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[h]SL 5.4 Tangents and normals at a given point, and their equations

[q] **Tangent/Normal**

We have seen that the tangent is what is actually being assessed when we are finding the slope at a point. Well, instead of just finding the slope of it, a natural extension would be to ask what the full equation of the tangent is.

[q]

TANGENT PROCESS: From (2.1), we know if we have an \((x_1, y_1)\) point on the line and its slope, we can use \(y – y_1 = m(x – x_1)\) to get the equation. So, if we are asked to find the equation of the tangent, given \(f(x)\) and an \(x\), we can:

1. Do \(f(x_1)\) to find \(y_1\)

2. Find \(f'(x)\) by differentiating

3. Do \(f'(x_1)\) to find \(m\)

4. Plug \(x_1\), \(y_1\), and \(m\) into \(y – y_1 = m(x – x_1)\) and rearrange if needed

[a]

NORMAL: The normal to a curve is a straight line, passing through a given point, but perpendicular to the tangent (see diagram).

PROCESS: We can still simply find an \((x_1, y_1)\) and the appropriate gradient, we just use the negative reciprocal \(\left[-\frac{1}{m}\right]\) instead, as it is perpendicular.

Example : With \(f(x)\) and point from Example 1, find the equation of the normal:

We still have \(f(2) = 4\) and \((x_1, y_1) = (2, 4)\). \(f'(2) = 2\) still, but we use \(\frac{-1}{m} = \frac{-1}{2}\) instead.

\(y – 4 = -\frac{1}{2}(x – 2)\) ⟶ \(y – 4 = -\frac{1}{2}x + 1\) ⟶ \(y = -\frac{1}{2}x + 5\)

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