High School

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Air is being pumped into a spherical balloon so that its volume increases at a rate of 60c(m³)/s. How fast is the surface area of the balloon increasing when its radius is 18cm?

Answer :

The surface area of a spherical balloon increases at a rate of 480 cm²/s when the volume is increasing at 60 cm³/s and the radius is 18 cm.

When air is being pumped into a spherical balloon, and its volume is increasing at a rate of 60 cm3/s, we can determine how fast the surface area is increasing using related rates in calculus.

The volume of a sphere is given by V = rac{4}{3}

ight)

ight) and r for the radius.

Since we're given dV/dt, the rate at which the volume is increasing, we want to find dA/dt, the rate at which the surface area is increasing, when the radius is 18 cm.

Step 1: Write down the known rates and the equations for volume and surface area of a sphere.
V = rac{4}{3}
)

Step 2: Differentiate the volume equation with respect to time (t) to find the relationship between the change in radius and time.
dV/dt = 4
ight)

Thus, the surface area of the balloon is increasing at 480 cm2/s when the radius is 18 cm.

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