Respuesta :
Answer:
The maximum error is [tex]dS = 25 \ cm^{2}[/tex]
Relative error is Relative error is = [tex]\frac{1}{78}[/tex]
Step-by-step explanation:
Circumference of a sphere C = 2[tex]\pi r[/tex]
Differentiate above equation with respect to r
[tex]\frac{dC}{dr} = 2\pi[/tex]
[tex]dr = \frac{dC}{2\pi}[/tex]
Given that [tex]dC = 0.5 cm[/tex]
[tex]dr = \frac{0.5}{2\pi}[/tex]
[tex]dr = \frac{1}{4\pi}[/tex]
Surface area of the sphere S = [tex]4\pi r^{2}[/tex]
Differentiate above equation with respect to r
[tex]\frac{dS}{dr} = 8 \pi r[/tex]
[tex]dS = 4 C dr[/tex]
Where C = circumference of the sphere
given that C = 78 cm & [tex]dr = \frac{1}{4\pi}[/tex]
So the maximum error is given by
[tex]dS = 4 (78)(\frac{1}{4\pi} )[/tex]
[tex]dS = \frac{78}{\pi}[/tex]
[tex]dS = 25 \ cm^{2}[/tex]
Now the relative error is given by
Relative error = [tex]\frac{dS}{S}[/tex]
⇒ [tex]\frac{\frac{78}{\pi} }{4\pi r^{2} }[/tex]
Since [tex]r = \frac{C}{2 \pi}[/tex]
Relative error is = [tex]\frac{1}{78}[/tex]
Therefore the maximum error is [tex]dS = 25 \ cm^{2}[/tex]
Relative error is Relative error is = [tex]\frac{1}{78}[/tex]
