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

• Oct 20th 2009, 07:25 PM
nch
Calculus:Derivatives:Differential (I am stuck on this question help PLEASE)
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.
• Oct 21st 2009, 01:00 PM
Soroban
Hello, nch!

Some of this is just arithmetic . . .

Quote:

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

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

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

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

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

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

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

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

Quote:

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

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

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

• Oct 21st 2009, 04:44 PM
nch
thankz a lot :)