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Given:
y = 8 sin(x)
y'(x) = 8 cos(x)
At the point(1/6, 4), obtain the slope as
y' = 8 cos(1/6) = 7.889
Parallel line:
A line parallel to the curve at (1/6, 4) will have the same slope.
Let the line be
y = 7.889x + c
Because the line passes through(1/6, 4), therefore
7.889(1/6) + c = 4 => c = 2.6852
The line is y = 7.889x + 2.6852
Two points on the line are (0, 2.6852) and (1, 10.5742)
The distance between the points is √(1 + 7.889²) = 7.952
The umit vector is (0.126, 0.992)
Perpendicular line:
The product of the slopes of the parallel and perpendicular line is -1.
Therefore the slope of the perpendicular line is - 1/7.889 = - 0.1268.
Let the perpendicular line be
y = -0.1268x + c
Because the line passes through (1/6, 4), therefore
-0.1268(1/6) + c = 4 => c = 4.0211
The line is y = -0.1268x + 4.0211
Two points on the line are (0, 4.0211) and (1, 3.8943).
The distance between the points is √(1 + (-0.1268)²) = 1.008
The unit vector is (0.992, -0.126)
A graph of the curve and the two lines is shown below.
Answer:
The unit vector for the parallel line is (0.126, 0.992)
The unit vector for the perpendicular line is (0.992, -0.126)
y = 8 sin(x)
y'(x) = 8 cos(x)
At the point(1/6, 4), obtain the slope as
y' = 8 cos(1/6) = 7.889
Parallel line:
A line parallel to the curve at (1/6, 4) will have the same slope.
Let the line be
y = 7.889x + c
Because the line passes through(1/6, 4), therefore
7.889(1/6) + c = 4 => c = 2.6852
The line is y = 7.889x + 2.6852
Two points on the line are (0, 2.6852) and (1, 10.5742)
The distance between the points is √(1 + 7.889²) = 7.952
The umit vector is (0.126, 0.992)
Perpendicular line:
The product of the slopes of the parallel and perpendicular line is -1.
Therefore the slope of the perpendicular line is - 1/7.889 = - 0.1268.
Let the perpendicular line be
y = -0.1268x + c
Because the line passes through (1/6, 4), therefore
-0.1268(1/6) + c = 4 => c = 4.0211
The line is y = -0.1268x + 4.0211
Two points on the line are (0, 4.0211) and (1, 3.8943).
The distance between the points is √(1 + (-0.1268)²) = 1.008
The unit vector is (0.992, -0.126)
A graph of the curve and the two lines is shown below.
Answer:
The unit vector for the parallel line is (0.126, 0.992)
The unit vector for the perpendicular line is (0.992, -0.126)