Question

Air is being pumped into a spherical balloon so that its volume increases at a rate of 60cm³/s. How fast is the surface area of the balloon Increasing when its radius is 14cm? Recall that a ball of radius r has volume V= Tr and surface area S- 4x3. 3​

Answer 1

8.57 cm²/s

Explanation:

To determine how fast the surface area (S) of a spherical balloon is increasing when its radius (r) is 14 cm, we need to find dS/dt when r = 14.

Given that the volume of the spherical balloon is increasing at a rate of 60 cm³/s, then:

`\frac{\text{d}V}{\text{d}t}=60\; \sf cm^3/s`

The volume (V) of a sphere with radius (r) is given by the formula:

`V = (4)/(3) \pi r^3`

Rearrange to isolate r:

`r=\sqrt[3]{(3V)/(4\pi)}`

If S is the surface area of the spherical balloon, then:

`S = 4\pi r^2`

Substitute the expression for r into the surface area equation:

`S = 4\pi \left(\sqrt[3]{(3V)/(4\pi)}\right)^2`

`S = 4\pi \left((3V)/(4\pi)\right)^{(2)/(3)}`

Differentiate S with respect to V:

`\frac{\text{d}S}{\text{d}V}=4\pi \cdot (2)/(3)\left((3V)/(4\pi)\right)^{-(1)/(3)}\cdot (3)/(4\pi)`

`\frac{\text{d}S}{\text{d}V}=2\left((3V)/(4\pi)\right)^{-(1)/(3)}`

Write out the chain rule for dS/dt using the information available:

`\frac{\text{d}S}{\text{d}t}=\frac{\text{d}S}{\text{d}V}}* \frac{\text{d}V}{\text{d}t}`

Substitute dS/dV and dV/dt to create an equation for the rate of change of the surface area of the spherical balloon at any value of r:

`\frac{\text{d}S}{\text{d}t}=2\left((3V)/(4\pi)\right)^{-(1)/(3)}* 60`

`\frac{\text{d}S}{\text{d}t}=120\left((3V)/(4\pi)\right)^{-(1)/(3)}`

Substitute the expression for r in terms of V:

`\frac{\text{d}S}{\text{d}t}=120\left((3\left((4)/(3) \pi r^3\right))/(4\pi)\right)^{-(1)/(3)}`

`\frac{\text{d}S}{\text{d}t}=(120)/(r)`

To calculate how fast the surface area of the balloon is increasing at the instant the radius is 14 cm, substitute r = 14 into dS/dt:

`\frac{\text{d}S}{\text{d}t}=(120)/(14)`

`\frac{\text{d}S}{\text{d}t}=8.57\; \sf cm^2/s`

Therefore, the surface area of the balloon is increasing at a rate of 8.57 cm²/s when its radius is 14 cm.