Applications of Derivatives
Angle Between Curves
Grade 12
Question:
<p>Consider the two graphs <span style="font-style: italic;">y</span> = 2<span style="font-style: italic;">x</span> and <span style="font-style: italic;">x</span><sup>2</sup> − <span style="font-style: italic;">xy</span> + 2<span style="font-style: italic;">y</span><sup>2</sup> = 28. The absolute value of the tangent of the angle between the two curves at the points where they meet, is ……….</p>
Step-by-Step Solution
Key Concept: Find intersection points and calculate slopes using implicit differentiation. Use the tangent angle formula between two curves.
<p><strong>Step 1:</strong> Find points of intersection by solving $y = 2x$ and $x^2 - xy + 2y^2 = 28$ simultaneously.</p><p>Substituting $y = 2x$ into the second equation: $x^2 - x(2x) + 2(2x)^2 = 28$</p><p>$x^2 - 2x^2 + 8x^2 = 28$ → $7x^2 = 28$ → $x = \pm 2$</p><p>The points of intersection are (2, 4) and (−2, −4).</p><p><strong>Step 2:</strong> Find slopes of both curves at the point of intersection.</p><p>For curve 1: $y = 2x$, $\frac{dy}{dx} = 2$ → $m_1 = 2$</p><p>For curve 2: $x^2 - xy + 2y^2 = 28$</p><p>Differentiating implicitly: $2x - y - x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0$</p><p>$\frac{dy}{dx}(4y - x) = y - 2x$ → $\frac{dy}{dx} = \frac{y - 2x}{4y - x}$</p><p>At (2, 4): $m_2 = \frac{4 - 4}{16 - 2} = 0$</p><p><strong>Step 3:</strong> Use the formula for angle between two curves.</p><p>$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \left|\frac{2 - 0}{1 + 0}\right| = 2$</p><p>∴ Answer is 2.</p>
Correct Answer: 2