CAIE IGCSE Additional Math 14.5 gradients tangents and normals

Use this tangents-and-normals page to differentiate a curve, substitute the point and form the required tangent or normal equation.

Exam points

Find dy/dx and substitute the given x-value to get the tangent gradient.
Use the negative reciprocal gradient when forming the normal equation.
Write tangent or normal equations using point-gradient form and curve coordinates.

[Maximum number: 7]

A curve has equation $y=\ln(5-3x)$ , where $x<\frac{5}{3}$ . The normal to the curve at the point where x=-5 cuts the x-axis at the point P.
Find the equation of the normal and the x-coordinate of P.

$\frac{dy}{dx}=\frac{-3}{5-3x}, \qquad \left.\frac{dy}{dx}\right|_{x=-5}=-\frac{3}{20}.$
At (x=-5), (y=\ln20), so the normal gradient is $\frac{20}{3}$ . Hence
$y-\ln20=\frac{20}{3}(x+5).$
At the (x)-axis, (y=0), giving
$x=-5-\frac{3\ln20}{20}=-5.449\ldots.$

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14.5. Find gradients, tangents and normals question 1