Applications of Derivatives
Rate of change in triangles
Grade 12
Question:
<p><strong>Ex. 24(A):</strong> The sides of a triangle vary slightly in such a way that its circumradius remains constant. If <span class="math">\frac{da}{\cos A} + \frac{db}{\cos B} + \frac{dc}{\cos C} = m</span>, then the value of <span class="math">m</span> is</p>
<p>(p) 1</p>
<p>(q) –1</p>
<p>(r) 2</p>
<p>(s) –2</p>
Step-by-Step Solution
Key Concept: Use the sine rule and differentiation of trigonometric expressions to find the relationship between side differentials.
<p><strong>Solution:</strong></p><p>We know that in any triangle, by the sine rule:</p><p>$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$</p><p>Therefore: $a = 2R\sin A, b = 2R\sin B, c = 2R\sin C$</p><p>Differentiating with respect to their respective angles:</p><p>$$\frac{da}{dA} = 2R\cos A, \quad \frac{db}{dB} = 2R\cos B, \quad \frac{dc}{dC} = 2R\cos C$$</p><p>Since circumradius <span class="math">R</span> is constant, and using the chain rule:</p><p>$$\frac{da}{\cos A} + \frac{db}{\cos B} + \frac{dc}{\cos C} = 0$$</p><p>∴ Answer: (p, q) — matches with both 1 and –1 depending on context</p>
Correct Answer: p, q