Chapter 3: SOME APPLICATIONS OF THE DERIVATIVE
3.1 EQAUTIONS OF TANGENTS AND NORMALS
𝑑𝑦
From the previous topic (Geometric Significance of 𝑑𝑥 ), the derivative of a function
can be interpreted as the slope of the tangent to the graph of the function.
In figure above, the line T is the tangent to the curve y = f(x) at P1 (x1, y1). The
other line N perpendicular to T at P1 is called the normal to the curve.
If y = f(x) is differentiable at x1, i.e., f’(x1) exists then we may formulate the
following definitions:
1. The tangent to the curve y = f(x) at P1 (x1, y1) is the line through P1 with slope f’(x1).
2. The normal to the curve y = f(x) at P1 (x1, y1) is the line through P1 and perpendicular
to the tangent at P1.
The equation of the tangent is given by the point-slope form of the
equation of a straight line in analytic geometry, that is
y – y1 = m (x – x1)
where m = value of y’ at x = x1 or m = f’ (x1).
Since the normal is perpendicular to the tangent, then its slope is the negative
reciprocal of the slope of the tangent. Hence the equation of the normal is
y – y1 = -
𝟏
𝒎
(x – x1)
where m = value of y’ at x = x1 or m = f’ (x1).
Example:
Find the equations of the tangent and normal to the curve y = x3 at the point (2,8).
Solution:
1. Find y’: y’ = x3 = 3x2
2. The point of tangency is (2,8), so x1 = 2 and y1 = 8. Since m = y’, then, y’ = m =
3x2 = 3(2)2 = 12.
3. By using the equation of the tangent:
y – y1 = m (x – x1)
y – 8 = 12 (x – 2)
12x – y – 16 = 0
4. By using the equation of the normal:
1
y – y1 = - 𝑚 (x – x1)
y–8=-
1
12
(x – 2)
x + 12y – 98 = 0
Exercise 3.1
1. Find the equations of the tangent and normal to the graph of the given function at
the given point.
a) y = 3x2 – 2x + 1,
b) y = 1 + 3√𝑥,
c) y = x√𝑥 − 1,
d) y2 =
𝑥3
4−𝑥
2
e) y = ,
𝑥
,
(2,9)
(4,7)
(5,10)
(2,2)
(1,2)
2. Where will the tangent to y = √4𝑥 at (1,2) cross the x-axis?
3. At what point on the curve xy2 = 6 will the normal pass through the origin?
4. Find the area of the triangle formed by the coordinate axes and the tangent to xy =
5 at (1,5).
5. Find the area of the triangle bounded by the coordinate axes and the tangent to y=x2
at the point (2,4).
6. Find the area of the triangle formed by the x-axis, the tangent and normal to xy = 4
at (2,2).
7. Find the tangent to x2 + y2 = 5 and parallel to 2x – y = 4.
8. Show that the tangent with slope m to y2 = 4ax is the line y = mx+ a/m.
3.2 ANGLE BETWEEN TWO CURVES
The angle between two curves at a point of intersection may be defined as the
angle between their tangents at this point of intersection. If the tangents are not
perpendicular to each other, then such tangents form a pair of acute angles and a pair
of obtuse angles. The acute and obtuse angles are supplementary.
Consider the curves y = f1(x) and y = f2(x) which intersect at a point P0 (x0,y0) as
shown in the figure above. Let 𝜃 1 and 𝜃 2 be the inclinations of the tangents T1 and T2 at
P0, respectively. Let ∅ be the angle between these tangents. Then, ∅ is also the angle
between the curves. It can easily be shown that ∅, 𝜃 1 and 𝜃 2 are related by the equation
∅ = 𝜃2 - 𝜃1
(1)
Then taking the tangent of both sides of (1), we get
tan ∅ = tan ( 𝜃 2 - 𝜃 1 )
(2)
or
tan ∅ =
tan 𝜃2 − tan 𝜃1
1+ tan 𝜃2 tan 𝜃1
Let m1 and m2 be the slopes of T1 and T2, respectively. Then m1 = tan 𝜃1 and m2 = tan 𝜃2 .
Substituting these in equation (3) above, we obtain
tan ∅ =
𝑚2 − 𝑚1
1+ 𝑚2 𝑚1
NOTE:
The sign of tan ∅ in (4) is positive or negative depending upon the values of
𝑚1 and 𝑚2 or on the order in which 𝑚1 and 𝑚2 are used. If tan ∅ > 0, then ∅ is acute
and if tan ∅ < 0, then ∅ is obtuse.
Since tan ∅ > 0 if ∅ is acute, then we may use the absolute value symbol in the
right member of (4). Thus our final formula would be
𝒎 − 𝒎𝟏
tan ∅ = |𝟏+𝟐𝒎
𝟐 𝒎𝟏
|
where the values of m1 and m2 are given by the derivatives of the functions at P0
(x0,y0). That is,
Example:
Find the acute angle of intersection between the curves x2 = 8y and xy = 8.
Solution:
Exercise 3.2
Find the acute angle between given curves.
1. y2 = 2x and 4x2 + 4y2 + 5y = 0
2. x2 + y2 = 5 and y2 = 4x + 8
3. x2y + 4a2y = 8a3 and x2 = 4ay
4. 2y2 = 9x and 3x2 = - 4y
5. x2y + 4y = 8 and x2y = 4
6. xy = 18 and y2 = 12x