A trench is being dug by a team of labourers who removes V cubic metres of soil in t minutes, where V=10*t-(t^2)/20.

Question

V=10*t-(t²)/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 ?

Answer 1

V is defined to be the total amount of soil removed, so we want V to be positive and not negative.

 

More importantly though, if the only operation going on is soil removal, then the rate of soil removal must always be positive.   So it is the derivative of V that determines the domain (values of t that are meaningful).

 

 

dV/dt = 10 - (t/10) > 0 

            10 > (t/10)

            100 > t

 

Note that 200 is smaller than the time over which V is positive as determined by the original answer.  So the domain is (0, 100)

 

So the domain is based on T being less than 100.  The rate of soil removal after 40 minutes is 10 - 40/100 = 9.6 cubic meters per minute.

 

With regards to using 41.  This is not correct.  The question was about the rate at the end of 40 minutes and not the 'end of the 40th minute.   Use 40 in the formula.

Answer 2

A) The logical domain would be when the volume and rate are both positive. If the rate of volume removal is negative, then we are filling the hole back in. We can find the domain after the next step.

 

B) The rate of digging is the derivative (or slope) of the function.

 

d/dt [10t - t^2/20] = 10 - 1/20 * d/dt [t^2]

 

rate = 10 - 2/20 * t

 

       = 10 - 1/10 * t

       = 10 - (1/10) * t

 

To find the rate at 40 min, we would plug in t=40

 

Also find when the rate is 0, this tells us the logical domain.

 

C) If the laborers are working at a constant rate, our derivative would be a constant. Is it? No, it is a function of time.

 

D) The initial rate of work is the rate when t=0

 

rate = 10 - (1 / 10) *t

 

if t=0, the initial rate is 10 m^3/min

 

E) This part asks "when is the rate equal to 5." So, we replace rate with 5

 

5 = 10 - (1/10) *t

 

Solve for t

 

Good luck and I hope this helps!