linear density at $x=2\text{m}$ is, $\rho \left(2\right)=12\text{kg/m}$.

Explanation of Solution

From part (a), the linear density of the rod is, $\rho \left(x\right)=6x$.

The linear density at $x=2\text{m}$ is,

$\begin{array}{c}\rho \left(2\right)=6\left(2\right)\\ =12\text{kg/m}\end{array}$

The linear density at $x=2\text{m}$ is, $\rho \left(2\right)=12\text{kg/m}$.

(c)

To determine

To find: The linear density at $x=3\text{m}$; and to tell the conclusion obtained from the parts (a), (b) and (c).

Answer to Problem 17E

The linear density at $x=3\text{m}$ is, $\rho \left(3\right)=18\text{kg/m}$.

Explanation of Solution

From part (a), the linear density of the rod is, $\rho \left(x\right)=6x$.

The linear density at $x=3\text{m}$ is, $\rho \left(3\right)=18\text{kg/m}$.

Notice that the density function is an increasing function. Therefore, the density of the rod is highest at its right end and lowest at its left end.