A sphere is initially measured as having a radius of 4 inches. A subsequent measurement shows that the radius is really 3.99. Find (a) the actual difference in volume between the initial measurement and the correction, and (b) the difference as estimated through differentials. Note- the formula for the volume of a sphere is V=4/3(pi)r^3

Question

A sphere is initially measured as having a radius of 4 inches. A subsequent measurement shows that the radius is really 3.99. Find (a) the actual difference in volume between the initial measurement and the correction, and (b) the difference as estimated through differentials. Note- the formula for the volume of a sphere is V=4/3(pi)r^3

I did question A, but I need help on question B...

thanks :)

Answer 1

V = (4/3) pi r^3

dV/dr = (4/3)3 pi r^2 = 4 pi r^2
(which is the surface area of the sphere of course)
so
dV = 4 pi r^2 dr
here dr = - .01
so dV = -.01 * 4 * pi * 16
dV = -2.01 in^3

Answer 2

Thank you soo muccchh

Answer 3

sorry actually i got that part which is part a. What I need is part (b). Find the delta(v)

Answer 4

To estimate the difference in volume using differentials, we can use the concept of differentials or differentiability.

Let's start by finding the volume of the sphere using the initial measurement with a radius of 4 inches. The formula for the volume of a sphere is V = (4/3)πr^3.

V₁ = (4/3)π(4^3)
V₁ = (4/3)π(64)
V₁ ≈ 268.08 cubic inches

Now, let's find the volume of the sphere using the corrected measurement with a radius of 3.99 inches.

V₂ = (4/3)π(3.99^3)
V₂ = (4/3)π(63.996039)
V₂ ≈ 267.94 cubic inches

To find the actual difference in volume between the initial measurement and the correction, we can subtract V₂ from V₁:

Actual Difference = V₁ - V₂
Actual Difference = 268.08 - 267.94
Actual Difference ≈ 0.14 cubic inches

Now, let's move on to estimating the difference using differentials. The differential of a function f(x) is given by df(x) = f'(x)dx, where f'(x) is the derivative of f(x) with respect to x and dx is the change in x.

In the case of the volume of a sphere, V = (4/3)πr^3, we can differentiate both sides with respect to r:

dV = d((4/3)πr^3)
dV = (4/3)π(3r^2)dr

Since we are interested in estimating the difference for a change in radius, we set dr = -0.01 (as the radius went from 4 to 3.99 inches or a decrease of 0.01 inches). Plugging this into the differential formula, we get:

dV = (4/3)π(3(4^2))(-0.01)
dV = (4/3)π(48)(-0.01)
dV ≈ -2.02 cubic inches

Therefore, using differentials, the estimated difference in volume is -2.02 cubic inches. Note that this estimate is an approximation and may not be exactly equal to the actual difference.