Use the approximation (1 + x)^k 1 + kx to estimate (1.0002)^28 and 6 root 1.008.
ID: 2865846 • Letter: U
Question
Use the approximation (1 + x)^k 1 + kx to estimate (1.0002)^28 and 6 root 1.008. (1.0002)^28 = 1.0056 [ ] The edge x of a cube is measured with an error of at most 0.4%. Complete parts(a) and (b) below. a. What is the maximum corresponding percentage error in computing the cube's surface area? The maximum corresponding percentage error in computing the cube's surface area is [ ]%. The volume of a solid can be expressed as V = 6x^3. The volume is to be calculated with an error of no more than 2% of the true value. Find approximately the greatest error that can be tolerated in the measurement of x, expressed as a percentage of x. the greatest tolerated error in the measurement of x is [ ] %. The diameter of sphere is measured as 60+ - 4 cm and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation. There is approximately [ ]% error in the volume calculation.Explanation / Answer
(1 + x)^k = 1 + kx
So, (1.0002)^28 is (1 + 0.0002)^28 becomes 1 + 0.0002*28 ---> 1.0056
(1.008)^(1/6) ---> (1 + 0.008)^(1/6) --> 1 + 0.008(1/6) --> 1.0013
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x has an error of 0.4%
A = 6x^2
dA = 12x * dx
dA = 12x * 0.4%
dA = 4.8x % ---> ANSWER
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V = 6x^3
dV = 18x^2 * dx
2 = 18x^2 * dx
dx = 1/(9x^2) % ---> ANSWER
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Actual diameter = 60
Error = +/-4
V = (4pi/3)*(d/2)^3
V = (4pi/3)(d^3/8)
V = (pi/6)*d^3
Deriving :
dV = (pi/6)*(3d^2)*dD
dV = (pi/6)*(3*60^2)*4
dV = 7200pi
Now, actual volume = (4pi/3)(60/2)^3 = 36000pi
So, percentage = (7200pi / 36000pi) * 100
20% --- >ANSWER
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