# NCEES Sample questions- Breadth- Q126



## New2WR (Mar 5, 2013)

Dear All,

I hope all is going well with you preparing for the exam! I have a difficulty to understand problem# 126 solution (the alternate solution part) per the Breadth (morning) exam in the WR-ENV NCEES sample questions book. It solves the problem using the following equation which is the same as equation 79.47 per CERM 12 ed:

Y= Yvpc + g1X+[(g2-g1)/2L]X^2

Then it adjusts the left hand side to be the slope by subtracting Yvpc from both sides and dividing by X as shown below:

Y’ (slope) = g1 + [(g2-g1)/ L]

However I think there is a typo per the denominator it should be 2L instead of L. I checked the book's errata at the NCEES website but nothing there in this regard.

Your input guys is truly appreciated!

Thanks!


----------



## ptatohed (Mar 6, 2013)

Which year NCEES Sample Q&amp;Ss are you referring to? I checked my 2011 copy (Transpo depth but all breadth questions are the same) and #126 is a vertical curve problem.


----------



## bradlelf (Mar 6, 2013)

ptatohed said:


> Which year NCEES Sample Q&amp;Ss are you referring to? I checked my 2011 copy (Transpo depth but all breadth questions are the same) and #126 is a vertical curve problem.


+1 give a year or post the problem for assistance


----------



## K19 (Mar 6, 2013)

This is indeed the vertical curve problem from the 2011 sample questions. @ NewtoWR, your second equation is missing an "X" at the end and should read:

Y’ (slope) = g1 + [(g2-g1)/ L] X

This is not an algebraic manipulation of the first equation (elevation Y as a function of X); rather it is taking the derivative of the first equation with respect to X to get a function for slope. The derivative of X^2 is 2X, and the 2 in the denominator is thus cancelled out. HTH


----------



## ptatohed (Mar 7, 2013)

You're right K19. I guess New2WR never said #126 was not a vertical curve problem. I'm just not used to the term Y and Y' used in vertical curve problems so I assumed it was some sort of open channel flow problem.

I just checked my notes. Before I go any further, I'd like to state that it drives me nuts that the question and solution use the terms "VPC, VPI and VPT" instead of the _*correct *_terms "PVC, PVI and PVT". Anyway, moving on...... 

I solved this problem in (what I believe) to be less steps and less time:

r = (g2 - g1) / L = (3% - -2.3%) / 3 sta = 1.767% [the rate of change per station]

r (x) = 1.767% (2 sta) = +3.534% [the rate of change at Sta 14+00]

r14 + g1 = +3.534% + -2.3% = 1.23% [the tangent slope at Sta 14+00]

Answer (B)


----------



## New2WR (Mar 10, 2013)

Thanks all for your inputs! Special thanks to K19 you nailed it man! yes that is exactly what was confusing me. I looked at it from the algebraic stand point and did not see it as a first derivative. Thanks again.


----------

