## 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 ?

1. Rahulmr

$V=10t-\frac{ t^2 }{ 20 }$

2. 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

3. anonymous

so we get $\frac{ dV }{ dt }=10-\frac{ t }{ 10 }$

4. FireKat97

5. anonymous

now initial rate of work occurs at t=0. therefore, $\frac{ dV }{ dt }_{t=0}=10 m^3/\min$

6. anonymous

@FireKat97 yes you are right. @Jadedry needs to take the derivative to answer question b

7. FireKat97

yup and then sub in t = 40, correct?

8. anonymous

yep. just like what i did for part d

I also needed to type 8m^3 rather than 80.. whoops...

10. FireKat97

and for part e you let the derivative equal to 5m^3

So I was making a mistake by taking the equation at face value? I needed to find the derivatives first?

12. anonymous

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 $5=10-\frac{ t }{ 10 }$ $-5=-\frac{ t }{ 10 }$ $t=50\min$

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

: )