Applications of Derivatives
Tangent line to curve
Grade 12
Question:
<p>In cartesian coordinates the point A is <span>\((x_1, y_1)\)</span>, where <span>\(x_1 = 1\)</span> on the curve <span>\(y = x^2 + x + 10\)</span>. Then the tangent at A cuts the X-axis at B. Find the value of the dot product <span>\(\vec{OA} \cdot \vec{AB}\)</span>.</p>
<p>(a) <span>\(-\frac{520}{3}\)</span></p>
<p>(b) <span>\(-148\)</span></p>
<p>(c) <span>\(140\)</span></p>
<p>(d) <span>\(12\)</span></p>
Step-by-Step Solution
Key Concept: Use calculus to find the slope of the tangent at a point on a curve, then use the tangent equation to find where it intersects the X-axis, and finally compute the dot product.
<p><strong>Step 1:</strong> Find point A on the curve.</p><p>Given curve: <span>\(y = x^2 + x + 10\)</span></p><p>When <span>\(x_1 = 1\)</span>: <span>\(y_1 = 1^2 + 1 + 10 = 12\)</span></p><p><span>\(\therefore A = (1, 12)\)</span></p><p><strong>Step 2:</strong> Find the equation of the tangent at A.</p><p><span>\(\frac{dy}{dx} = 2x + 1\)</span></p><p>At <span>\(x = 1\)</span>: <span>\(\frac{dy}{dx} = 2(1) + 1 = 3\)</span></p><p>Equation of tangent: <span>\(y - 12 = 3(x - 1)\)</span></p><p><span>\(y = 3x + 9\)</span></p><p><strong>Step 3:</strong> Find point B where tangent cuts the X-axis.</p><p>Set <span>\(y = 0\)</span>: <span>\(0 = 3x + 9\)</span></p><p><span>\(x = -3\)</span></p><p><span>\(\therefore B = (-3, 0)\)</span></p><p><strong>Step 4:</strong> Calculate <span>\(\vec{OA} \cdot \vec{AB}\)</span>.</p><p><span>\(\vec{OA} = (1, 12)\)</span></p><p><span>\(\vec{AB} = B - A = (-3, 0) - (1, 12) = (-4, -12)\)</span></p><p><span>\(\vec{OA} \cdot \vec{AB} = (1)(-4) + (12)(-12) = -4 - 144 = -148\)</span></p><p>∴ Answer is (b).</p>
Correct Answer: B