Applications of Derivatives
Tangent to a curve
Grade 12
Question:
<p>If \(\frac{dy}{dx} = (2x^2 + 1)e^{x^2}\), find the equation of the tangent at <em>x</em> = 1. Which point does the tangent pass through?</p>
<p>(a) \(\left(\dfrac{2}{3}, e\right)\)</p>
<p>(b) \(\left(\dfrac{4}{3}, 2e\right)\)</p>
<p>(c) \(\left(\dfrac{1}{3}, e\right)\)</p>
<p>(d) \(\left(\dfrac{4}{3}, e\right)\)</p>
Step-by-Step Solution
Key Concept: Integrate the derivative to find y(x), then use point-slope form at x=1. The tangent line equation will reveal which specific point it passes through by substituting values.
<p><strong>Step 1:</strong> Find y(x) by integrating the given derivative.</p><p>Given: dy/dx = (2x² + 1)e^(x²)</p><p>Notice that d/dx(e^(x²)) = 2xe^(x²), so we can rewrite:</p><p>dy/dx = 2x² e^(x²) + e^(x²) = d/dx(x²e^(x²) - e^(x²)) + e^(x²)</p><p>Using substitution or recognizing the pattern: ∫(2x² + 1)e^(x²)dx = x²e^(x²) + C</p><p>Therefore: <strong>y = x²e^(x²) + C</strong></p><p><strong>Step 2:</strong> Find the slope at x = 1.</p><p>dy/dx|_(x=1) = (2(1)² + 1)e^(1²) = 3e</p><p><strong>Step 3:</strong> Find y-coordinate at x = 1 (assuming C = 0 for the specific curve).</p><p>y(1) = 1² · e^(1²) = e</p><p>Point of tangency: <strong>(1, e)</strong></p><p><strong>Step 4:</strong> Write the tangent line equation using point-slope form.</p><p>y - e = 3e(x - 1)</p><p>y = 3ex - 3e + e</p><p><strong>y = 3ex - 2e</strong></p><p>∴ The tangent passes through <strong>(0, -2e)</strong> and <strong>(1, e)</strong>. Answer: B</p>
Correct Answer: B