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June 2002 p1 q9
1406
A curve is such that \(\frac{dy}{dx} = \frac{12}{(2x+1)^2}\) and \(P(1, 5)\) is a point on the curve.
(i) The normal to the curve at \(P\) crosses the x-axis at \(Q\). Find the coordinates of \(Q\).
(ii) Find the equation of the curve.
(iii) A point is moving along the curve in such a way that the \(x\)-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the \(y\)-coordinate when \(x = 1\).
Solution
(i) At \(P(1,5)\), the gradient \(m\) of the tangent is \(\frac{4}{3}\). The gradient of the normal is \(-\frac{3}{4}\) (since \(m_1 \cdot m_2 = -1\)). The equation of the normal is \(y - 5 = -\frac{3}{4}(x - 1)\). Setting \(y = 0\) to find where it crosses the x-axis, we solve \(0 - 5 = -\frac{3}{4}(x - 1)\), giving \(x = \frac{23}{3}\). Thus, \(Q \left( \frac{23}{3}, 0 \right)\).
(ii) Integrate \(\frac{dy}{dx} = \frac{12}{(2x+1)^2}\) to find \(y\). Let \(u = 2x+1\), then \(du = 2dx\), so \(dx = \frac{du}{2}\). The integral becomes \(\int \frac{12}{u^2} \cdot \frac{du}{2} = -\frac{6}{u} + C\). Substituting back, \(y = -\frac{6}{2x+1} + C\). Using \(P(1,5)\), \(5 = -\frac{6}{3} + C\), so \(C = 7\). Therefore, \(y = -\frac{6}{2x+1} + 7\).
(iii) Given \(\frac{dx}{dt} = 0.3\), find \(\frac{dy}{dt}\) using \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). At \(x = 1\), \(\frac{dy}{dx} = \frac{4}{3}\). Thus, \(\frac{dy}{dt} = \frac{4}{3} \times 0.3 = 0.4\) units per second.
