A spherical balloon is filled with air at the constant rate of 200cm2s.Calculate the rate at which the radius is increasing when the radius is10cm.1

The problem involves the concept of related rates in calculus. Here's how you can solve it:

Step 1: Understand the problem The problem is asking for the rate at which the radius of the balloon is increasing (dr/dt) when the radius is 10 cm. We know that the balloon is being filled at a constant rate of 200 cm^3/s.

Step 2: Use the formula for the volume of a sphere The volume V of a sphere with radius r is given by the formula V = 4/3πr^3.

Step 3: Differentiate both sides with respect to time t We get dV/dt = 4πr^2 * dr/dt.

Step 4: Substitute the given values into the equation We know that dV/dt = 200 cm^3/s and we want to find dr/dt when r = 10 cm. Substituting these values into the equation gives us 200 = 4π(10)^2 * dr/dt.

Step 5: Solve for dr/dt Solving the equation for dr/dt gives us dr/dt = 200 / (4π(10)^2) = 0.159 cm/s.

So, the radius of the balloon is increasing at a rate of

The problem involves the concept of related rates in calculus. Here's how you can solve it:

Step 1: Understand the problem The problem is asking for the rate at which the radius of the balloon is increasing (dr/dt) when the radius is 10 cm. We know that the balloon is being filled at a constant rate of 200 cm^3/s.

Step 2: Use the formula for the volume of a sphere The volume V of a sphere with radius r is given by the formula V = 4/3πr^3.

Step 3: Differentiate both sides with respect to time t We get dV/dt = 4πr^2 * dr/dt.

Step 4: Substitute the given values We know that dV/dt = 200 cm^3/s and we want to find dr/dt when r = 10 cm. Substituting these values into the equation gives us 200 = 4π(10)^2 * dr/dt.

Step 5: Solve for dr/dt Solving the equation for dr/dt gives us dr/dt = 200 / (4π(10)^2) = 0.159 cm/s.

So, the radius of the balloon is increasing at a rate of approximately 0.159 cm/s when the radius is 10 cm.