Q. 53.7( 3 Votes )
If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.
Given the error in the measurement of the radius of a sphere is 0.1%.
Let x be the radius of the sphere and Δx be the error in the value of x.
Hence, we have
∴ Δx = 0.001x
The volume of a sphere of radius x is given by
On differentiating V with respect to x, we get
Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as
Here, and Δx = 0.001x
⇒ ΔV = (4πx2)(0.001x)
∴ ΔV = 0.004πx3
The percentage error is,
⇒ Error = 0.003 × 100%
∴ Error = 0.3%
Thus, the error in calculating the volume of the sphere is 0.3%.
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