08Aequationtangentsnormals

toc = Tangents and Normals =


 * Recall, a tangent is a straight line that just touches the curve only at the point of contact.


 * This point is called the ** point of tangency **.


 * At this point, the tangent line and the curve have the same gradient.


 * The __ **derivative** __ of a function is a gradient function that gives the measure 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=f '(x_1)(x-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

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
 * __Solution:__**

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:
 * the normal will have gradient -¼.
 * (See the graph on the right)

Therefore the equation of the normal is:

math \\ . \qquad y-y_1=-\dfrac{1}{f'(x_1)}(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 the 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.

** Example 2 **
... ... Find the equation of the normal to the curve with equation //y = x sin// (//x//) at the point on the curve where //x =// p.


 * Solution:**

math \\ . \qquad \text{When } x = \pi, \quad y = \pi \, \sin (\pi) \\. \\ . \qquad \qquad \qquad \; y = 0 math

math . \qquad \dfrac{dy}{dx}=sin(x)+xcos(x) \qquad \{\textit{using the product rule} \} math

math . \qquad \text{At } x=\pi, \quad \dfrac{dy}{dx}=-\pi math

math \\ . \qquad \text{The gradient of the tangent is } -\pi \\. \\ . \qquad \text{So, the gradient of the normal is } \dfrac{1}{\pi} math

math . \qquad y-0=\dfrac{1}{\pi}(x-\pi) \\.\\ . \qquad \pi{y}=x-\pi math

** Example 3 **
... ... Find the equation of the tangent to the curve with equation //y = e 2x–4 // that is parallel to the line //y = 5 + ½x//.


 * __Solution:__**


 * Parallel lines have equal gradient.
 * So the tangent will have the same gradient as the line y = 5 + ½x.


 * So gradient of tangent is ½.


 * Need to find the coordinates of the point where the gradient of the curve is ½.

math . \qquad \text{i.e. } \dfrac{dy}{dx}=\dfrac{1}{2} \\.\\ . \qquad \dfrac{dy}{dx}=2e^{2x-4}\qquad\text{ using the chain rule} \\.\\ . \qquad 2e^{2x-4}=\dfrac{1}{2} \\.\\ . \qquad \;e^{2x-4}=\dfrac{1}{4} \\.\\ . \qquad 2x-4=ln \left( \dfrac{1}{4} \right) \\.\\ . \qquad 2x=4+ln \big( {4^{-1}} \big) \\.\\ math

math . \qquad \;x=2-\dfrac{1}{2}ln(4) \\.\\ . \qquad \;x=2-ln(4^{\frac{1}{2}}) \\.\\ . \qquad \;x=2-ln(2) \\.\\ math

math . \qquad \text{When } x=2-ln(2) \\.\\ . \qquad y=e^{2(2-ln(2)) - 4} \\.\\ . \qquad y=e^{ln(2^{-2})} \\.\\ . \qquad \; y=e^ { ln \left( \frac{1}{4} \right) } \\.\\ . \qquad\;y=\dfrac{1}{4} math


 * Therefore the equation of the tangent is:

math . \qquad y-\dfrac{1}{4}=\dfrac{1}{2} \Big( x- \big(2-ln(2) \big) \Big) \\.\\ . \qquad y -\dfrac{1}{4} = \dfrac{x}{2} -1 + \dfrac{ln(2)}{2} \\.\\ . \qquad y = \dfrac{x}{2} -\dfrac{3}{4} + \dfrac{ln(2)}{2} \\.\\ . \qquad \big\{ or \big\} \\.\\ . \qquad 4y = 2x - 3 - 2ln(2) \\.\\ . \qquad 4y - 2x = -3 - 2ln(2) \qquad \big\{ or \big\} \\.\\ . \qquad 2x - 4y = 3 + 2ln(2) math Go to top of page flat

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