Question

The radius of a sphere is measured to be 3.0 inches. If the measurement is correct within 0.01 inches, use differentials to estimate the error in the volume of sphere.

Answer 1

Using differentials, we estimated the error in the volume of a sphere with a radius error of ±0.01 inches to be approximately 1.13 cubic inches by applying the formula dV = 4πr^2 × dr, with r being 3.0 inches.

Step-by-step explanation:

The question involves using differentials to estimate the error in the volume of a sphere given a measurement error in the radius. The formula for the volume of a sphere is V = (4/3)πr^3. To estimate the error in volume, we will use the concept of differentials, which is given by dV = ∂V/∂r × dr, where dV is the differential change in volume and dr is the differential change in radius.

The radius of the sphere is given as 3.0 inches with an error of ±0.01 inches. Applying this to the formula, the differential of the volume, dV, approximates how much the volume changes with a small change in radius, dr.

The partial derivative of V with respect to r is dV/dr = 4πr^2. Plugging in the values we get:

dV = 4π(3.0)^2 × 0.01

Substituting 3.0 for r and 0.01 for dr, we get:

dV = 4π(9) × 0.01 = 4π× 0.09 = 0.36π ≈ 1.13 in^3 (since π≈ 3.14)

Therefore, the estimated error in the volume of the sphere is approximately 1.13 cubic inches.

Answer 2

ΔV = 0.36π in³

Step-by-step explanation:

Given that:

The radius of a sphere = 3.0

If the measurement is correct within 0.01 inches

i.e the change in the radius Δr = 0.01

The objective is to use differentials to estimate the error in the volume of sphere.

We all know that the volume of a sphere

`V = (4)/(3) \pi r^3`

The differential of V with respect to r is:

`(dV)/(dr )= 4 \pi r^2`

dV = 4 πr² dr

which can be re-written as:

ΔV = 4 πr² Δr

ΔV = 4 × π × (3)² × 0.01

ΔV = 0.36π in³