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Lesson 6, Topic 1
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# Tangent & Normal

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For a curve given by y = f(x), $$\frac{dy}{dx}$$ is the gradient function. If given the coordinates of a point as on the curve, The value of $$\frac{dy}{dx}$$ and the equation of the tangent at the point can be found.

Note: if two lines with gradients m1, m2 are perpendicular, then m1, m2 = -1

Hence the gradient of the line perpendicular to the tangent, the normal to the curve is $$\scriptsize m_2 = \normalsize \frac{-1}{m}$$

m2 = $$\frac {-1}{\frac{dy}{dx}}$$

Examples:

1. Find the equation of the tangent and normal to the curve y = (1 – x)(2 + x), at the point where x = 2.

Solution:

y = (1 – x)(2 + x)

y = 2 – x – x2

$$\frac{dy}{dx}$$ = -1 -2x

x = 2

y = 2 – 2 – (22)

y = -4

Gradient of the tangent at the point (2, -4)

$$\frac {dy}{dx} \scriptsize = -1 \; – \; 2x \\ \scriptsize = \; -1 \; – \; 2(2) \\ \scriptsize = 5$$

y + 4 = -5(x – 2)

So its equation is y + 5x = 6

Gradient of the normal at the point = $$\frac{1}{5}$$ and its equation is

y + 4 = $$\frac {1}{5} \scriptsize(x \; – \; 2) \\ \scriptsize 5y = x \; – \; 22$$

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