09Dtangents

= Tangents and Normals =


 * Tangents **


 * Recall, a ** tangent ** is a straight line that just touches the curve only at the point of contact.
 * The tangent line has the same gradient as the curve at the point of contact


 * The __**derivative**__ of a function is a gradient function that gives the value of the gradient at each point on the curve.
 * This means that if f '(2) = 5, then f(x) has a gradient of 5 at x = 2


 * Having the gradient, we can find the equation of the tangent for a given point on the curve.


 * If (x, y) is a point on the curve, f (x), and f is differentiable at x = x 1 ,
 * then the equation of the __tangent__ at (x 1, y 1 ) is given by:

math . \qquad\qquad y-y_1=m(x-x_1) \qquad \text{ where } \; m = f '(x_1) math

** Example 1 **
... ... Find the equation of the tangent of the curve:

math . \qquad \qquad y = x^3 + \dfrac{1}{2}x^2 \quad \text{ at the point where } \;\; x = 1 math


 * Solution:**

math \\ . \qquad \text{When } x = 1 \\.\\ . \qquad y = 1^3 + \dfrac{1}{2} \times 1^2 \\. \\ . \qquad y = 1 + \dfrac{1}{2} \\.\\ . \qquad y = \dfrac{3}{2} \\.\\.\\ math math . \qquad \dfrac{dy}{dx}=3x^2+x math

math \\ . \qquad \text{When }x = 1 \\.\\ . \qquad \dfrac{dy}{dx} = 3 \times 1^2 + 1 \\. \\ . \qquad \dfrac{dy}{dx} = 4 math


 * Therefore, the equation of the tangent is:

math \\ . \qquad y - y_1 = m(x - x_1) \\. \\ .\\ . \qquad y-\dfrac{3}{2}=4(x-1) \\. \\ . \qquad y-\dfrac{3}{2}=4x-4 \\. \\ . \qquad y=4x-2\frac{1}{2} math

Normals

 * The **normal** to a curve at a point is the __**straight line**__ that passes through the point
 * and is __**perpendicular**__ to the tangent at that point.


 * If m 1 is the gradient of the tangent
 * and m 2 the gradient of the normal
 * then m 1 × m 2 = –1.

So for Example 1 above: Therefore the equation of the normal is:
 * the normal will have gradient of -¼.
 * (See the graph on the right)

math \\ . \qquad y - y_1 = m(x - x_1) \\. \\ . \\ . \qquad y-\dfrac{3}{2}= -\dfrac{1}{4}(x-1) \\. \\ . \qquad y - \dfrac{3}{2} = -\dfrac{1}{4}x + \dfrac{1}{4} \\. \\ . \qquad y = -\dfrac{1}{4}x + \dfrac{7}{4} math

or

math \\ . \qquad 4y = -x + 7 \\. \\ . \qquad 4y+x=7 math

Using Classpad

 * We can use a CAS calculator to find an expression for the equation of the tangent and normal.


 * On the Main page type:

math . \qquad \qquad x^3+\dfrac{1}{2}x^2 math


 * Highlight it and tap:INTERACTIVE, CALCULATION, tanLine.
 * In the point box enter x = 1.


 * Repeat the process to obtain the equation of the normal,
 * choosing normal at x = 1.


 * Note that the CAS calculator shows an expression instead of the equation.
 * You will need to write your answer as an equation, as indicated in the answers above.

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