DIFFERENTIATION IT LESSON 2 - EQUATIONS OF
TANGENT AND NORMAL (SEC 4 ADDITIONAL MATHEMATICS)
Software: LiveMath
(previously known as MathView)
Thinking Skills: Induction
and Deduction At the end of the lesson, the students should be able to find the equations
of the tangent and the normal to a curve at a given point.
The 1st LiveMath document (IT4AMDiffnTangent1.thp)
shows the graph of y = ax3 + bx2 + cx + d (blue
curve) where a=0.5, b = 1, c = –3
and d = 3. The red line is the tangent
to the curve at P(x1, y1) and the pink
line is the normal to the curve at the same point. In this case,
the point P is (1, 1.5).
2.
A normal is a line which is perpendicular
to the tangent at the point of contact P.
3.
To find the equation of the tangent, you have to find the
gradient of the curve at P(1, 1.5). Show your working in the space provided
in the Worksheet. Please note that gradient of curve at P = gradient of
tangent to the curve at P.
4.
Double click on the Comment Icon beside "Show
Derivative of y" to check your answer for para 3. Please note
that LiveMath uses a different symbol for dy/dx. [Actually, the symbol
is for partial derivatives.] To hide the answer, double click on the same
icon.
Note:
If you are using Internet Explorer,
you may need to do a triple click
instead of a double click.
5.
Since we know the gradient of the tangent and the point
P on the tangent, we use the formula for the equation of a straight line:
y – y1 = m (x – x1). Show how you find the equation
of the tangent in the space provided in the Worksheet.
6.
Double click on the Comment Icon beside "Show
Equation of Tangent" to check your answer in para5.
7.
To find the equation of the normal to the curve at P, just remember
that the normal is perpendicular to the tangent and so the product of their
gradients is –1. Therefore the equation of the normal is y – y1
= (–1/m) (x – x1). Show how you find the equation of the normal
in the space provided in the Worksheet.
8.
Double click on the Comment Icon beside "Show
Equation of Normal" to check your answer in para 7.
B. EXERCISE
1.
Change the x-coordinate of the point P and the equation
of the curve to find the equations of the tangent and the normal of the
following curves at the given points. You may need to adjust the graph
to view the point of contact P. The first one has been done for you.
No.
Curve
Point P
Equation of Tangent
Equation of Normal
1.
y = 0.5x3 + x2 – 3x + 3
x = 1
y = 0.5x + 1
y = – 2x + 3.5
2.
y = – x3 + 2x2 + 3x + 2
x = 2
3.
y = 2x2 – 3x + 3
x = 0
4.
y = – x2 + 3x + 6
x = – 0.5
5.
y = 2x3 + 3x2 + 2x + 5
x = – 1
6.
y = 0.5x2 – x + 3
x = 1
2.
For No. 6, the gradient of the tangent
is zero and so the gradient of the normal is not defined. By looking at
the tangent in the graph, deduce the equation of the normal which is a
vertical line.
The 2nd document (IT4AMDiffnTangent2.thp) shows the graph
of y = 0.5x3 + x2 – 3x + 3, the tangent and the normal
to the curve at the point P where the x-coordinate is h.
2.
Click on the graph and it will show the animation of the tangent and
the normal at different points on the curve. The x-coordinates of these
points, h, will increase from –3.5 to 3 in steps of 0.5.
Note:
The animation will stop at 2.5. It
will not include the last value.
3.
Change the values of a, b, c and d and observe the animation.
You may need to adjust the graph to view the point of contact P. The instructions
for zooming in and out are given in the first LiveMath document above.