Mathematics Solving The Waveform In The Figure - 275 Words

Mathematics Solving The Waveform In The Figure (Math Problem Sample) Content: Solutions.To determine the equation for the waveform in the figure. The point at which the sine is turned is on . The answer is in 3 parts.Part 1The gradient for line y= 0 is 0 therefore,0d= Part 2Sin d=-cos+cos=1+cosPart 32d=Hence, the equation of the area of the covered by the curve+(1+cos)+ #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+cis the equation of a straight line where m=gradient and c= y-intercept?M1=2ITPAnd henceY1= 2ITPXAnd therefore, for negative gradient,M2= -2IPTy intercept = 2IPhence,Y2 = -2IPTX+2IPTherefore, area under the triangular wave is given by;0TY1+Y2.dtThen substitute the values of y1 and y2 above you get;0T2ITPX-2IPTX+2IPThen integrate by parts to get;2IPX0T#3 * From...