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Mathematics Solving The Waveform In The Figure (Math Problem Sample)

Instructions:

solving

source..
Content:
Solutions.
To determine the equation for the waveform in the figure. The point at which the sine is turned is on φ . The answer is in 3 parts.
Part 1
The gradient for line y= 0 is 0 therefore,
0φdθ= φ
Part 2
πφSinθ dθ=-cosθ+cosφ=1+cosφ
Part 3
π2πdθ=π
Hence, the equation of the area of the covered by the curve
φ+(1+cosφ)+ π
#2 Determine the equation for the waveform in the figure. This is a symmetric triangle wave. The equation has two parts. Each segment is of the form y=mx+b. The second segment has a negative slope and a y intercept of 2Ip.
y=mx+c
is the equation of a straight line where m=gradient and c= y-intercept?
M1=2ITP
And hence
Y1= 2ITPX
And therefore, for negative gradient,
M2= -2IPT
y intercept = 2IP
hence,
Y2 = -2IPTX+2IP
Therefore, area under the triangular wave is given by;
0TY1+Y2.dt
Then substitute the values of y1 and y2 above you get;
0T2ITPX-2IPTX+2IP
Then integrate by parts to get;
2IPX0T
#3
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