Application of Derivatives: Related Rates (Inflating Balloon, Draining Water on a Tank) - Calculus

Summary

This video explains the concept of related rates using two examples. In the first example, it demonstrates how to calculate the rate at which the radius of a spherical balloon is increasing when the volume is inflating at a specific rate. The second example involves finding the rate at which the height of water is dropping in a cylindrical tank being drained. Both problems involve applying implicit differentiation to solve for unknown rates, highlighting the importance of geometry formulas and understanding the relationships between different quantities over time.

Takeaways

• 😀 A spherical balloon is inflated at a rate of 10 cubic centimeters per second, and the problem is to find how fast the radius is increasing when the radius is 5 cm.
• 😀 The volume of a sphere is given by the formula V = (4/3)πr³, and the task involves related rates to find the rate of change of the radius.
• 😀 Implicit differentiation is used to find the derivative of the volume with respect to time (dV/dt), leading to the equation (dV/dt) = 4πr²(dr/dt).
• 😀 After substituting the given values (V = 10, r = 5), the equation simplifies to 10 = 100π(dr/dt), and solving for dr/dt gives the result 1/(10π).
• 😀 The rate of change of the radius of the balloon is 1/(10π) centimeters per second when the radius is 5 cm.
• 😀 For the second example, a cylindrical tank with a radius of 5 feet is being drained at a rate of 30 cubic feet per second, and the goal is to find how fast the height of the water is decreasing.
• 😀 The formula for the volume of a cylinder is V = πr²h, where r is the radius and h is the height of the water in the tank.
• 😀 Differentiating the volume of the cylinder with respect to time (dV/dt), and applying implicit differentiation, leads to the equation (dV/dt) = 25π(dh/dt).
• 😀 By substituting the given values (dV/dt = -30, r = 5), we find that 25π(dh/dt) = -30, and solving for dh/dt gives the result -6/(5π).
• 😀 The rate of change of the height of the water in the cylindrical tank is -6/(5π) feet per second, indicating that the water height is decreasing over time.
• 😀 These examples highlight the importance of understanding related rates and using implicit differentiation to solve real-world problems involving changing volumes and dimensions.

The video is abnormal, and we are working hard to fix it.
Please replace the link and try again.