Question

The circumference of a sphere was measured to be 82 cm with a possible error of 0.5 cm.

Answer 1

Answer:

A) Maximum error = 170.32 cm³

B)Relative error = 0.0575

Step-by-step explanation:

A) Formula for circumference is: C = 2πr

Differentiating with respect to r, we have;

dC/dr = 2π

r is small, so we can write;

ΔC/Δr = 2π

So, Δr = ΔC/2π

We are told that ΔC = 0.5.

Thus; Δr = 0.5/2π = 0.25/π

Now, formula for Volume of a sphere is;

V(r) = (4/3)πr³

Differentiating with respect to r, we have;

dV/dr = 4πr²

Again, r is small, so we can write;

ΔS/Δr = 4πr²

ΔV = 4πr² × Δr

Rewriting, we have;

ΔV = ((2πr)²/π) × Δr

Since C = 2πr, we now have;

ΔV = (C²/π)Δr

ΔV will be maximum when Δr is maximum

Thus, ΔV = (C²/π) × 0.25/π

C = 82 cm

Thus;

ΔV = (82²/π) × 0.25/π

ΔV = 170.32 cm³

B) Formula for relative error = ΔV/V

Relative error = 170.32/((4/3)πr³)

Relative error = 170.32/((4/3)C³/8π³)

Relative errror = 170.32/((4/3)82³/8π³)

Relative error = 170.32/2963.744

Relative error = 0.0575