How much toilet paper is left on the roll? The volume of a cylinder with length...

 

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How much toilet paper is left on the roll? The volume of a cylinder with length and radius r is given by the formula: Пер2 You've got some time on your hands, so you carefully measure your toilet roll and discover its dimensions to the nearest mm: • length = 94 mm, • radius of the inner cardboard cylinder r = 26 mm, and • radius of the outer cylinder R = 57 mm. The volume of toilet paper on the roll is thus given by the formula: According to your measurements, the volume of toilet paper is (to the nearest cubic mm): V (94, 26, 57) = mm³. However, since your initial measurements were only accurate to the nearest mm, the true volume of toilet paper may be different. V(l,r, R) = πl(R² -p²). The total differential approximation can be used to estimate how the measured volume V (94, 26, 57) differs from the true value of V (l, r, R) as: Əl ᏭᏙ ᎧᏙ V(l,r, R) V(94, 26, 57) + (94, 26, 57) (94) + (94, 26, 57) (r - 26) + Əl Ꮩ OR (94, 26, 57) (R-57) We calculate the partial derivatives (to the nearest integer): av -(94, 26, 57) = 数字 (94, 26, 57) ** Ər av (94, 26, 57) * ƏR Since we measured l, r and R to the nearest mm, ᎧᏙ Ər av Əl |l94|< 0.5, |r – 26| < 0.5 and R — 57| < 0.5. Using the integer approximations above together with the triangle inequality, |V (l, r, R) – V (94, 26, 57)| ≤ (94, 26, 57)||(l — 94)| + | (94, 26, 57)||(r - 26)| + av Ər And so our measured volume V (94, 26, 57) differs from the true volume by |V (l,r, R) – V (94, 26, 57)| < To two decimal places, this represents an approximate error of no more than mm³. |V(l,r, R)-V(94,26,57)| V (94,26,57) Are you satisfied with this level of precision? (this part is not worth marks) O No, we need more precise measurements. O Yes, this level of precision is sufficient for my needs. O Better go buy some more toilet paper to account for this potentially significant error. <3.76%. av OR {(94, 26, 57)||(R - 57)| How much toilet paper is left on the roll? The volume of a cylinder with length and radius r is given by the formula: Пер2 You've got some time on your hands, so you carefully measure your toilet roll and discover its dimensions to the nearest mm: • length = 94 mm, • radius of the inner cardboard cylinder r = 26 mm, and • radius of the outer cylinder R = 57 mm. The volume of toilet paper on the roll is thus given by the formula: According to your measurements, the volume of toilet paper is (to the nearest cubic mm): V (94, 26, 57) = mm³. However, since your initial measurements were only accurate to the nearest mm, the true volume of toilet paper may be different. V(l,r, R) = πl(R² -p²). The total differential approximation can be used to estimate how the measured volume V (94, 26, 57) differs from the true value of V (l, r, R) as: Əl ᏭᏙ ᎧᏙ V(l,r, R) V(94, 26, 57) + (94, 26, 57) (94) + (94, 26, 57) (r - 26) + Əl Ꮩ OR (94, 26, 57) (R-57) We calculate the partial derivatives (to the nearest integer): av -(94, 26, 57) = 数字 (94, 26, 57) ** Ər av (94, 26, 57) * ƏR Since we measured l, r and R to the nearest mm, ᎧᏙ Ər av Əl |l94|< 0.5, |r – 26| < 0.5 and R — 57| < 0.5. Using the integer approximations above together with the triangle inequality, |V (l, r, R) – V (94, 26, 57)| ≤ (94, 26, 57)||(l — 94)| + | (94, 26, 57)||(r - 26)| + av Ər And so our measured volume V (94, 26, 57) differs from the true volume by |V (l,r, R) – V (94, 26, 57)| < To two decimal places, this represents an approximate error of no more than mm³. |V(l,r, R)-V(94,26,57)| V (94,26,57) Are you satisfied with this level of precision? (this part is not worth marks) O No, we need more precise measurements. O Yes, this level of precision is sufficient for my needs. O Better go buy some more toilet paper to account for this potentially significant error. <3.76%. av OR {(94, 26, 57)||(R - 57)|

Answer rating: 100% (QA)

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