We have studied the derivative of functions such as polynomial, rational, exponential, logarithmic and trigonometric functions. We now are ready to apply
these derivatives in:
finding the equations of tangent and normal lines,
finding the critical points to analyze the maximum, minimum and inflection points,
testing the concavity,
finding the maxima and minima,
solving the approximation problems.
9.1 Tangent Line and Normal Line
We first introduce the inclination of the line and its slope. The inclination of a line that intersects the x-axis is the measure of the smallest
nonnegative angle which the line makes with the positive end of the x-axis. We shall use the symbol θ to represent the angle of inclination.
We here discuss the expression related to the inclination, namely, the slope of a line, say m, is the tangent of the inclination,
m = tan θ .
We see that the vertical lines have inclination 90° but no slope. A horizontal line has a slope and that slope is the number 0.
If θ is as shown in either of the two positions in the following figure, then
tan θ =
y⁄x .
Suppose we have a line with a pair of points, P1 = (x1, y1) and P2 = (x2,
y2), on it.
From the figure, we can write as m =
y2 - y1⁄x2 - x1
= tan φ
Since θ = φ, we get m = tan θ.
From the figure, we can write as m =
y1 - y2⁄x2 - x1 = -
y2 - y1⁄x2 - x1
= - tan θ .
Next we recall the notion of equations of tangent and normal line.
For the curve y = f(x), the gradient of the tangent ℓ1 at the point (x1, y1) is the value of
- y'(x) at x = x1, hence the equation of the tangent at (x1, y1) is y - y1 = y' (x1)(x - x1) .
The line ℓ2 which is perpendicular to the tangent ℓ1 at (x1, y1) is called the normal
to the curve at (x1, y1). Hence its gradient is the value of -(y' (x1) )-1 where
y'(x1) ≠ 0 and the equation of the normal at (x1, y1) is y - y1 = -
1⁄y'(x1) (x - x1).
Example 1.
Find the equations of tangent and normal line to the curve y = ln x at the point y = -1. Solution y = ln x, y' = 1⁄x .
When y = -1, ln x = -1, x = e-1 =
1⁄e .
The equation of tangent at
(1⁄e , -1) is y - (-1) = e(x -
1⁄e) y = ex - 2.
The equation of normal at
(1⁄e , -1) is y - (-1) = -
1⁄e (x -
1⁄e) y = -
1⁄ex +
1⁄e2 - 1.
Example 2.
Find the equation of tangent to y = sin x at the origin. Solution y = sin x, y' = cos x.
At (0,0), the gradient of tangent is y' (0) = cos 0 = 1.
The equation of tangent is y - 0 = 1(x - 0) y = x.
Example 3.
Consider the curve f(x) = x2 + ax + b where a and b are constants. The tangent to this curve at the
point x = 2 is y = 2x + 1. Find the values of a and b. Solution f(x) = x2 + ax + b, f'(x) = 2x + a.
At x = 2, f(2) = 2a + b + 4 and f'(2) = a + 4.
By problem, the gradient of tangent line is 2, so a + 4 = 2 which gives a = -2.
The tangent line equation at (2, 2a + b + 4) is y - (2a + b + 4) = (a + 4) (x - 2).
Substitute the value a = -2 in above equation, then y - b = 2(x - 2) y = 2x + b - 4.
Since the tangent line equation at x = 2 is y = 2x + 1, it follows that b = 5
Therefore a = -2 and b = 5.
Exercise 9.1
1. Find the equations of tangent and normal line to the curve y = 2 ln x at (1, 0). Solution y = 2 ln x y' = 2 ⋅ 1 ⁄x =
2 ⁄x
At (1, 0), the gradient of tangent is m = y'(1) =
2 ⁄ 1 = 2
The equation of tangent line is y - y1 = m(x - x1) y - 0 = 2(x - 1) y = 2x - 2
The gradient of normal line = - 1 ⁄m =
- 1 ⁄ 2
The equation of normal line is y - y1 = -
1 ⁄m (x - x1) y - 0 = - 1 ⁄ 2 (x - 1) y = - 1 ⁄ 2 x +
1 ⁄ 2
2. Find the equation of the line which passes through the origin and tangent to y = ex. Solution y = ex y' = ex
The tangent line at (0, 0) is y - y0 = y' (x - x0) y - 0 = y' (x - 0) y = y'x y = ex ⋅ x ex = ex ⋅ x (∵ y = ex is given) ex⁄ex = x x = 1 y = ex = e1 = e
The point at (x1, y1) is (1, e)
The gradient of tangent line at (1, e) is
m = y' = ex = e1 = e
Equation of tangent line at (1, e) is y - y1 = m(x - x1) y -e = e(x - 1) y - e = ex - e y = ex
3. Find the points of contact where horizontal tangent meet the curve y = x4 - 2x2 + 2. Solution y = x4 - 2x2 + 2 dy⁄dx = 4x3 - 4x
When
dy⁄dx = 0,
4x3 - 4x = 0
4x(x2 - 1) = 0 x(x2 - 1) = 0 x = 0 or x2 - 1 = 0 x = 0, 1, -1
When x = 0, y = 04 - 2(0)2 + 2 = 2
When x = 1, y = 14 - 2(1)2 + 2 = 1
When x = -1, y = (-1)4 - 2(-1)2 + 2 = 1
The point of contacts are (0,2), (1,1) and (-1,1).
4. Consider the curve y = a
√x +
b⁄
√x where a and b are constants. The normal to this curve at the point at x = 4 is 4x + y = 22. Find the values of a and b. Solution y = a
√x +
b⁄
√x
= ax 1 ⁄ 2 +
b⁄x 1 ⁄ 2
= ax 1 ⁄ 2 + b x -
1 ⁄ 2 dy⁄dx = (a ⋅
d⁄dxx 1 ⁄2 +
x 1 ⁄ 2 ⋅
da⁄dx) +
(b ⋅
d⁄dxx -1 ⁄2 + x-
1 ⁄ 2 ⋅
db⁄dx)
= (a ⋅
1 ⁄ 2 x -1 ⁄2 +
x 1 ⁄ 2 ⋅ 0) +
(b ⋅
(- 1 ⁄ 2 ) x -3 ⁄2 + x-
1 ⁄ 2 ⋅ 0 )
= a⁄2 x1⁄2
- b⁄2 x3⁄2
At x = 4, dy⁄dx =
a⁄2 ⋅ 4
1⁄2
- b⁄2 ⋅ 4
3⁄2
= a⁄ 2 ⋅ 2
- b⁄ 2 ⋅ 23
= a⁄ 4
- b⁄ 16 m = dy⁄dx =
a⁄ 4
- b⁄ 16
The normal line is
4x + y = 22 y = 22 - 4x dy⁄dx =
0 - 4 = -4
Gradient of the normal line is - 1 ⁄m = -4
Gradient of the tangent line is
-1 = -4m m = 1 ⁄ 4
Hence, a⁄ 4
- b⁄ 16 =
1 ⁄ 4 4a - b⁄ 16 =
- 1 ⁄ 4
4a - b = 1 ⁄ 4 ⋅ 16
4a - b = 4 __________(1)
At x = 4, y = 22 - 4x
= 22 - 4⋅4
= 22-16 = 6
(4, 6) lie on y = a
√x +
b⁄
√x
6 = a
√ 4 +
b⁄
√
4
6 = 2a + b⁄2
6 - 2a = b⁄2
(6 - 2a) 2 = b
12 - 4a = b
12 = 4a + b
4a + b = 12 __________(2)
eq______(1) + (2)
4a - b = 4
4a + b = 12
___________
8a = 16 a = 2
Substituting a = 2 in equation (1)
4a - b = 4
4(2) -b = 4
8 - b = 4
-b = -4 b = 4
