The edge of a cube was found to be 15 cm with a possible error in measurement of
ID: 2841338 • Letter: T
Question
The edge of a cube was found to be 15 cm with a possible error in measurement of 0.4 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round your answers to four decimal places.)
a.) The volume of the cube
maximum possible error(in cm^3):____
relative error:____
percentage error(in %):____
b.) The surface are of the cube
maximum possible error(in cm^2):____
relative error:____
percentage error(in %):____
Explanation / Answer
x is the edge length, so V = x^3 and dV= 3x^2 * dx
Thus, when x = 15cm, possible error, or dx = 0.4cm
dV=3(15)^2 * (0.4) = 270 = maximum possible error in cm^3 ----->(270cm^3)
Relative error is calculated by dividing the change in V by V... Change in V here is dV.
so dV/V = 3x^2*dx/x^3 = 3(dx/x) = 3(0.4/15) = 0.08
Percentage error is simply relative error times 100 to receive the percentage 0.08 * 100% = 8%
b)
surface area of the cube is =A = 6a2
Given: A= 1350
Taking it's derivative
=>dA = 12 a da
solving: 2(15)(0.4)
=72 will be possible error
Now to find relative error = dA / A
= (72 /1350)
= 0.0533
For percentage error = (dA /A) . 100
= (0.0533).100
=5.33 % Ans
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