Static Fields: Fall-2025
Proportional Reasoning Practice (SOLUTION): Due W5 D3

1. Mass Density S1 5451S Consider a rod of length \(L\) lying on the \(z\)-axis. Find an algebraic expression for the mass density of the rod if the mass density at \(z=0\) is \(\lambda_0\) and at \(z=L\) is \(7\lambda_0\) and you know that the mass density increases

   1. like the square of the distance along the rod;

      In the equation for \(\lambda(z)\), "name the things you don't know". \(\alpha\) is the initial value (for \(z=0\)) and \(\beta\) is the proportionality constant. \begin{align*} \lambda(z) &=\alpha +\beta z^2\\ \lambda(0) &=\alpha +\beta 0^2=\lambda_0\\ &\Rightarrow\alpha=\lambda_0\\ \lambda(L) &=\lambda_0+\beta L^2=7\lambda_0\\ &\Rightarrow\beta=\frac{6\lambda_0}{L^2}\\ \lambda(z) &=\lambda_0\left(1+\frac{6}{L^2} z^2\right) \end{align*}

   2. exponentially.

      In the equation for \(\lambda(z)\), "name the things you don't know". There are two parameters in an exponential function, \(\alpha\) is an overall multiplicative constant. Since the exponential function is equal to one when the argument is zero, \(\alpha\) gives the initial value (for \(z=0\)). The exponential function also has a parameter \(k\) that makes the exponent dimensionless. \begin{align*} \lambda(z) &=\alpha\, e^{kz}\\ \lambda(0) &=\alpha\, e^{k0}=\lambda_0\\ &\Rightarrow \alpha=\lambda_0\\ \lambda(z) &=\lambda_0\, e^{kz}\\ \lambda(L) &=\lambda_0\, e^{kL}=7\lambda_0\\ &\Rightarrow e^{kL}=7\\ &\Rightarrow kL=\ln 7\\ &\Rightarrow k=\frac{\ln 7}{L}\\ \lambda(z) &=\lambda_0\, e^{z\ln 7 /L} \end{align*}