Calculus:Derivatives:Differential (I am stuck on this question help PLEASE)

nch · #1 $Old$ October 20th, 2009, 08:25 PM

In a manufacturing process, ball bearings must be made with radius of 0.4 mm, with a maximum error in the radius of ±0.013 mm. Estimate the maximum error in the volume of the ball bearing.
Solution: The formula for the volume of the sphere is________. If an error ∆r is made in measuring the radius of the sphere, the maximum error in the volume is ∆V=__________.
Rather than calculating ∆V, approximate ∆V with dV, where dV=__________.
Replacing r with___ and dr=∆r with ±____ gives dV= ±____.
The maximum error in the volume is about____mm^3.

Soroban · #2 $Old$ October 21st, 2009, 02:00 PM

Hello, nch!

Some of this is just arithmetic . . .

Quote:

In a manufacturing process, ball bearings must be made with $r \,=\,0.4\text{ mm,}$
with a maximum error in the radius of: . $\Delta r = dr = \pm0.013\text{ mm.}$

Estimate the maximum error in the volume of the ball bearing.

Solution: The formula for the volume of the sphere is: . $\boxed{V \:=\:\tfrac{4}{3}\pi r^3}$

If an error $\Delta r$ is made in measuring the radius of the sphere,
the maximum error in the volume is: . $\Delta V \:=\:\boxed{ {\color{white}XXX} }$

The volume of the correct sphere is: . $V_o \:=\:\tfrac{4}{3}\pi (0.4)^3 \:=\:\frac{0.256}{3}\pi\text{ mm}^3$

With an error of $\Delta r = 0.013$ , the radius is: . $r \:=\:0.4 + 0.013 \:=\:0.413\text{ mm.}$
And the volume is: . $V_1 \;=\;\tfrac{4}{3}\pi(0.413)^3 \;\approx\;\frac{0.28178}{3}\pi\text{ mm}^3$

The error in volume is: . $\Delta V \;=\;\frac{0.28178}{3}\pi - \frac{0.256}{3}\pi$

. . Hence: . $\Delta V \;=\;0.00859\pi\text{ mm}^3$

Quote:

Rather than calculating $\Delta V$ , approximate $\Delta V$ with $dV$ , where: . $dV \:=\:\boxed{4\pi r^2dr}$

Replacing $r$ with $0.4$ ,and $dr =\Delta R$ with $\pm0.013$ gives: . $dV \:=\:\boxed{{\color{white}XX}}$

$dV \;=\;4\pi(0.4)^2(0.13) \;=\;0.00832\pi\text{ mm}^3$

nch · #3 $Old$ October 21st, 2009, 05:44 PM

thankz a lot