The diameter of a shpere is measured to be 6 inches with a possible error of 0.05 inches. Use differentials to estimate the maximum error in the calculated surface area.
I've gotten this far:
A=4pir^2
dA=8(pi)(r)(dr)
Since the problem gives me the diameter and the equation A=4pir^2 uses the radius, do I divide both the radius and dr (.05) by two, giving me:
dA=8(pi)(3)(.025)
dA=1.88
the error (in D) is ±.05
A = π D²
dA = 2 π D dD = .6 π
same difference...
You're on the right track! However, there are a couple of adjustments you need to make.
First, since the diameter is given as 6 inches, you correctly recognized that you need to divide it by 2 to get the radius. So the radius, in this case, is 6/2 = 3 inches.
Secondly, you have the formula for the differential of the surface area correct: dA = 8πr(dr).
To estimate the maximum error in the calculated surface area, you need to find the maximum value that dr could be. Here, the possible error in the diameter is given as 0.05 inches. So the possible error in the radius, dr, is half of that, which is 0.05/2 = 0.025 inches.
Plugging in the numbers, you get:
dA = 8π(3)(0.025)
dA = 0.6π
Therefore, the maximum error in the calculated surface area is approximately 0.6π square inches.