Error Estimation with Differentials

$A = s^2$

$dA = 2s \cdot ds = 2(12)(0.3) = 7.2$ cm²

The maximum error in the area is approximately 7.2 cm².

(a) $A = 12^2 = 144$ cm²

Relative error: $\frac{dA}{A} = \frac{7.2}{144} = 0.05$

(b) Percentage error: $0.05 \times 100\% = 5\%$

(c) Percentage error in side: $\frac{0.3}{12} \times 100\% = 2.5\%$

The percentage error in the area (5%) is twice the percentage error in the side (2.5%).

This makes sense! Since $A = s^2$, the relative error formula gives: $$\frac{dA}{A} = 2\frac{ds}{s}$$

(a) Maximum error in volume:

$V = \frac{4}{3}\pi r^3$

$\frac{dV}{dr} = 4\pi r^2$

$dV = 4\pi r^2 \, dr = 4\pi (15)^2 (0.1) = 4\pi (225)(0.1) = 90\pi \approx 283$ cm³

(b) Relative error:

$V = \frac{4}{3}\pi (15)^3 = \frac{4}{3}\pi (3375) = 4500\pi$ cm³

$\frac{dV}{V} = \frac{90\pi}{4500\pi} = \frac{90}{4500} = 0.02 = 2\%$

(c) Direct approach using the rule:

For $V = kr^3$, relative error in $V$ = $3 \times$ relative error in $r$.

If $\frac{dr}{r} = 0.67\%$, then $\frac{dV}{V} = 3(0.67\%) = 2\%$. ✓

(a) Rewriting: $T = 2\pi\sqrt{\frac{L}{g}} = \frac{2\pi}{\sqrt{g}} \cdot L^{1/2}$

$\frac{dT}{dL} = \frac{2\pi}{\sqrt{g}} \cdot \frac{1}{2}L^{-1/2} = \frac{\pi}{\sqrt{gL}}$

More elegantly: $$\frac{dT}{T} = \frac{1}{2}\frac{dL}{L}$$

So: $dT = \frac{T}{2L} \cdot dL$

(b) If percentage error in $L$ is 0.5%, then: $$\text{Percentage error in } T = \frac{1}{2}(0.5\%) = 0.25\%$$

(c) Comparing:

• Period ($T \propto L^{1/2}$): Error multiplied by $\frac{1}{2}$
• Volume ($V \propto r^3$): Error multiplied by $3$

The period is much less sensitive to length errors than volume is to radius errors.

The square root “dampens” errors, while the cube “amplifies” them.

(a) Maximum error calculation:

$dV = 2\pi rh\,dr + \pi r^2\,dh$

With $r = 3$, $h = 10$, $dr = dh = 0.05$:

Error from $dr$: $2\pi(3)(10)(0.05) = 3\pi \approx 9.42$ m³

Error from $dh$: $\pi(3)^2(0.05) = 0.45\pi \approx 1.41$ m³

Maximum total error: $dV = 3.45\pi \approx 10.84$ m³

(b) Comparison:

• Error from radius: $3\pi \approx 9.42$ m³
• Error from height: $0.45\pi \approx 1.41$ m³

The radius measurement contributes about 6.7 times more to the volume error!

(c) Reducing error in the radius would have a much bigger impact on reducing volume error, since the radius term dominates.

(d) Explanation:

In $V = \pi r^2 h$, the radius is squared. This means:

• A 1% error in $r$ causes a 2% error in $V$ (from the $r^2$ term)
• A 1% error in $h$ causes only a 1% error in $V$ (linear term)

Even though both measurements have the same absolute error, the radius error is “amplified” by the squared relationship. This is why precision in $r$ matters more than precision in $h$.

General principle: When a quantity appears with a higher power, its measurement error has a proportionally larger effect.

For $y = x^n$, relative error in $y$ = $\vert n\vert \times$ relative error in $x$.

• (A) $y = x^2$: Error multiplied by 2 → 4% error (worse)
• (B) $y = x^3$: Error multiplied by 3 → 6% error (worse)
• (C) $y = \sqrt{x} = x^{1/2}$: Error multiplied by $\frac{1}{2}$ → 1% error (better!)
• (D) $y = 5x$: Error multiplied by 1 → 2% error (same)

Answer: (C)

The square root dampens errors because $n = \frac{1}{2} < 1$.

$f(x) = x^2 - 4x + 5$

$f'(x) = 2x - 4$

$f'(x) = 0$ when $x = 2$.

At a critical point, the derivative is zero, so $dy = f'(x) \cdot dx = 0 \cdot dx = 0$.

Answer: (B) $x = 2$

Near a local minimum or maximum, the function is “flat,” so small changes in $x$ cause almost no change in $y$. This is one reason why optimization problems focus on critical points!