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secondary 4 | A Maths
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Part 2
Date Posted: 5 years ago
Views: 421
when x = 0, dy/dx = 2
So,
2 = p sec²(0 x p)
2 = p sec²0
= p(1/cos²0)
= p/(1/1²)
= p
When x = π/8, dy/dx = q
q = 2sec²(π/8 x 2)
q = 2/(cos²(π/4))
q = 2/½
q = 4
So,
2 = p sec²(0 x p)
2 = p sec²0
= p(1/cos²0)
= p/(1/1²)
= p
When x = π/8, dy/dx = q
q = 2sec²(π/8 x 2)
q = 2/(cos²(π/4))
q = 2/½
q = 4
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For Q6a (ii), we read off the coordinates of the graph as usual, except that in (a, b), a is the horizontal x-axis while b is the vertical dy/dx-axis.
Note that the vertical intercept contains dy/dx instead of y, and hence we call this vertical intercept the 'dy/dx-intercept' instead of the 'y-intercept'.
Note that the vertical intercept contains dy/dx instead of y, and hence we call this vertical intercept the 'dy/dx-intercept' instead of the 'y-intercept'.
Date Posted:
5 years ago
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