Suppose you have a cube of length \( s \). The volume of that cube is \[ V=s^{3} \text {. } \] Now ler's suppose the dimensions of that cube (and nence is volume) depend on time. We are wondering about the relationship between the growth of the length versus the growth of the volum Suppose \[ s(t)=t \text { and } V^{\prime}(t)=\square \] Then \( s^{\prime}(t)=\square \) Next, suppose \[ V^{\prime}(t)=t . \] Then \( a^{\prime}(t)=\square V^{\prime}(t)=\square \)

To analyze the relationship between the growth of the length of the cube and the growth of its volume, we start with the given information. 1. The volume of the cube is given by: \[ V = s^3 \] 2. If the side length of the cube is a function of time, \( s(t) = t \), then we can find the volume as a function of time: \[ V(t) = (s(t))^3 = t^3 \] 3. To find the derivative of the volume with respect to time, we differentiate \( V(t) \): \[ V'(t) = \frac{d}{dt}(t^3) = 3t^2 \] So, we have: \[ V'(t) = 3t^2 \] 4. Now, we need to find the derivative of the side length \( s(t) \): \[ s'(t) = \frac{d}{dt}(t) = 1 \] Now we have: - \( V'(t) = 3t^2 \) - \( s'(t) = 1 \) Next, let's consider the case where we are given that \( V'(t) = t \). 5. If \( V'(t) = t \), we can integrate this to find \( V(t) \): \[ V(t) = \int t \, dt = \frac{t^2}{2} + C \] (where \( C \) is a constant of integration). 6. Since \( V = s^3 \), we can set \( V(t) = s(t)^3 \). Thus, we have: \[ s(t)^3 = \frac{t^2}{2} + C \] 7. To find \( s'(t) \), we differentiate \( s(t)^3 \) with respect to \( t \): \[ 3s(t)^2 s'(t) = \frac{d}{dt}\left(\frac{t^2}{2} + C\right) = t \] 8. Solving for \( s'(t) \): \[ s'(t) = \frac{t}{3s(t)^2} \] Now, we have: - \( V'(t) = t \) - \( s'(t) = \frac{t}{3s(t)^2} \) In summary, we have established the relationships: - If \( s(t) = t \), then \( V'(t) = 3t^2 \) and \( s'(t) = 1 \). - If \( V'(t) = t \), then \( s'(t) = \frac{t}{3s(t)^2} \).