Nebwork AsSignment (1 point) Your neighbor is growing a slightly different watermelon. It also has a rind whose thickness is one tenth of the radius of that watermelon. However, the rind of your neighbor's water melons grows at a constant rate of 20 cubic centimeters a week. The radius of your neighbor's watermelon after 5 weeks is and at that time it is growing at centimeters per week.

To solve the problem, we need to determine the radius of your neighbor's watermelon after 5 weeks and the rate at which it is growing at that time. 1. **Understanding the growth of the watermelon:** - The rind thickness is one-tenth of the radius of the watermelon. - The volume of the watermelon (without the rind) can be expressed as \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the watermelon. 2. **Volume growth:** - The volume of the watermelon is increasing at a constant rate of 20 cubic centimeters per week. - Therefore, after 5 weeks, the total volume increase is: \[ \text{Total Volume Increase} = 20 \, \text{cm}^3/\text{week} \times 5 \, \text{weeks} = 100 \, \text{cm}^3 \] 3. **Initial volume:** - Let’s assume the initial radius of the watermelon is \( r_0 \). The initial volume is: \[ V_0 = \frac{4}{3} \pi r_0^3 \] 4. **Volume after 5 weeks:** - The volume after 5 weeks will be: \[ V = V_0 + 100 = \frac{4}{3} \pi r^3 \] 5. **Setting up the equation:** - We can set up the equation: \[ \frac{4}{3} \pi r^3 = \frac{4}{3} \pi r_0^3 + 100 \] - Dividing through by \( \frac{4}{3} \pi \): \[ r^3 = r_0^3 + \frac{100 \cdot 3}{4 \pi} = r_0^3 + \frac{300}{4 \pi} = r_0^3 + \frac{75}{\pi} \] 6. **Finding the radius after 5 weeks:** - To find the radius \( r \) after 5 weeks, we need to know the initial radius \( r_0 \). However, since we don't have that information, we can express \( r \) in terms of \( r_0 \): \[ r = \sqrt[3]{r_0^3 + \frac{75}{\pi}} \] 7. **Finding the growth rate:** - The volume of the watermelon is related to the radius by the formula \( V = \frac{4}{3} \pi r^3 \). The rate of change of volume with respect to time is given by: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] - Since we know \( \frac{dV}{dt} = 20 \, \text{cm}^3/\text{week} \), we can set up the equation: \[ 20 = 4 \pi r^2 \frac{dr}{dt} \] - Solving for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{20}{4 \pi r^2} = \frac{5}{\pi r^2} \] 8. **Calculating the growth rate at \( r \):** - Substitute the value of \( r \) we found earlier into this equation to find the growth rate at that time. In conclusion, without the initial radius \( r_0 \), we cannot provide a numerical answer for the radius after 5 weeks or the growth rate at that time. If you have the initial radius, please provide it, and we can calculate the specific values.