I’ve shown my work and the textbooks solution… did I do something wrong? Or is the textbook giving me the wrong answer. Handwritten work only please. Messages File Edit View Conversations Window HelpTransportation engineering textbook.pdfEXAMPLE 3.1SOLUTIONChapter 3VERTICAL CURVE STATIONS AND ELEVATIONSA 600-ft equal-tangent sag vertical curve has the PVC at station 170+ 00 and elevation1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationingand elevation of the PVI, the PVT, and the lowest point on the curve.1000-3.5 ft/station X (3 stations) = 989.5 ftSimilarly, with the PVI at elevation 989.5 ft, the elevation of the PVT isSince the curve is equal tangent, the PVI will be 300 ft or three stations (measured in ahorizontal plane) from the PVC, and the PVT will be 600 ft or six stations from the PVC.Therefore, the stationing of the PVI and PVT is 173 +00 and 176 +00, respectively. For theelevations of the PVI and PVT, it known that a -3.5% grade can be equivalently writtenas -3.5 ft/station (a 3.5 ft drop per 100 ft of horizontal distance). Since the PVI is threestations from the PVC, which is known to be at elevation 1000 ft, the elevation of the PVI isGeometric Design of Highways7,280989.5 +0.5 ft/station X (3 stations) = 991.0 ftIt is clear from the values of the initial and final grades that the lowest point on the verticalcurve will occur when the first derivative of the parabolic function (Eq. 3.1) is zero becausethe initial and final grades are opposite in sign. When initial and final grades are notopposite in sign, the low (or high) point on the curve will not be where the first derivative iszero because the slope along the curve will never be zero. For example, a sag curve with aninitial grade of -2.0% and a final grade of -2.0% will have its lowest elevation at the PVT,and the first derivative of Eq. 3.1 will not be zero at any point along the curve. However, inAiMessage25Open wtv SMacBook Pro a 600 ft equal tangent sag vertical curve has the PVC at Station 170+00 and elevation1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationingAnd elevation of the PVI, the PVT, and the lowest point on the curve.L=600 ftT=300 ftOptions-3.5%PVC = PVT-L17000 = PVT -60017600 = PVT176 +00= PVTPVC170+00PUI = PVC + T54°FCloudyElevation1000= 17,000 + 300- 17300PVI = 173+00-y = 2ax + bO SearchPVI173+00Elevation =a989.5'Elevation PUI = E pvc + 9²/12= 1000 + (-0.035) (600)=989.5'G₂+0.5%PVT1776+00elevation = 974.5¹Elevation PVT = EPI-9²/2/2b== 989.5 - (+0.05)(b)= 974.5'a== G₂-G₁₂₁ = 0.05-(-0.035)2L2(600)= 7.0833•10-50.0354:54 PM

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I have tried to solve it in fresh with handwritten as per your demand plz go through it