Mat 1475

This topic contains 5 replies, has 4 voices, and was last updated by  HenryLam 2 years, 6 months ago.

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  • #42044

    randymedina
    Participant

    Leader: Randy Medina
    Abdullah Zafar
    Julia Chang
    Henry Lam
    Abdullah Saleh
    Ahmed Taher
    Joel Silva
    We’ll be discussing the problems of problem 2 to agree upon a response

    #42047

    randymedina
    Participant

    Alright so what I’ve understood from these problems was that for
    a) The tangent line to the graph at the lowest point of the dive is that it’s located between the interval of 4 and 6 and that the lowest point stays constant for a bit before the submarine goes upward.
    b) To find how many minutes till the submarine reaches its lowest point just substitute the t with the numbers between 4 and 6, as it shows it looks like the lowest point is near 5, so I divided each minute into 4 segments (5.25- for 5 minutes and 15 seconds, 5.5 for 5 min and 30 sec etc.) and I got 5.25 as the lowest point: a(5.25)=500cos(5.25/2)+125(5.25)-564, a(5.25)=-434.75+656.25-564, a(5.25)=-434.75+92.25=-342.5, while a(5)=-339.57 and a(5.5)=-338.65 which both are less than the value of a(5.25)
    c) To find the greatest depth of which there submarine traveled we need to know at which time it traveled the deepest which was solved at the previous question (b), a(5.25)= -342.5 rounded to the nearest foot is -343ft.
    d) I don’t understand this question, from what I can assume from reading the question I’m thinking it has to do with instantaneous rate.

    If ya disagree let me know where you think I’ve messed up

    #42091

    jsilva10
    Participant

    I agree with your answers. I got similar answers

    #42210

    Jonathan
    Participant

    It’s Jonathan Camacho
    My name is not posted, but i would still like to join the group. I also didn’t understand part d you included, but I also got around the same answers as you did.

    #42262

    randymedina
    Participant

    Sure thing jonathan you can join the group, and if anyone wants to contact me, I’m going to try and be more frequent here, but in case I’m not my personal email is randym0203@gmail.com, also I still haven’t looked up the answer for number 4, I’ll look into it tonight and post anything I was able to understand. I’ll review my answers from the notes and research and update anything that needs clarification,

    #42263

    HenryLam
    Participant

    C) What i did was get the value from problem B and plugged it in for t.
    a(t)=500cos(t/2)+125t-564
    a(5.24)=500COS(t/2)+125t-564
    a(5.24)=500COS(5.24/2)+125(5.24)-564
    a(5.24)= -433.513+655-564
    a(5.24)=-342.513

    D.I think i got the answer for the last one.
    Basically what i did was 0-(342.513)/7.634-5.24 = 50.1067 Rate of change
    How i got 0, i used the original formula a(t)=500cos(t/2)+125t-564 and finding the t value.

    • This reply was modified 2 years, 6 months ago by  HenryLam.
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