Applications of Derivatives
Related Rates
Grade 12
Question:
<p>A rod of length 5 has ends A and B sliding along the curve \(y = 2x^2\). Let \(x_A\) and \(x_B\) be the x-coordinates of the ends. When A is at \((0, 0)\) and B is at \((1, 2)\), find \(\frac{dx_B}{dx_A}\).</p>
<p>(a) \(\frac{1}{9}\)</p>
<p>(b) \(\frac{1}{7}\)</p>
<p>(c) \(\frac{1}{5}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the constraint equation for rod length and differentiate implicitly with respect to one variable to find the relationship between the rates of change.
<p><strong>Solution:</strong> We have $y = 2x^2$. The constraint is:</p><p>$$(x_B - x_A)^2 + (2x_B^2 - 2x_A^2)^2 = 25$$</p><p>Expanding: $(x_B - x_A)^2 + 4(x_B^2 - x_A^2)^2 = 25$</p><p>Differentiating w.r.t. $x_A$ and denoting $\frac{dx_B}{dx_A} = D$:</p><p>$$2(x_B - x_A)(D - 1) + 8(x_B - x_A)(2x_B D - 2x_A) = 0$$</p><p>When $x_A = 0$ and $x_B = 1$:</p><p>$$2(1)(D - 1) + 8(1)(2D - 0) = 0$$</p><p>$$2D - 2 + 16D = 0$$</p><p>$$18D = 2$$</p><p>$$D = \frac{1}{9}$$</p><p>∴ Answer is (a).</p>
Correct Answer: A