Answer :
Although part of your question is missing, you might be referring to this full question: The radius of a spherical ball is increasing at a rate of 2 cm/min. At what rate is the surface area of the ball increasing when the radius is 8 cm? The surface area of a sphere with radius r is given by 4πr².
The rate at which the surface area of the ball increasing when the radius is 8 cm is 128 cm²/min.
Given A = 4πr²
So, we have:
dA/dt = d/dt (4πr²)
= 4π * d/dt * (r²)
= 4π * 2r * dr/dt
= 8πr * dr/dt
Now, we have that dr/dt = 2 cm/min. So, when the radius of the ball is 8 cm, we have that:
dA/dt = 8πr * dr/dt
= 8π * 8 * 2
= 128π
Since we are measuring area, our unit is cm²/min, not cm/min.
Thus, the rate at which the surface area of the ball increasing when the radius is 8 cm is 128 cm²/min.
Learn more about surface area of sphere at: https://brainly.com/question/14692728
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