<p>The gradient of a curve at any point is \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \frac{1}{2x^2}\). The line \(y = 3x + 1\) is a tangent to the curve at the point \(P\), where the \(x\)-coordinate of \(P\) is positive.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find the coordinates of \(P\).</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find the equation of the curve.</span></div></li></ol>

The gradient of a curve at any point is \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \frac{1}{2x^2}\). The line \(y = 3x + 1\) is a tangent to the curve at the point \(P\), where the \(x\)-coordinate of \(P\) is positive.

Definite Integrals and Area under graph - Finding the Equation of the Curve

Question 045e319d

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