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Timepoint crossed

  • 15-06-2020 6:25pm
    #1
    ShyWho12
    Registered Users, Registered Users 2 Posts: 8 ✭


    I want to find when pH 6 hits.

    I am using the eqn y=a+bx

    The info I have is difference between time7 (07hrs 00 mins) and time5 (05hrs 05 mins) is 125 minutes
    ph at time5 is 6.16 and the ph at time7 is 5.94
    I filled in as follows
    6=6.16+b(125)
    I got an answer of 0.00128
    Is 0.00128 (slope) the change in mins from one point to the next, therefore ph 6 occured at 5.05128?
    Am I doing this right, what time does 6 actually hit?
    Tagged:


Comments

  • #2
    murphm45
    Registered Users, Registered Users 2 Posts: 507 ✭✭✭murphm45


    You probably already figured this out (or don;t care anymore) but a couple of things wrong.

    First off your time difference is 115 minutes not 125 minutes.

    y = a+bx is the equation of the line (assuming x is the time 'a' signifies where the y axis is crossed, this is the value at time 0, and 'b' is the slope of line) but the problem is none of this is directly observable from the information you have. What you know is:
    6.16 = a + 5.0833b &
    5.94 = a + 7b

    If you want you can use simultaneous equations to solve for a and b but you don't really need to, an easier way to do this is to look at the rate of change. you know that between time5 & time7 the ph drops 0.22 (6.16 - 5.94). This means for every minute the ph is dropping 0.001913 (0.22 / 115) and you want to find out how long it takes the ph to drop by 0.16 (6.16 - 6) which you can get by dividing this by the per minute change rate to give 90.91 minutes (0.16 / 0.001913).

    Hope this makes sense


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