Applications of Derivatives
Angle of Intersection Between Curves
Grade 12
Question:
<p>Let <span>\(f(x) = 1 + \int_0^1 (xe^y + ye^x)f(y)\,dy\)</span> where <span>\(x\)</span> and <span>\(y\)</span> are independent variables.</p><p>If the acute angle of intersection of the curves <span>\(\frac{x}{2} + \frac{y}{3} + \frac{1}{3} = 0\)</span> and <span>\(y = f(x)\)</span> is <span>\(\theta\)</span>, then <span>\(\tan\theta\)</span> equals to:</p>
<p>(a) <span>\(\frac{8}{25}\)</span></p>
<p>(b) <span>\(\frac{16}{25}\)</span></p>
<p>(c) <span>\(\frac{14}{25}\)</span></p>
<p>(d) <span>\(\frac{4}{5}\)</span></p>
Step-by-Step Solution
Key Concept: Find both curves, compute their slopes at the intersection point, and use the angle formula between two lines.
<p>First, find <span>$f(x)$</span> from the integral equation.</p><p>The line <span>$\frac{x}{2} + \frac{y}{3} + \frac{1}{3} = 0$</span> can be rewritten as <span>$y = -\frac{3x}{2} - 1$</span>, so its slope is <span>$m_1 = -\frac{3}{2}$</span>.</p><p>Find <span>$f'(x)$</span> to get the slope <span>$m_2$</span> of the curve <span>$y = f(x)$</span> at the point of intersection.</p><p>The acute angle <span>$\theta$</span> between two lines with slopes <span>$m_1$</span> and <span>$m_2$</span> is given by:</p><p><span>$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$</span></p><p>Computing with the derived values yields <span>$\tan\theta = \frac{8}{25}$</span>.</p><p>∴ Answer is (a).</p>
Correct Answer: A