Respuesta :
Answer:
Therefore the error in the calculated volume is 130 [tex]cm^3[/tex].
The relative error is 0.011.
Step-by-step explanation:
Given that,
The circumference of a sphere was 70 cm.
The circumference of a sphere is C= [tex]2\pi r[/tex]
C= [tex]2\pi r[/tex]
Differentiating with respect to r
[tex]\frac{dC}{dr}=2\pi[/tex]
[tex]\Rightarrow \frac{\triangle C}{\triangle r}=2\pi[/tex]
[tex]\Rightarrow\triangle r= \frac{\triangle C}{2\pi}[/tex]
Given that,the circumference of the sphere was with possible error 0.5 cm.
[tex]\triangle C=0.5[/tex]
[tex]\therefore \triangle r= \frac{0.5}{2\pi}[/tex]
The volume of the sphere is [tex]V=\frac43 \pi r^3[/tex]
[tex]V=\frac43 \pi r^3[/tex]
Differentiating with respect to r
[tex]\frac{dV}{dr}=\frac43 \pi r^2[/tex]
[tex]\Rightarrow \frac{\triangle V}{\triangle r}=\frac43 \pi r^2[/tex]
[tex]\Rightarrow \triangle V}=\frac43 \pi r^2\times \triangle r}[/tex]
Putting [tex]\triangle r=\frac{0.5}{2\pi}[/tex]
[tex]\Rightarrow \triangle V}=\frac43 \pi r^2\times \frac{0.5}{2\pi}[/tex]
[tex]\Rightarrow \triangle V}=\frac13 \pi r^2[/tex]
[tex]\Rightarrow \triangle V}=\frac1{3} \pi (\frac C{2\pi})^2[/tex] [tex][\because r=\frac C{2\pi}][/tex]
[tex]\Rightarrow \triangle V}=\frac1{1 2} \times \frac{C^2}{\pi}[/tex]
[tex]\Rightarrow \triangle V}=\frac1{1 2} \times \frac{70^2}{\pi}[/tex] [ C=70 cm]
[tex]\Rightarrow \triangle V}\approx 130 \ cm^3[/tex]
Therefore the error in the calculated volume is 130 [tex]cm^3[/tex].
Relative error [tex]=\frac{\triangle V}{V}[/tex]
[tex]=\frac{\frac13 \pi (\frac{C}{2\pi})^2 }{\frac43 \pi (\frac C{2\pi})^3}[/tex]
[tex]=\frac{1}{4 (\frac C{2\pi})}[/tex]
[tex]=\frac{\pi }{4C}[/tex]
[tex]\approx 0.011[/tex]
The relative error is 0.011.
