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Unit 4 – Polynomial Curves
General Objective:
At the end of the unit, the student should be able to differentiate polynomial curves.
Specific Objectives:
At the end of the unit, the student is expected to:
- Determine the tangents and normal to plane curves;
- Familiarize with the behavior of the increasing and decreasing functions;
- Determine the maxima and minima, concavity, and inflections points of the plane curves; and
- Sketch the graph of plane polynomial curves.
Content:
Learning Activity 4.1: Determining tangents and normal to plane curves
The equation of the line of slope m through the point x 1 , y 1 is
y y 1 m x x 1
Hence, to find the tangent at any point of a plane curve we have to find the slope of the curve at that point, and substitute for m in the above formula.
The normal to a curve at the point x 1 , y 1 is defined to be the line through
that point and perpendicular to the tangent line. The equation of this normal can be found by recalling that the slope of it is the negative reciprocal of the slope of the tangent.
Example 1: Find the equations of the tangent and normal to the curve
y 3 x^2 2 x 1 at point 1 , 2 .
Solution: Evaluate the derivative of to obtain slope of the curve at 1 , 2 ,
then
y m x
y x x
Use point 1 , 2 to find the tangent,
x y
y x
Then, use 4
m 1 m to have the equation of the normal:
x y
y x
Example 2: Find the tangents to the ellipse x^2 4 y^2 8 parallel to the line x 2 y 6.
Solution: Solving for the slope of the given line and then obtain for the derivative of the ellipse and determine the points of tangency. Thus, we get
2
1
2
m
y x
x y
x y
x y
x yy
x y
2
1
2 2
The points on the ellipse where the tangents have slope of 21 can be
determined by substituting x 2 y to the equation of the given ellipse. This is
shown as follows:
(^22)
2 2
x y
y
y y
x y
Equation of the tangents:
y 1 21 x 2 or x 2 y 4 0 and x 2 y 4 0
Example 3: Find the tangents to the curve y x^4 14 x^2 17 x 40
perpendicular to the line x 7 y 4.
Solution: x 7 y 4 ; 7
m 1 ; m 7
These results are also useful in demonstrating the validity of inequalities which are valuable in many phases of advanced mathematics.
Learning Activity 4.3: Determining the maxima and minima, concavity and inflection points
At a point such as B (Fig. 4.1) where the function is algebraically greater than the neighboring point, the function is said to have the maximum value, and the point is called the maximum point. Similarly, at D the function has a minimum value. At such points the tangent is parallel to Ox. Thus,
y ' 0
But the vanishing of the derivative does not mean that the function is at maximum or minimum; the tangent is parallel Ox at F , yet the function is neither a maximum nor a minimum. From the figure, we deduce the following test:
At point where y ' 0 , if y ' changes from positive to negative (as x
increases), y is a maximum; if y ' changes from negative to positive, y is a
minimum; if y ' does not change sign, y is neither a maximum nor a minimum.
The points at which y ' 0 are called critical points. Maxima and minima
collectively are called extremes.
Example: Locate and clarify the critical points of y 31 x^3 21 x^2 2 x 2.
We find y ' x^2 x 2 x 1 x 2 0
We get the critical values x 1 or 2 and the critical points are 1 , 196 , 2 , 34 ,
maximum value y ^196 , and minimum value y 34.
The second derivative is the rate of change of the first derivative. it follows that when y '' is positive, y ' is increasing; as x increases, the tangent turns in a
counterclockwise sense and the curve is concave upward. When y ''is negative,
y ' decreases; the curve is concave downward.
Fig. 4.
At a maximum point the curve is concave downward, and hence y '' if it is
not zero, must be negative. At a minimum, if not zero, (^) y '' must be positive.
In summary, the test for concavity is as follows:
At a point where y ' 0 , if y ' ' 0 , y is a maximum; if y '' 0 , y is
minimum; if y ' ' 0 , the test fails.
Example: Examine the function y 31 x^3 21 x^2 2 x 2
We can have
y x
y x x x x
At x 1 , y ' ' 3 ; y is a maximum. At x 2 , y ' ' 3 ; y is a minimum.
Learning Activity 4.4: Sketching polynomial curves
The steps in sketching polynomial curves may be summarized as follows:
- Find the points of intersection with the axis.
- Determine the behavior of y for larges values of x.
- Locate the points where y ' 0 , and determine the maxima and minima.
- Locate the points where y ' ' 0 (points of inflection, in most cases), and draw the tangent at each of those points.
- If necessary, plot a few additional points.
However, any step that leads to serious algebraic difficulties may be omitted, provided sufficient information is obtainable without it.
Example 1: Sketch the curve y 13 x^3 21 x^2 2 x 2.
Solution:
- When x 0 , y 0. The intercepts are irrational and will not be determined.
- When x is numerically large, the sign of y is the same as the sign of the highest-degree term in x. Hence, when x is large and negative, y is large and negative; when x is large and positive, y is large and positive.
- y ' x^2 x 2 ^ x 1 ^ x 2 ; critical points ^ ^1 ,^196 the maximum and 2 , 34 the minimum.
