Ask Question
20 October, 23:08

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 to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm2 What is the relative error? (Round your answer to three decimal places.) (b) Use differentials to estimate the maximum error in the calculated volume. (Round your answer to the nearest integer.) cm3 What is the relative error? (Round your answer to three decimal places.)

+2
Answers (2)
  1. 20 October, 23:46
    0
    A) For the area; maximum error is 28.65 cm² while relative error is 0.011

    B) For the volume; maximum error is 205.18 cm³ while relative error is 0.239

    Step-by-step explanation:

    Circumference; C = 2πr

    Differentiating both sides with respect to r;

    dc/dr = 2π

    When r is small, we can write;

    Δc/Δr = 2π

    Thus, Δr = Δc/2π

    Now, we are given that, Δc = 0.5

    So, Δr = 0.5/2π = 1/4π

    A) For the area, the formula for surface area of sphere is 4πr²

    Thus; S (r) = 4πr²

    Differentiating both sides with respect to r; ds/dr = 4 (2πr)

    When r is small, we can write;

    Δs/Δr = 4 (2πr)

    So, Δs = 4 (2πr) Δr

    From earlier Circumference (C) = 2πr

    Thus, Δs = 4CΔr

    Now, our Circumference is 90cm and we have established Δr to be 1/4π.

    Δs will be maximum when Δr is maximum,

    Thus, maximum error in S is;

    Δs = 4 x 90 x 1/4π = 90/π = 28.65 cm²

    Relative error is given by;

    R. E = Δs/s

    Now, s = surface area of sphere which 4πr²

    We don't have r, so let's attempt simplify it to reflect C.

    s = 4π (2πr/2π) ² = 4π (C²/4π²) = C²/π

    s = 90²/π

    Relative Error = Δs/s = (90/π) / (90²/π)

    = 1/90 = 0.011

    B) For the volume, the formula for volume of a sphere is (4/3) πr³

    Thus; V (r) = (4/3) πr³

    Differentiating both sides with respect to r; ds/dr = 4πr²

    When r is small, we can write;

    Δs/Δr = (2πr) ²/π

    So, Δs = [ (2πr) ²/π]Δr

    From earlier Circumference (C) = 2πr

    Thus, Δs = (C²/π) Δr

    Now, our Circumference is 90cm and we have established Δr to be 1/4π.

    Δv will be maximum when Δr is maximum,

    Thus, maximum error in v is;

    Δv = (90²/π) x (1/4π) = 8100/4π² = 205.18 cm³

    Relative error is given by;

    R. E = Δv/v

    Now, v = volume of sphere which (4/3) πr³

    We don't have r, so let's attempt to simplify it to reflect C.

    v = (1/3π) (2πr) ² = (1/3π) (C²) = C²/3π

    v = 90²/3π = 8

    Relative Error = Δv/v = (8100/4π²) / (90²/3π)

    = 3/4π = 0.239
  2. 21 October, 01:19
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 to estimate ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers