 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/14/a/solution

c^{2} = a^{2} + b^{2}

2 c d c = 2 a d a + 2 b d b

, so

\frac{d c}{c} = \frac{a d a + b d b}{a ^{2} + b ^{2}} = \frac{a ^{2}}{a ^{2} + b ^{2}} \frac{d a}{a} + \frac{b ^{2}}{a ^{2} + b ^{2}} \frac{d b}{b} .

A = ab /2

, so the product rule gives

d A = (b d a + a d b) /2

and

\frac{d A}{A} = \frac{d a}{a} + \frac{d b}{b} .

The prescribed small changes therefore give

d A / A \approx 0.01 - 0.02 = - 0.01

: the area decreases by approximately $1%$ . For comparison, retaining the product of the two finite percentage changes gives

(1.01) (0.98) - 1 = - 0.0102

, an exact decrease of

1.02%

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