|
1. Tangent and normal to a curve
Tangent to a curve is a straight line that touches the curve at one point. Normal to the curve at that point is another straight line that is perpendicular to the tangent.
2. Gradient/slope of tangent
As you know, the gradient/slope formula is
| \(y_2-y_1 \over x_2-x_1\) |
Note that in the above formula, the numerator describes the changes in the dependent variable y and the denominator describes the changes in the independent variable x.
In terms of derivatives, the gradient of tangent is given by
Example
Find the gradient of tangent to y=x2-x+3 at point (1,3).
|
\(m={dy \over dx}=2x-1 \)
\( At (1,3), x=1; \) \( m=2(1)-1 =1 \)
|
Example
Find the gradient of tangent to y=3x2+7x-5 at point (-2,-7).
|
\( {dy \over dx}=6x+7 \)
\(At \text { } x=-2; \) \( m=6-2+7 =-5 \)
|
Example
Find the gradient of tangent to y=2x3-9x2 at point (-1,-11).
|
\( {dy \over dx}=6x^2-18x \)
\( At \text { }x=-1; \) \( m=6-12-18(-1) =24 \)
|
3. Gradient/slope of normal
The relationship between the gradients of two perpendicular lines is
which gives
Since the normal line is perpendicular to the tangent line, the gradient of the normal line is given by
| \(m=-{1 \over {dy \over dx}} \) |
Example
Find the gradient of normal to y=x2-3x at point (2,-2).
|
\({dy \over dx}=2x-3 \)
\( At \text { } x=2; \) \( {dy \over dx}=2(2)-3 =1 \)
\( ∴ m_{normal}=-11=-1 \)
|
Example
Find the gradient of normal to y=5x+x2-3 at point (-1,-7).
|
\({dy\over dx}=2x+5 \)
\( At \text { } x=-1; \) \( {dy \over dx}=2-1+5 =3 \)
\( ∴ m_{normal}=-13 \)
|
Example
Find the gradient of normal to y=x3-x2-2 at point (1,-2).
|
\( {dy \over dx}=3x^2-2x \)
\( At \text { } x=1; \) \( {dy \over dx}=3-2 =1 \)
\( ∴ m_{normal}=-1 \)
|
Example
Find the equation of normal to the curve y=2x3-4 at point (1,-2).
|
\( {dy \over dx}=6x^2 \)
\( At \text { } x=1; \) \( m=6 \)
\( Then, m_1=-16 \)
\( Normal: y-(-2)=-16x-1 \)
\( 6y+2=-1(x-1) \)
\( 6y+12=-x+1 \)
\( 6y=-x-11 \)
|
|