THE TANGENT AND NORMAL LINE
A tangent line is a line that “brushes” a point on a curve. It passes through
the function only once near that point.
The slope of the tangent line allows us to calculate an instantaneous rate of
change of the function. The slope of the tangent of 𝑓(𝑥) at 𝑥 = 𝑎 is calculated
using the derivative of 𝑓(𝑥):
𝑚 ( = 𝑓′(𝑎)
We can find the equation of the tangent line the same way we do any line by
using the slope of the line and a point on the line to find the 𝑦-intercept, 𝑏
and putting the equation of the line in slope 𝑦-intercept form: 𝑦 = 𝑚𝑥 + 𝑏.
A normal line that is perpendicular to the curve and perpendicular to the
tangent line. Since it is perpendicular, the slope of the tangent and the slope
of the normal, 𝑚. , are negative reciprocals of each other:
𝑚. =
−1
𝑚(
EXAMPLE: Find the equation of the
tangent line and normal line for the
parabola 𝑦 = 0.5𝑥 4 when 𝑥 = 2.
SOLUTION: First we need the slope of
the tangent by finding 𝑓 6 2 :
𝑓6 𝑥 = 𝑥
𝑚 ( = 𝑓 6 2 = 2
Now we need a point. So we’ll find the
𝑦-value corresponding to 𝑥 = 2:
𝑓 2 = 0.5 2
4
=2
Now we can plug into 𝑦 = 𝑚𝑥 + 𝑏 and solve for 𝑏:
2=2 2 +𝑏
𝑏 = −2
𝑦 = 2𝑥 − 2
To find the equation of the normal line, we use the reciprocal of 𝑚 ( to find
8
𝑚. = − . Then plug in 2,2 and solve again:
4
2=−
1
2 +𝑏
2
𝑏=3
1
𝑦=− 𝑥+3
2
So the equation of the tangent line is 𝑦 = 2𝑥 − 2 and the equation of the
8
normal line is: 𝑦 = − 𝑥 + 3.
4
Applications of Derivatives
Equation of the Tangent and Normal Line
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2
Level 8
EXAMPLE: Find the equation of the tangent to the curve 3𝑥 4 𝑦 + 2𝑥 = 1
when 𝑥 = −1.
SOLUTION: First, we’ll determine our point (since we’ll need to use the 𝑦value later) by substituting into the original function:
3(−1)4 𝑦 + 2 −1 = 1
𝑦 = 1
We can find the derivative implicitly to determine 𝑚 ( :
𝑑𝑦
+2=0
𝑑𝑥
𝑑𝑦 −6𝑥𝑦 − 2
=
𝑑𝑥
3𝑥 4
−6 −1 1 − 2 4
𝑚( =
= 3 −1 4
3
6𝑥𝑦 + 3𝑥 4
Now that we have a point on the tangent line a the slope of the tangent line,
determine the 𝑦-intercept:
1=
4
3
−1 + 𝑏
7
𝑏= 3
So the equation of the tangent is:
𝑦=
4
7
𝑥+ 3
3
EXAMPLE: Find the equation of the tangent(s) to the curve: 𝑓 𝑥 =
is parallel to: 𝑥 + 16𝑦 + 3 = 0.
?
A
@
that
SOLUTION: In this question, we’ll use the derivative to help find our point
instead. This is because we already know 𝑚 ( . It has the same slope as the
line: 𝑥 + 16𝑦 + 3 = 0. So, 𝑚 ( = −1/16.
The derivative of 𝑓(𝑥) is:
1
D
𝑓 6 𝑥 = −𝑥 C? = −
𝑥 D/?
Substitute in 𝑚 ( :
−
1
1
= − D/? 16
𝑥
Flip both sides and divide by −1:
𝑥 D/? = 16
𝑥 = 16
?/D
= 8
Applications of Derivatives
Equation of the Tangent and Normal Line
#BEEBETTER
at www.tutorbee.tv
3
Level 8
Substitute 𝑥 = 8 into the original function to get:
𝑓 8 =
3
A
3
= 𝑥 2
Now solve for 𝑏:
3
1
= −
2
16
The equation of the tangent is:
𝑦=−
8 +𝑏
𝑏 = 2
1
𝑥+2
16
Applications of Derivatives
Equation of the Tangent and Normal Line
#BEEBETTER
at www.tutorbee.tv
4
Level 8