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**Unformatted text preview: **Section 2.8: Linear approximation and differentials (concluded): Section 2.8, Problem 24: Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 cm thick to a hemispherical dome with diameter 50 m. Solution: For a hemispherical dome, V ( r ) = (2/3) π r 3 . The paint needs to fill a spherical shell of inner radius r and outer radius r + dr , where r = 25 m and dr = 0.05 cm. The volume of this spherical shell is ∆ V = V ( r + dr ) – V ( r ). We can approximate this by dV = V ′ ( r ) dr = 2 π r 2 dr = 2 π (25 m) 2 (0.0005 m) = 5 π /8 m 3 1.96350 m ≈ 3 . (You can check that ∆ V = 0.62501 π m 3 1.96353 m ≈ 3 , so dV and ∆ V are quite close in this case. That’s because dr is much less than r .) One application of differentials is to the estimation of errors. Problem: Suppose we know that a hemispherical dome has radius 25 m plus or minus about 0.05 cm. How accurately do we know the volume of the dome? Solution: Write the true radius as r + dr , where r = 25 m and dr = the measurement error, with | dr | no more than 0.05 cm. If we estimate the volume of the hemisphere by V ( r ) = (2/3) π r 3 = (2/3) π (25 m) 3 32,724.9 m ≈ 3 , our error will be ∆ V = V ( r + dr ) – V ( r ), which we can approximate by dV = 2 π r 2 dr , which is no greater than 2 π (25 m) 2 (0.0005 m) 1.96 m ≈ 3 as before. That is, the relative error ∆ V / V will be no more than about (1.96 m 3 ) / (32,700 m 3 ) 0.00006 = 0.006%. ≈ (For this particular problem, you could also compute the volumes of hemispheres of radius 25 m ± 0.05 cm.) Section 3.1:Section 3....

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