Approximate the change in the volume of a night circular cylinder of foxed radius \( \mathrm{r}=21 \mathrm{~cm} \) when its height decreases from \( \mathrm{h}=12 \mathrm{~cm} \) to \( \mathrm{h}=11.9 \mathrm{~cm}\left(\mathrm{~V}(\mathrm{~h})=\pi^{2} \mathrm{~h}\right) \). Use a linear approximation.

Given: - The radius of the circular cylinder is fixed at \( r = 21 \, \text{cm} \). - The height of the cylinder decreases from \( h = 12 \, \text{cm} \) to \( h = 11.9 \, \text{cm} \). - The volume of the cylinder is given by the formula \( V(h) = \pi r^2 h \). To approximate the change in the volume of the cylinder using a linear approximation, we need to find the derivative of the volume function with respect to the height and then use it to approximate the change in volume. 1. Find the derivative of the volume function with respect to the height: \[ \frac{dV}{dh} = \frac{d}{dh}(\pi r^2 h) = \pi r^2 \] 2. Use the linear approximation formula to approximate the change in volume: \[ \Delta V \approx \frac{dV}{dh} \cdot \Delta h \] Substitute the values of \( r \), \( h \), and \( \Delta h \) into the formula to find the approximate change in volume. Calculate the value by following steps: - step0: Calculate: \(\pi \times 21^{2}\left(11.9-12\right)\) - step1: Subtract the numbers: \(\pi \times 21^{2}\left(-0.1\right)\) - step2: Rewrite the expression: \(-\pi \times 21^{2}\times 0.1\) - step3: Multiply the terms: \(-44.1\pi \) - step4: Simplify: \(-138.544236\) The approximate change in the volume of the night circular cylinder when its height decreases from \( h = 12 \, \text{cm} \) to \( h = 11.9 \, \text{cm} \) is approximately \( -138.544236 \, \text{cm}^3 \).