Applications of Derivatives
Tangent to a curve
Grade 12
Question:
<p>The tangent to the curve \(y = xe^{x^2}\) at the point \((1, e)\), also passes through the point:</p>
<p>\(\left(\dfrac{5}{3}, 2e\right)\)</p>
<p>\(\left(\dfrac{4}{3}, 2e\right)\)</p>
<p>\((3, 6e)\)</p>
<p>\((2, 3e)\)</p>
Step-by-Step Solution
Key Concept: Find the equation of the tangent line using the derivative, then check which point satisfies this linear equation. The derivative gives the slope, and point-slope form gives the tangent equation.
<p><strong>Step 1:</strong> Find the derivative of y = xe^(x²)</p><p>Using the product rule: dy/dx = e^(x²) + x · e^(x²) · 2x = e^(x²)(1 + 2x²)</p><p><strong>Step 2:</strong> Evaluate the slope at x = 1</p><p>dy/dx|ₓ₌₁ = e¹(1 + 2(1)²) = e(1 + 2) = 3e</p><p><strong>Step 3:</strong> Write the equation of tangent line at point (1, e)</p><p>Using point-slope form: y - e = 3e(x - 1)</p><p>y = 3ex - 3e + e</p><p>y = 3ex - 2e</p><p><strong>Step 4:</strong> Identify which given point satisfies this equation</p><p>Substitute the coordinates of each option into y = 3ex - 2e and verify which one is true. The point that satisfies this equation is the answer.</p><p>∴ Answer: B</p>
Correct Answer: B