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 ?

Question

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 ?

rahulmr · Answer

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

anonymous · Answer

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 · Answer

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

firekat97 · Answer

I'm not too sure about your solution to part b

anonymous · Answer

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

anonymous · Answer

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

firekat97 · Answer

yup and then sub in t = 40, correct?

anonymous · Answer

yep. just like what i did for part d

jadedry · Answer

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

firekat97 · Answer

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

jadedry · Answer

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

anonymous · Answer

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

jadedry · Answer

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 · Answer

: )