Complete question is;
(a) The edge of a cube was measured to be 6 cm, with a maximum possible error of 0.5 cm. Use a differential to estimate the maximum possible error in computing the volume of the cube.
(b) Using a calculator, find the actual error in measuring volume if the radius was really 6.5 cm instead of 6 cm, and find the actual error if the radius was actually 5.5 cm instead of 6 cm. Compare these errors to the answer you got using differentials.
Answer:
A) Maximum possible error = 54 cm³
B) i) The actual error of 58.625 cm³ is more than that gotten from the differential, so we will reject this radius of 6.5 cm.
(ii) The actual error of 49.625 cm³ gotten falls within range of the maximum error gotten from the differential so we can accept the radius of 5.5cm.
Explanation:
The formula for volume of a cube is;
V(x) = x³
Where x is the length of the edge of the cube.
So, dV/dx = 3x²
When Δx is small like in this case, we can say that;
ΔV/Δx ≈ 3x²
So, ΔV ≈ 3x²•Δx
We are given;
Δx = 0.5 and x = 6.
Thus;
Maximum possible error = 3x²•Δx = 3 × 6² × 0.5 = 54 cm³
B) (i) If radius was 6.5,then;
Volume = 6.5³ = 274.625 cm³
Comparing to radius of 6cm;
Volume = 6³ = 216 cm³
So, actual error = 274.625 cm³ - 216 cm³ = 58.625 cm³
This actual error is more than that gotten from the maximum error from the Differential. So we will not accept this radius of 6.5cm.
ii) If radius was 5.5,then;
Volume = 5.5³ = 166.375 cm³
Comparing to radius of 6cm;
Volume = 6³ = 216 cm³
So, actual error = 216 cm³ - 166.375 cm³ = 49.625 cm³
This actual error gotten falls within range of the maximum error gotten from the differential so we can accept the radius of 5.5 cm.