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secondary 4 | A Maths
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Pls help me with this qn.. Thank you!
Date Posted: 4 years ago
Views: 260
note:
sin (2x - π) = -sin2x [3rd quad]
cos(x+π) = -cosx [3rd quad]
(i)
so
y = -sin2x - 4cosx
dy/dx = -cos2x . 2 - 4(-sinx)
= -2cos2x + 4sinx (shown)
(ii)
given dy/dt = -0.2 units/s
dy/dt = dy/dx . dx/dt
-0.2 = [-2cos2(-π) + 4sin(-π)] . dx/dt
-0.2 = [-2cos(-2π) + 4 x 0] . dx/dt
-0.2 = [-2 x 1 + 0] . dx/dt
-0.2 = [-2 x 1 + 0] . dx/dt
-0.2 = -2 dx/dt
dx/dt = 0.1 units/s
x is increasing at 0.1 units per sec.
...
sin (2x - π) = -sin2x [3rd quad]
cos(x+π) = -cosx [3rd quad]
(i)
so
y = -sin2x - 4cosx
dy/dx = -cos2x . 2 - 4(-sinx)
= -2cos2x + 4sinx (shown)
(ii)
given dy/dt = -0.2 units/s
dy/dt = dy/dx . dx/dt
-0.2 = [-2cos2(-π) + 4sin(-π)] . dx/dt
-0.2 = [-2cos(-2π) + 4 x 0] . dx/dt
-0.2 = [-2 x 1 + 0] . dx/dt
-0.2 = [-2 x 1 + 0] . dx/dt
-0.2 = -2 dx/dt
dx/dt = 0.1 units/s
x is increasing at 0.1 units per sec.
...
See 1 Answer
done
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Here is one way to do this question. I presume that you have learnt the addition/subtraction angle formula for sine and cosine.
Date Posted:
4 years ago
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