Answer:
A) For the area; maximum error is 28.65 cm² while relative error is 0.011
B)For the volume; maximum error is 205.18 cm³ while relative error is 0.239
Step-by-step explanation:
Circumference; C = 2πr
Differentiating both sides with respect to r;
dc/dr = 2π
When r is small, we can write;
Δc/Δr = 2π
Thus, Δr = Δc/2π
Now, we are given that, Δc = 0.5
So, Δr = 0.5/2π = 1/4π
A) For the area, the formula for surface area of sphere is 4πr²
Thus; S(r) = 4πr²
Differentiating both sides with respect to r; ds/dr = 4(2πr)
When r is small, we can write;
Δs/Δr = 4(2πr)
So, Δs = 4(2πr)Δr
From earlier Circumference(C) = 2πr
Thus, Δs = 4CΔr
Now, our Circumference is 90cm and we have established Δr to be 1/4π.
Δs will be maximum when Δr is maximum,
Thus, maximum error in S is;
Δs = 4 x 90 x 1/4π = 90/π = 28.65 cm²
Relative error is given by;
R.E = Δs/s
Now, s = surface area of sphere which 4πr²
We don't have r, so let's attempt simplify it to reflect C.
s = 4π(2πr/2π)² = 4π(C²/4π²) = C²/π
s = 90²/π
Relative Error = Δs/s = (90/π)/(90²/π)
= 1/90 = 0.011
B) For the volume, the formula for volume of a sphere is (4/3)πr³
Thus; V(r) = (4/3)πr³
Differentiating both sides with respect to r; ds/dr = 4πr²
When r is small, we can write;
Δs/Δr = (2πr)²/π
So, Δs = [(2πr)²/π]Δr
From earlier Circumference(C) = 2πr
Thus, Δs = (C²/π)Δr
Now, our Circumference is 90cm and we have established Δr to be 1/4π.
Δv will be maximum when Δr is maximum,
Thus, maximum error in v is;
Δv = (90²/π) x (1/4π) = 8100/4π² = 205.18 cm³
Relative error is given by;
R.E = Δv/v
Now, v = volume of sphere which (4/3)πr³
We don't have r, so let's attempt to simplify it to reflect C.
v = (1/3π)(2πr)² = (1/3π)(C²) = C²/3π
v = 90²/3π = 8
Relative Error = Δv/v = (8100/4π²)/(90²/3π)
= 3/4π = 0.239
