Respuesta :
The surface area of a sphere of radius r is
A(r) = 4πr²
The rate of change of the surface area with respect to time is
[tex] \frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} [/tex]
The radius increases at the constant rate of 2 cm/min, therefore
[tex] \frac{dA}{dt} = 2 \frac{dA}{dr}=2*(8 \pi r) =16 \pi r [/tex]
When r = 100 cm, the rate of change of the surface area is
16π(100) cm²/min
= 1600π cm²/min
= 5026.5 cm²/min
Answer: 1600π or 5026.5 cm²/min
A(r) = 4πr²
The rate of change of the surface area with respect to time is
[tex] \frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} [/tex]
The radius increases at the constant rate of 2 cm/min, therefore
[tex] \frac{dA}{dt} = 2 \frac{dA}{dr}=2*(8 \pi r) =16 \pi r [/tex]
When r = 100 cm, the rate of change of the surface area is
16π(100) cm²/min
= 1600π cm²/min
= 5026.5 cm²/min
Answer: 1600π or 5026.5 cm²/min
