The surface area, S, of a sphere of radius r feet is S = S(r) = 4πr2. Find the instantaneous rate of change of the surface area with respect to the radius r at r = 2. (4 points)

s = 4π r^2

ds/dr = 4π(2r) = 8πr

To find the instantaneous rate of change of the surface area with respect to the radius at r = 2, we need to find the derivative of the surface area function with respect to r and evaluate it at r = 2.

The surface area function is given by S(r) = 4πr^2.

To find the derivative of S(r), we can use the power rule for derivatives. According to the power rule, if f(x) = x^n, then the derivative of f(x) with respect to x is given by f'(x) = nx^(n-1).

Applying the power rule, we get:

S'(r) = dS(r)/dr = d(4πr^2)/dr = 8πr

Now, we can evaluate the derivative at r = 2:

S'(2) = 8π(2) = 16π

Therefore, the instantaneous rate of change of the surface area with respect to the radius at r = 2 is 16π.

To find the instantaneous rate of change of the surface area with respect to the radius at r = 2, we need to take the derivative of the surface area function with respect to r and evaluate it at r = 2.

Let's find the derivative first:

S(r) = 4πr^2

To find dS/dr, we can use the power rule for differentiation. According to the power rule:

d/dx(x^n) = nx^(n-1)

Applying this rule to our function S(r), where n = 2:

dS/dr = 2 * 4πr^(2-1)
= 8πr

Now we have the derivative of S(r) with respect to r. To find the instantaneous rate of change of the surface area at r = 2, we substitute r = 2 into the derivative:

dS/dr = 8π(2)
= 16π

Therefore, the instantaneous rate of change of the surface area with respect to the radius at r = 2 is 16π.

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