Najis the Best

Topics: Calculus, Phase, Maxima and minima Pages: 3 (472 words) Published: March 5, 2013
1.| Define each term and write the formula if needed|
a) Amplitude
Amplitude is half the distance between the minimum and maximum values of the range of a periodic function with a bounded range.

b) Period
A function whose value is repeated at constant intervals, such as sin x.

c) Area of a sector
Area = (1/2 )(r^2)(θ) radians
= (θ/360)( π r^2) degrees
d) Micron
A unit of length equal to one millionth of a meter.
e) Area of a minor segment
For finding the area of a segment you need to know the length of the chord, the radius and the angle subtended by the arc at the centre. Then, the area of the minor segment will be-:

The area = (1/2)r²(θ - sinθ)
f) Example of an indefinite integral

g) π= 180 degrees
h) Radian is a way to measure angles.
2) a)

b) i) Record the maximum and minimum values.
Y=sin(x)
Max value = 1 and Min value = -1
Y=2sin(x)
Max value = 2 and Min value = -2
Y=0.5sin(x)
Max value = 0.5 and Min value = -0.5
Y=-sin(x)
Max value = 1 and Min value = -1

ii) State the period and amplitude.
Y=sin(x):- Amplitude = 1 and Period = 2π
Y=2sin(x):- Amplitude = 2 and Period = 2π
Y=0.5sin(x):- Amplitude =0.5 and Period = 2π Y=-sin(x):- Amplitude =-1 and Period = 2π c) ‘a’ in the function y= a sin(x) is the amplitude.

3) a)

b) i) Record the maximum and minimum values.
Y=sin(x)
Max value = 1 and Min value = -1
Y=sin2x
Max value = 1 and Min value = -1
Y=sin(x/2)
Max value = 1 and Min value = -1
Y=sin(-x)
Max value = 1 and Min value = -1
ii) State the period and amplitude.
Y=sin(x):- Amplitude = 1 and Period = 2π
Y=2sin(x):- Amplitude = 1 and Period = π
Y=0.5sin(x):- Amplitude =1 and Period = 4π Y=-sin(x):- Amplitude =1 and Period = -2π
c) ‘b’ in...
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