Montserrat is training for her first race. In her first training week she runs 3 km, and in the second training week she runs 3.
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The maximum error in the calculated surface area is 24.19cm² and the relative error is 0.0132.
Given that the circumference of a sphere is 76cm and error is 0.5cm.
The formula of the surface area of a sphere is A=4πr².
Differentiate both sides with respect to r and get
dA÷dr=2×4πr
dA÷dr=8πr
dA=8πr×dr
The circumference of a sphere is C=2πr.
From above the find the value of r is
r=C÷(2π)
By using the error in circumference relation to error in radius by:
Differentiate both sides with respect to r as
dr÷dr=dC÷(2πdr)
1=dC÷(2πdr)
dr=dC÷(2π)
The maximum error in surface area is simplified as:
Substitute the value of dr in dA as
dA=8πr×(dC÷(2π))
Cancel π from both numerator and denominator and simplify it
dA=4rdC
Substitute the value of r=C÷(2π) in above and get
dA=4dC×(C÷2π)
dA=(2CdC)÷π
Here, C=76cm and dC=0.5cm.
Substitute this in above as
dA=(2×76×0.5)÷π
dA=76÷π
dA=24.19cm².
Find relative error as the relative error is between the value of the Area and the maximum error, therefore:
As above its found that r=C÷(2π) and r=dC÷(2π).
Substitute this in the above
Hence, the maximum error in the calculated surface area with the circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm is 24.19cm² and the relative error is 0.0132.
Learn about relative error from here brainly.com/question/13106593
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