The tangent line (or simply tangent) to a plane curve at a given point is the straight line that just touches the curve at that point.
To find equation of a tangent to a curve, we need the point of tangency (where tangent is touching the curve) and slope of the tangent.
How to find slope of the tangent ?
1. Find the first derivative f'(x) or dy/dx of the function/equation which represents the curve.
2. Substitute the point of tangency in the first derivative to get the slope of tangent.
Example 1 :
Find the equation of tangent to the curve y = x^{2} – x – 2 at x = 1.
Solution :
Substitute x = 1 into the equation of curve.
y = 1^{2} – 1 – 2
= 1 - 1 - 2
= -2
Point of tangency is (1, -2).
y = x^{2} – x – 2
dy/dx = 2x - 1
Substitute x = 1.
= 2(1) - 1.
= 2 - 1
= 1
Slope of the tangent is 1.
Equation of the tangent :
y - y_{1} = m(x - x_{1})
Substitute (x_{1}, y_{1}) = (1, -2) and m = 1.
y - (-2) = 1(x - 1)
y + 2 = x - 1
y = x - 3
Example 2 :
Find the equation of tangent to the curve y = x^{2} – 4x – 5 at x = -2.
Solution :
Substitute x = -2 into the equation of curve.
y = (-2)^{2} – 4(-2) – 5
= 4 + 8 - 5
= 7
Point of tangency is (-2, 7).
y = x^{2} – 4x – 5
dy/dx = 2x - 4
Substitute x = -2.
= 2(-2) - 4
= -4 - 4
= -8
Slope of the tangent is -8.
Equation of the tangent :
y - y_{1} = m(x - x_{1})
Substitute (x_{1}, y_{1}) = (-2, 7) and m = -8.
y - 7 = -8[x - (-2)]
y - 7 = -8(x + 2)
y -7 = -8x - 16
y = -8x - 9
Example 3 :
Find the equation of tangent to the curve y = x^{4} + 1 and it is parallel to the line 32x - y = 15.
Solution :
y = x^{4} + 1 ----(1)
Slope of the tangent :
dy/dx = 4x^{3 }----(1)
Write the equation 32x - y = 15 in slope intercept form.
32x - y = 15
y = 32x - 15
Slope of the line 32x - y = 15 :
m = 32
Since tangent is parallel to the line 32x - y = 15, the slopes are equal.
4x^{3 }= 32
x^{3 }= 8
x^{3 }= 2^{3}
x = 2
Substitute x = 2 in (1).
y = 2^{4} + 1
= 16 + 1
= 17
Point of tangency is (2, 17).
Substitute x = 2 in (2).
dy/dx = 4(2^{3})
= 4(8)
= 32
Slope of the tangent is 32.
Equation of the tangent :
y - y_{1} = m(x - x_{1})
Substitute (x_{1}, y_{1}) = (2, 17) and m = 32.
y - 17 = 32(x - 2)
y - 17 = 32x - 64
y = 32x - 47
Example 4 :
Find the equations of tangent lines to the curve
y = x^{3 }- 3x^{2 }+ 3x - 3
and are parallel to the line 3x - y = 15.
Solution :
y = x^{3 }- 3x^{2 }+ 3x - 3 ----(1)
Slope of the tangent :
dy/dx = 3x^{2 }- 6x + 3 ----(2)
Write the equation 3x - y = 15 in slope intercept form.
3x - y = 15
y = 3x - 15
Slope of the line 3x - y = 15 :
m = 3
Since tangent is parallel to the line 3x - y = 15, the slopes are equal.
3x^{2 }- 6x + 3 = 3
3x^{2 }- 6x = 0
Divide each side by 3.
x^{2 }- 2x = 0
x(x^{ }- 2) = 0
x = 0 or x = 2
Substitute x = 0 and 2 in (1).
y = 0^{3 }- 3(0^{2})^{ }+ 3(0) - 3 = 0 - 0 + 0 - 3 = -3 (0, -3) |
y = 2^{3 }- 3(2^{2})^{ }+ 3(2) - 3 = 8^{ }- 12^{ }+ 6 - 3 = -1 (2, -1) |
Points of tangency are (0, -3) and (2, -1).
Substitute x = 0 and 2 in (2).
dy/dx = 3(0^{2})^{ }- 6(0) + 3 = 0 - 0 + 3 = 3 |
dy/dx = 3(2^{2})^{ }- 6(2) + 3 = 12 - 12 + 3 = 3 |
Slope of the tangent is 3.
Equation of the tangent at (0, -3) with slope 3.
y - y_{1} = m(x - x_{1})
Substitute (x_{1}, y_{1}) = (0, -3) and m = 3.
y - (-3) = 3(x - 0)
y + 3 = 3x
y = 3x - 3
Equation of the tangent at (2, -1) with slope 3.
y - y_{1} = m(x - x_{1})
Substitute (x_{1}, y_{1}) = (2, -1) and m = 3.
y - (-1) = 3(x - 2)
y + 1 = 3x - 6
y = 3x - 7
Example 5 :
Find a cubic function in the form y = ax^{3 }+ bx^{2 }+ cx + d whose graph has horizontal tangents at the points (-2, 6) and (2, 0).
Solution :
Slope of the tangent :
dy/dx = 3ax^{2 }+ 2bx^{ }+ c
Slope at (-2, 6) :
dy/dx = 3a(-2)^{2 }+ 2b(-2) + c
dy/dx = 12a - 4b + c
Slope at (2, 0) :
dy/dx = 3a(2)^{2 }+ 2b(2) + c
dy/dx = 12a^{ }+ 4b + c
Since the tangents are horizontal, slope = 0
12a - 4b + c = 0 ----(1) |
12a^{ }+ 4b + c = 0 ----(2) |
The curve passes through the point (-2, 6).
a(-2)^{3 }+ b(-2)^{2 }+ c(-2) + d = 6
-8a + 4b - 2c + d = 6 ----(3)
The curve passes through the point (2, 0).
a(2)^{3 }+ b(2)^{2 }+ c(2) + d = 0
8a + 4b + 2c + d = 0 ----(4)
(1) - (2) :
(12a - 4b + c) - (12a + 4b + c) = 0
12a - 4b + c - 12a - 4b - c = 0
-8b = 0
b = 0
(3) + (4) :
(-8a + 4b - 2c + d) + (8a + 4b + 2c + d) = 6
8b + 2d = 6
Substitute b = 0.
2d = 6
d = 3
Substitute b = 0 in (2).
12a^{ }+ 4(0) + c = 0
12a + c = 0 ----(5)
Substitute b = 0 and d = 3 in (4).
8a + 4(0) + 2c + 3 = 0
8a + 2c + 3 = 0 ----(6)
(6) - 2(5) :
(8a + 2c + 3) - 2(12a + c) = 0
8a + 2c + 3 - 24a - 2c = 0
-16a + 3 = 0
a = 3/16
Substitute a = 3/16 in (5).
12(3/16) + c = 0
9/16 + c = 0
c = -9/16
Substitute the values of a, b c in y = ax^{3 }+ bx^{2 }+ cx + d.
y = (3/16)x^{3 }+ 0x^{2 }- (9x/16) + 3
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Jan 17, 22 10:45 AM
Trigonometry Word Problems Worksheet with Answers
Jan 17, 22 10:41 AM
Trigonometry Word Problems with Solutions
Jan 16, 22 11:56 PM
Writing Numbers in Words Worksheet