Calculating Error of Volume of a Sphere Using Differentials

[Drakkith]

Homework Statement

The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated volume.

Homework Equations

Volume of sphere: V=4/3πR³
Circumference of Sphere: C=2πR
ΔC = 0.5 cm

The Attempt at a Solution

Stating R in terms of C:

R=C/2π

Inserting this new term into the volume equation:

V=4/3π(C/2π)³

Now, before taking the derivative of both sides, I went ahead and expanded the right side:

V=C³/6π²

Then, the derivative of both sides:

dV/dC = C³/2π²(dC)

Now, is dC = ΔC? I'm not quite sure what to do next.

 

[Mark44]

Homework Statement

The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated volume.

Homework Equations

Volume of sphere: V=4/3πR³
Circumference of Sphere: C=2πR
ΔC = 0.5 cm

The Attempt at a Solution

Stating R in terms of C:

R=C/2π

Inserting this new term into the volume equation:

V=4/3π(C/2π)³

Now, before taking the derivative of both sides, I went ahead and expanded the right side:

V=C³/6π²

Then, the derivative of both sides:

dV/dC = C³/2π²(dC)

Now, is dC = ΔC? I'm not quite sure what to do next.

You can approximate dC by ΔC. The differential quantities here -- dr, dC, dV -- are infinitesimals, which is why it says to "Use differentials to estimate the maximum error..." Since the independent variable here is r (for radius, lowercase r is usually used, not R), if Δr is "small" relative to r, the approximation will be better.

As for your work, you're taking the long way around, plus it doesn't seem that you're getting close to the answer that is called for here.

You know that $C = 2\pi r$, so $r = \frac C {2\pi}$ (which you have).
From your volume equation, $V = \frac 4 3 \pi r^3$, so $\Delta V \approx dV = 4\pi r^2 dr$. Now do the substitution to replace r using the equation just above, and you're just about home free.

 

[Drakkith]

From your volume equation, V=43πr3V = \frac 4 3 \pi r^3, so ΔV≈dV=4πr2dr\Delta V \approx dV = 4\pi r^2 dr. Now do the substitution to replace r using the equation just above, and you're just about home free.

What happens to dr?

 

[SteamKing]

What happens to dr?

You already know how C relates to r. You can find out how ΔC relates to Δr, and use Δr ≈ dr.

 

[Drakkith]

You already know how C relates to r. You can find out how ΔC relates to Δr, and use Δr ≈ dr.

It appears that ΔC relates to Δr by: Δr = ΔC/2π, which would make dr ≈ΔC/2π.

 

[SteamKing]

It appears that ΔC relates to Δr by: Δr = ΔC/2π, which would make dr ≈ΔC/2π.

Now, you should be able to relate ΔV to ΔC for the sphere.

 

[Drakkith]

Plugging in 0.5 for ΔC gives me: dV = 4π(90/2π)²(0.5/2π) = 205.175 cm³.

Checking work:

Using 90 for C: V=4/3π(90/2π)3 = 12,310.523 cm³
Using 89.5 for C: V=4/3π(89.5/2π)3 = 12,106.486 cm³
Using 90.5 for C: V=4/3π(90.5/2π)3 = 12,516.841 cm³

12,310.523-12,106.486 = 204.037
12,516.841-12,310.523 = 206.318

I'd say 205.175 is a pretty good answer since it's fits right in the middle of the two differences.

Thanks guys. You're awesome and you should feel awesome.