Rahulmr
  • Rahulmr
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 ?
Mathematics
schrodinger
  • schrodinger
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Rahulmr
  • Rahulmr
\[V=10t-\frac{ t^2 }{ 20 }\]
anonymous
  • anonymous
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
anonymous
  • anonymous
so we get \[\frac{ dV }{ dt }=10-\frac{ t }{ 10 }\]

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FireKat97
  • FireKat97
I'm not too sure about your solution to part b
anonymous
  • anonymous
now initial rate of work occurs at t=0. therefore, \[\frac{ dV }{ dt }_{t=0}=10 m^3/\min\]
anonymous
  • anonymous
@FireKat97 yes you are right. @Jadedry needs to take the derivative to answer question b
FireKat97
  • FireKat97
yup and then sub in t = 40, correct?
anonymous
  • anonymous
yep. just like what i did for part d
Jadedry
  • Jadedry
I also needed to type 8m^3 rather than 80.. whoops...
FireKat97
  • FireKat97
and for part e you let the derivative equal to 5m^3
Jadedry
  • Jadedry
So I was making a mistake by taking the equation at face value? I needed to find the derivatives first?
anonymous
  • anonymous
so lets answer this: a) 00. 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 \[5=10-\frac{ t }{ 10 }\] \[-5=-\frac{ t }{ 10 }\] \[t=50\min\]
Jadedry
  • Jadedry
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.
anonymous
  • anonymous
: )

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