The gradient of a curve at any point is $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \frac{1}{2x^2}$. Th

Question

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. Find the coordinates of $P$. Hence find the equation of the curve.

Accepted Answer

(a) $P=\left(\frac{1}{2},\frac{5}{2}\right)$ (b) $y=x-\frac{1}{2x}+3$

Differentiation - Tangents and Normals at a Point

Differentiation - Tangents and Normals at a Point

Video Solution:

Solution:

Solution locked