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.)

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

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