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Rahulmr

  • one year ago

A trench is being dug by a team of labourers who removes V cubic metres of soil in t minutes, where V=10t-t^2/20. a) State the logical domain of the function, i.e. the values of t during which soil is being removed. b) At what rate is the soil is being removed at the end of 40 minutes ? c) Are the labourers working at a constant rate? Provide a reason for your reason. d) What is their initial rate of work ? e) At what time they are removing soil at the rate of 5m^3 per minute ?

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  1. Rahulmr
    • one year ago
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    \[V=10t-\frac{ t^2 }{ 20 }\]

  2. anonymous
    • one year ago
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    d) What is their initial rate of work perhaps take the derivative of the function, which gives you the rate at which soil is being removed

  3. anonymous
    • one year ago
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    so we get \[\frac{ dV }{ dt }=10-\frac{ t }{ 10 }\]

  4. FireKat97
    • one year ago
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    I'm not too sure about your solution to part b

  5. anonymous
    • one year ago
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    now initial rate of work occurs at t=0. therefore, \[\frac{ dV }{ dt }_{t=0}=10 m^3/\min\]

  6. anonymous
    • one year ago
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    @FireKat97 yes you are right. @Jadedry needs to take the derivative to answer question b

  7. FireKat97
    • one year ago
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    yup and then sub in t = 40, correct?

  8. anonymous
    • one year ago
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    yep. just like what i did for part d

  9. Jadedry
    • one year ago
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    I also needed to type 8m^3 rather than 80.. whoops...

  10. FireKat97
    • one year ago
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    and for part e you let the derivative equal to 5m^3

  11. Jadedry
    • one year ago
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    So I was making a mistake by taking the equation at face value? I needed to find the derivatives first?

  12. anonymous
    • one year ago
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    so lets answer this: a) 0<t<200 b)\[\frac{ dV }{ dt }=10-\frac{ t }{ 10 }\] At t=40mins \[\frac{ dV }{ dt }_{t=40}=10-\frac{ 40 }{ 10 }=10-4=6 m^3/\min\] c) They are working at a rate of \[\frac{ dV }{ dt }=10-\frac{ t }{ 10 }\] Since there is a linear term, with a negative slope, it says they are reducing their work load as t>0. hence they do not work at a constant rate. d) as per my solution above. e) \[\frac{ dV }{ dt }=10-\frac{ t }{ 10 }\] Sub dV/dt=5 \[5=10-\frac{ t }{ 10 }\] \[-5=-\frac{ t }{ 10 }\] \[t=50\min\]

  13. Jadedry
    • one year ago
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    deleting my reply, no point in having bad working @chris00 @FireKat97 Thanks for showing your working" You cleared up the problem for me as well.

  14. anonymous
    • one year ago
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    : )

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