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miskamm [114]
3 years ago
11

-10,-17, -24, -31, -38, ...

Mathematics
1 answer:
ra1l [238]3 years ago
4 0
Not too sure maybe A?
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Ask Your Teacher The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm. (a) Use differentials t
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Answer:  

a)  28,662 cm²  max error

    0,0111     relative error

b) 102,692 cm³  max error

   0,004     relative error

   

Step-by-step explanation:

Length of cicumference is: 90 cm

L = 2*π*r

Applying differentiation on both sides f the equation

dL  =  2*π* dr    ⇒  dr = 0,5 / 2*π

dr =  1/4π

The equation for the volume of the sphere is  

V(s) =  4/3*π*r³     and for the surface area is

S(s) = 4*π*r²

Differentiating

a) dS(s)  =  4*2*π*r* dr    ⇒  where  2*π*r = L = 90

Then    

dS(s)  =  4*90 (1/4*π)

dS(s) = 28.662 cm²   ( Maximum error since dr = (1/4π) is maximum error

For relative error

DS´(s)  =  (90/π) / 4*π*r²

DS´(s)  = 90 / 4*π*(L/2*π)²      ⇒   DS(s)  = 2 /180

DS´(s) = 0,0111 cm²

b) V(s) = 4/3*π*r³

Differentiating we get:

DV(s) =  4*π*r² dr

Maximum error

DV(s) =  4*π*r² ( 1/  4*π*)   ⇒  DV(s) = (90)² / 8*π²

DV(s)  =  102,692 cm³   max error

Relative error

DV´(v) =  (90)² / 8*π²/ 4/3*π*r³

DV´(v) = 1/240

DV´(v) =  0,004

3 0
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