ASK YOUR TEACHER PRACTICE ANOTHER When a certain polyatomic gas undergoes adiabatic expansion, its pressure \( p \) and volume \( V \) satisfy the equation \( p V^{1.6}=k \), where \( k \) is a constant. Find the relationship between the related rates \( \frac{d p}{d t} \) and \( \frac{d^{\prime} V}{d t} \) in terms of \( p \) and \( V \). \[ \frac{d V}{d t}=(\square) \frac{d p}{d t} \]

To find the relationship between the rates of change of pressure and volume in terms of \( p \) and \( V \), we start from the given equation for pressure and volume, \[ p V^{1.6} = k. \] Differentiating both sides with respect to time \( t \), we apply the product rule: \[ \frac{d}{dt}(p) \cdot V^{1.6} + p \cdot \frac{d}{dt}(V^{1.6}) = 0. \] For \( \frac{d}{dt}(V^{1.6}) \), we use the chain rule: \[ \frac{d}{dt}(V^{1.6}) = 1.6 V^{0.6} \frac{dV}{dt}. \] Substituting this back into our differentiation gives: \[ \frac{dp}{dt} V^{1.6} + p \cdot 1.6 V^{0.6} \frac{dV}{dt} = 0. \] Rearranging the equation to isolate \(\frac{dV}{dt}\): \[ p \cdot 1.6 V^{0.6} \frac{dV}{dt} = -\frac{dp}{dt} V^{1.6}. \] Thus, \[ \frac{dV}{dt} = -\frac{V^{1.6}}{p \cdot 1.6 V^{0.6}} \frac{dp}{dt} = -\frac{V}{1.6 p} \frac{dp}{dt}. \] Therefore, we can express the relationship as: \[ \frac{d V}{d t} = -\frac{V}{1.6 p} \frac{d p}{d t}. \] So the final expression is: \[ \frac{d V}{d t} = -\frac{V}{1.6 p} \frac{d p}{d t}. \] In the form requested, we can fill in the square: \[ \frac{d V}{d t} = -\frac{V}{1.6 p} \frac{d p}{d t}. \]