The radius of a sphere was measured to be cm with a possible error of cm.
Use linear approximation to estimate the maximum error in the calculated surface area. Leave your answer in terms of .
Use linear approximation to estimate the maximum error in the calculated volume. Leave your answer in terms of .
To estimate the maximum error in the calculated surface area and volume of a sphere, we can use linear approximation.
The surface area of a sphere is given by the formula:
A = 4πr²
where r is the radius of the sphere.
To estimate the maximum error in the calculated surface area, we can use the formula for linear approximation:
ΔA ≈ 2πrΔr
where ΔA is the maximum error in the surface area, and Δr is the maximum error in the radius.
Given that the actual radius is cm with a possible error of cm, we can substitute the values into the formula:
ΔA ≈ 2π( cm)( cm)
Simplifying this expression, we get:
ΔA ≈ 2π( cm²)
Therefore, the maximum error in the calculated surface area is 2π cm².
Next, let's calculate the maximum error in the calculated volume using linear approximation.
The volume of a sphere is given by the formula:
V = (4/3)πr³
To estimate the maximum error in the calculated volume, we can use the formula for linear approximation:
ΔV ≈ 4πr²Δr
where ΔV is the maximum error in the volume, and Δr is the maximum error in the radius.
Substituting the values:
ΔV ≈ 4π( cm)²( cm)
Simplifying the expression:
ΔV ≈ 4π( cm³)
Therefore, the maximum error in the calculated volume is 4π cm³.