Respuesta :
Answer:
Therefore the maximum error in the surface area of the sphere is 22.27 cm².
Therefore the relative error is 0.014 (approx).
Step-by-step explanation:
Given that, The circumference of a sphere was 70 cm with the possible error 0.5 cm.
The circumference of the sphere is C [tex]=2 \pi r[/tex]
∴C [tex]=2 \pi r[/tex] [tex]\Rightarrow r =\frac{C}{2\pi}[/tex]
Differentiating with respect to r
[tex]\frac{dC}{dr}= 2 \pi[/tex]
[tex]\Rightarrow dr=\frac{dC}{2\pi}[/tex]
[tex]\Rightarrow dr= \frac{0.5}{2\pi}[/tex] [ relative error = dC= 0.5]
The surface area of the sphere is S= [tex]4\pi r^2[/tex]
∴S= [tex]4\pi r^2[/tex]
Differentiating with respect to r
[tex]\frac{dS}{dr}=4\pi \times 2r[/tex]
[tex]\Rightarrow dS=8\pi r dr[/tex]
dS will be maximum when dr is maximum.
Putting the value of r and dr
[tex]\Rightarrow dS= 8\pi \times \frac{C}{2\pi} \times \frac {0.5}{2\pi}[/tex]
[tex]\Rightarrow dS=\frac{C}{\pi}[/tex]
[tex]\Rightarrow dS= \frac{70}{\pi}[/tex] [ ∵ C= 70 ]
⇒dS= 22.27 (approx)
Therefore the maximum error in the surface area is 22.27 cm².
Relative error [tex]=\frac{dS}{S}[/tex]
[tex]=\frac{\frac{C}{\pi}}{4\pi r^2}[/tex]
[tex]=\frac{\frac{C}{\pi}}{4\pi (\frac{C}{2\pi})^2}[/tex]
[tex]=\frac{1}{C}[/tex]
[tex]=\frac{1}{70}[/tex]
=0.014 (approx)
Therefore the relative error is 0.014 (approx).
