How much toilet paper is left on the roll? The volume of a cylinder with length ℓℓ and radius r𝑟 is given by the formula:πℓr2𝜋ℓ𝑟2  You've got some time on your hands, so you carefully measure your toilet roll and discover its dimensions to the nearest mm:length ℓ=100ℓ=100 mm,radius of the inner cardboard cylinder r=30𝑟=30 mm, andradius of the outer cylinder R=51𝑅=51 mm.              The volume of toilet paper on the roll is thus given by the formula:V(ℓ,r,R)=πℓ(R2−r2)𝑉(ℓ,𝑟,𝑅)=𝜋ℓ(𝑅2−𝑟2) .  According to your measurements, the volume of toilet paper is (to the nearest cubic mm):V(100,30,51)=𝑉(100,30,51)= mm3.However, since your initial measurements were only accurate to the nearest mm, the true volume of toilet paper may be different. The total differential approximation can be used to estimate how the measured volume V(100,30,51)𝑉(100,30,51) differs from the true value of V(ℓ,r,R)𝑉(ℓ,𝑟,𝑅) as: V(ℓ,r,R)≈V(100,30,51)+∂V∂ℓ(100,30,51)(ℓ−100)+∂V∂r(100,30,51)(r−30)+∂V∂R(100,30,51)(R−51)𝑉(ℓ,𝑟,𝑅)≈𝑉(100,30,51)+∂𝑉∂ℓ(100,30,51)(ℓ−100)+∂𝑉∂𝑟(100,30,51)(𝑟−30)+∂𝑉∂𝑅(100,30,51)(𝑅−51) . We calculate the partial derivatives (to the nearest integer):∂V∂ℓ(100,30,51)=∂𝑉∂ℓ(100,30,51)= ∂V∂r(100,30,51)=∂𝑉∂𝑟(100,30,51)= ∂V∂R(100,30,51)=∂𝑉∂𝑅(100,30,51)= .Since we measured ℓ,rℓ,𝑟 and R𝑅 to the nearest mm,|ℓ−100|<0.5,|r−30|<0.5|ℓ−100|<0.5,|𝑟−30|<0.5 and |R−51|<0.5|𝑅−51|<0.5 .Using the integer approximations above together with the triangle inequality, |V(ℓ,r,R)−V(100,30,51)|≤∣∣∂V∂ℓ(100,30,51)∣∣|(ℓ−100)|+∣∣∂V∂r(100,30,51)∣∣|(r−30)|+∣∣∂V∂R(100,30,51)∣∣|(R−51)||𝑉(ℓ,𝑟,𝑅)−𝑉(100,30,51)|≤|∂𝑉∂ℓ(100,30,51)||(ℓ−100)|+|∂𝑉∂𝑟(100,30,51)||(𝑟−30)|+|∂𝑉∂𝑅(100,30,51)||(𝑅−51)| . And so our measured volume  V(100,30,51)𝑉(100,30,51) differs from the true volume by|V(ℓ,r,R)−V(100,30,51)|≤|𝑉(ℓ,𝑟,𝑅)−𝑉(100,30,51)|≤ mm3.To two decimal places, this represents an approximate error of no more than|V(ℓ,r,R)−V(100,30,51)|V(100,30,51)≤5.26%|𝑉(ℓ,𝑟,𝑅)−𝑉(100,30,51)|𝑉(100,30,51)≤5.26% .

Primero, calculemos el volumen de papel higiénico en el rollo usando la fórmula dada:

$V(\ell, r, R) = \pi \ell (R^2 - r^2)$

Sustituyendo los valores medidos:

$V(100, 30, 51) = \pi \cdot 100 \cdot (51^2 - 30^2)$

Calculamos los cuadrados de los radios:

$51^2 = 2601$ $30^2 = 900$

Restamos estos valores:

$2601 - 900 = 1701$

Multiplicamos por la longitud y por $\pi$:

$V(100, 30, 51) = \pi \cdot 100 \cdot 1701$ $V(100, 30, 51) \approx 3.1416 \cdot 100 \cdot 1701$ $V(100, 30, 51) \approx 534070.56 \, \text{mm}^3$

Ahora, calculemos las derivadas parciales:

1. Derivada parcial con respecto a $\ell$:

$\frac{\partial V}{\partial \ell} = \pi (R^2 - r^2)$ $\frac{\partial V}{\partial \ell}(100, 30, 51) = \pi (51^2 - 30^2)$ $\frac{\partial V}{\partial \ell}(100, 30, 51) = \pi \cdot 1701$ $\frac{\partial V}{\partial \ell}(100, 30, 51) \approx 5340.71$

2. Derivada parcial con respecto a $r$:

$\frac{\partial V}{\partial r} = -2\pi \ell r$ $\frac{\partial V}{\partial r}(100, 30, 51) = -2\pi \cdot 100 \cdot 30$ $\frac{\partial V}{\partial r}(100, 30, 51) = -6000\pi$ $\frac{\partial V}{\partial r}(100, 30, 51) \approx -18849.56$

3. Derivada parcial con respecto a $R$:

$\frac{\partial V}{\partial R} = 2\pi \ell R$ $\frac{\partial V}{\partial R}(100, 30, 51) = 2\pi \cdot 100 \cdot 51$ $\frac{\partial V}{\partial R}(100, 30, 51) = 10200\pi$ $\frac{\partial V}{\partial R}(100, 30, 51) \approx 32044.39$

Dado que las mediciones fueron precisas al milímetro más cercano, tenemos:

$|\ell - 100| < 0.5$ $|r - 30| < 0.5$ $|R - 51| < 0.5$

Usando la desigualdad del triángulo:

$|V(\ell, r, R) - V(100, 30, 51)| \leq \left| \frac{\partial V}{\partial \ell}(100, 30, 51) \right| |\ell - 100| + \left| \frac{\partial V}{\partial r}(100, 30, 51) \right| |r - 30| + \left| \frac{\partial V}{\partial R}(100, 30, 51) \right| |R - 51|$

Sustituyendo los valores:

$|V(\ell, r, R) - V(100, 30, 51)| \leq 5340.71 \cdot 0.5 + 18849.56 \cdot 0.5 + 32044.39 \cdot 0.5$ $|V(\ell, r, R) - V(100, 30, 51)| \leq 2670.36 + 9424.78 + 16022.20$ $|V(\ell, r, R) - V(100, 30, 51)| \leq 28117.34 \, \text{mm}^3$

Finalmente, el error aproximado en porcentaje es:

$\frac{|V(\ell, r, R) - V(100, 30, 51)|}{V(100, 30, 51)} \leq \frac{28117.34}{534070.56} \approx 0.0526$ $\frac{|V(\ell, r, R) - V(100, 30, 51)|}{V(100, 30, 51)} \leq 5.26\%$

Por lo tanto, el volumen medido $V(100, 30, 51)$ difiere del volumen verdadero en no más de 28117.34 mm³, lo que representa un error aproximado de no más del 5.26%.