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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

* From...

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

* From...

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