From your high-school mathematics, you should know that differentiating the equation of a curve gives you a formula for the gradient (m) of the curve. The gradient of a curve at a point is equal to the gradient of the tangent at that point. For example, in order to find the equation of the tangent to the curve y = x3at the point (2, 8), you first have to determine the derivative: m = y = 3x2.
The gradient of the tangent at the point (2, 8) when x = 2 is 3 (2)2= 12.
The equation for the tangent line is: y = 12x − 16 since y x =
8 − y
2 − x = 12 .
The normal to the curve is the line which is perpendicular (at right angles) to the tangent to the curve at that point.
If two lines are perpendicular, then the product of their gradients is −1. So if the gradient of the tangent at the point (2, 8) of the curve y = x3is 12, then the gradient of the normal is − 1
12, since − 1
12 ×12 = −1. Now try to work through the following problems involving tangents and normal lines.
Examples:
83. Two curves are said to be orthogonal at a point where the curves intersect if their tangents are perpendicular to each other at the point of intersection. Show that the curves y = sin(2x) and y = − sin(x/2) are orthogonal at the origin.
84. Find the x- and y-intercepts of the line that is tangent to the curve y = x3at the point (−2, −8).
85. Does the graph of the function
y = 2x + sin x
have any horizontal tangents in the interval 0 ≤ x ≤ 2π? If so, where? If not, why not?
86. Find equations for the lines that are tangent and normal to the curve f (x) = 1 + cos x at the point (π/2, 1).
87. (a) Find the equation for the tangent to the graph of
h(x) = 4 (1 + x)2 which makes an angle of 45◦with the x-axis.
[NB Assume that the scales along the x- and y-axes are the same. Angles are measured counterclockwise from the positive x-axis.]
(b) What is the equation of the normal to the curve of h at the point of tangency for the tangent considered in (a) above?
88. Find the points on the curve y = 2x3− 3x2− 12x + 20 where the tangent is parallel to the x-axis. What are the equations for these tangents?
89. Determine the equation for the normal to the curve of f (x) = 1
x2+ 1 at the point x = 1.
90. Determine the equations for the tangents to the curve of y = x3− 6x + 2 which are parallel to the line y = 6x − 2.
91. The line normal to the curve y = x2+ 2x − 3 at (1, 0) intersects the curve at what other point?
92. What are the equations for the tangents to the curve f (x) = x2+ 5x + 9 which pass through the origin?
Solutions:
83.
y = sin(2x)
⇒ dy
dx = (cos(2x)) d dx(2x)
= 2 cos(2x),
so dy
dx x=0= 2 × 1 = 2.
Also,
y = − sin(x 2)
⇒ dy
dx = −cos(x 2) d
dx x x
= −1 2cos(x
2), 2
so dy
dx x=0= −1
2 ×1 = −1 2. Since −1
2 (2) = −1, the given curves are orthogonal at the origin.
84. Let
f (x) = x3. Then
f (x) = 3x2,
so the gradient at the point (−2, −8) is f (−2) , that is 12. The equation of the tangent at the point (−2, −8) is therefore given by
y − (−8) = 12(x − (−2))
or
y = 12x + 16
= 4(3x + 4).
The x-intercept of this tangent is −4
3 and its y-intercept is 16.
85. Let
f (x) = 2x + sin x.
Then
f (x) = 2 + cos x.
Now the gradient of a horizontal line is 0, so if the graph of f is to have a horizontal tangent at the point (x, f (x)), then the value of f (x) must be 0. We must therefore solve the equation
f (x) = 0
for all values of x occurring in the interval 0 ≤ x ≤ 2π. But if 2 + cos x = 0 then
cos x = −2,
which is impossible. Hence the graph of f has no horizontal tangents.
86.
f (x) = 1 + cos x
⇒ f (x) = − sin x.
Hence the gradient at the point π
2, 1 is f π
2 , that is −1. The equation of the tangent at the point π 2, 1 is therefore
y − 1 = −1 x − π 2 or
y = −x + π 2 +1.
The gradient of the normal at the point π
2, 1 is 1. The equation for the normal at the point π 2, 1 is therefore
y − 1 = 1 x −π 2 or
y = x −π 2 +1.
87. (a) The tangent which makes an angle of 45◦with the x-axis has a gradient of 1. To find the x-coordinate of the corresponding point of tangency we let
h (x) = 1 and solve for x. Now
h(x) = 4(1 + x)−2
⇒ h (x) = −8(1 + x)−3, and
− 8
(1 + x)3 = 1
⇔ x = −3.
We have that
h(−3) = 4
(1 − 3)2 = 1.
The tangent thus passes through the point (−3, 1) and has a gradient of 1. Its equation is y = 1.(x + 3) + 1
that is
y = x + 4.
(b) We obtain the equation for the required normal:
y = −1.(x + 3) + 1 that is
y = −x − 2.
88.
y = 2x3− 3x2− 12x + 20
⇒ y = 6x2− 6x − 12.
Now
6x2− 6x − 12 = 0
⇒ 6(x2− x − 2) = 0
⇒ 6(x − 2)(x + 1) = 0
⇒ x = 2 or x = −1.
If x = 2 then y = 2 × 8 − 3 × 4 − 24 + 20 = 0.
If x = −1 then y = −2 − 3 + 12 + 20 = 27.
The required points are (2, 0) and (−1, 27). The equations for the tangents are y = 0 and y = 27 respectively.
89.
f (x) = 1
x2+ 1 = x2+ 1 −1
⇒ f (x) = −(x2+ 1)−2· 2x.
∴ f (1) = −1 2, so the gradient of the normal at x = 1 is 2. Now
f (1) = 1 2, so the equation for the normal at x = 1 is
y −1
2 =2(x − 1).
90. The slope of the tangent to the curve of y = x3− 6x + 2 is given by y = dy dx. Now y = 3x2− 6.
Since the tangent must be parallel to y = 6x −2, we must have that y is equal to the slope of y = 6x −2, that is
y = 6
⇔ 3x2− 6 = 6
⇔ 3x2− 12 = 0
⇔ x2− 4 = 0
⇔ x = ±2.
The corresponding y-values are
y = (−2)3− 6(−2) + 2
= −8 + 12 + 2 = 6 for x = −2, and
y = (2)3− 6(2) + 2
= 8 − 12 + 2 = −2 for x = 2.
Therefore the tangent points are (−2, 6) and (2, −2).
The equation for a line with slope m which passes through (xo, yo) is given by y − yo= m(x − xo).
The tangent through (−2, 6) therefore has the equation
y − 6 = 6(x − (−2))
= 6x + 12 that is
y = 6x + 18.
The tangent through (2, −2) has the equation
y − (−2) = 6(x − 2)
that is
y + 2 = 6x − 12 that is
y = 6x − 14.
91. The normal to the curve y = x2+ 2x − 3 at the point (1, 0) is perpendicular to the tangent at that point.
Hence, if we know the gradient of the tangent we can calculate the gradient of the normal. Now if f (x) = x2+ 2x − 3
then
f (x) = 2x + 2.
Thus
f (1) = 4,
which is the gradient of the tangent. Hence the gradient of the normal is −1
4. Since the normal passes through the point (1, 0), its equation is
y − 0 = −1
4(x − 1) that is
y = −1 4x + 1
4.
To find the points of intersection of the curve y = x2+ 2x − 3 and the line y = −1 4x + 1
4, let x2+ 2x − 3 = −1
4x + 1 4. Then
4x2+ 8x − 12 = −x + 1 so
4x2+ 9x − 13 = 0,
that is
(x − 1)(4x + 13) = 0.
Hence x = 1 or x = −13
4 . If x = −13 4 then
−1 4x + 1
4 = 13 16 +
1 4 =
17 16.
The other point of intersection of the normal and the curve is therefore −13 4 ,17
16 . 92. The equation of the tangent at a point (α, f (α)) is
y − f (α) = f (α)(x − α),
that is
y = f (α) · x − α · f (α) + f (α).
The tangent will pass through the origin only if the y–intercept is 0, that is f (α) − α · f (α) = 0,
and then the equation of the tangent becomes
y = f (α) · x. ...(∗)
Hence the solutions of the equation
f (x) − x · f (x) = 0
will give the x-coordinates α of the points of contact between the tangents and the curve, and then the equa-tions of the corresponding tangents can be determined from (∗).
Now
f (x) = x2+ 5x + 9
⇒ f (x) = 2x + 5;
and
f (x) − x · f (x) = 0
⇒ x2+ 5x + 9 − x(2x + 5) = 9 − x2= 0
⇒ x = 3 or x = −3.
We have that f (3) = 11 and f (−3) = −1. Hence the required tangents have the formulae y = 11x and y = −x.