NZ Level 7 (NZC) Level 2 (NCEA)
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Functions from rate of change
Lesson

When we were looking at differentiation we learned of a special relationship that exists between Displacement and Velocity, (to remind yourself you can go here)

As we have been looking at the reverse relationship, going the other way, we can rephrase this relationship to say that antidifferentiating velocity will give us the displacement. 

That's great.  So if we have the velocity function we can find the displacement function then? Well nearly. Just knowing the velocity function isn't enough because as we saw here, finding the antiderivative gives us a family of antiderivative functions.

Narrowing down the exact function is pretty important if we want to be able to answer specific questions about the displacment of a specific object.  

Let's look at an example.

Example 1

A particle moves with a velocity function of $v(t)=4t^3-6t^2-12t+9$v(t)=4t36t212t+9 for the first $10$10 seconds of travel.  Find its displacement at $t=5$t=5s and $t=10$t=10s if at $t=0$t=0, the displacement is $1$1cm.

Think: So what can we tell from this information?

  • Firstly, we are given the function $v(t)$v(t).
  • Then we are told that this function maps the movement of the particle for the domain $0\le t\le10$0t10, and that time is measured in seconds.  It is common for questions like this for the function to operate on a restricted domain and this should always be taken into account when answering questions about the situation later. 
  • We are told that the displacement is measured in centimeters
  • We also know that the initial position (at $t=0$t=0) is $1$1cm. 

Do: To find the $s(t)$s(t) function, we need to integrate $v(t)$v(t), because $s'(t)=v(t)$s(t)=v(t). That is, $\int v(t)dt=s(t)$v(t)dt=s(t).

Now since

$v(t)=4t^3-6t^2-12t+9$v(t)=4t36t212t+9

we integrate this to find

$s(t)=\frac{4t^4}{4}-\frac{6t^3}{3}-\frac{12t^2}{2}+9t+C$s(t)=4t446t3312t22+9t+C

$s(t)=t^4-2t^3-6t^2+9t+C$s(t)=t42t36t2+9t+C

To find the particular $s(t)$s(t) function we are interested in here we need a value for $C$C.  To do this we use the initial condition:

When $t=0,s=1$t=0,s=1, and substituting this into

$s(t)=t^4-2t^3-6t^2+9t+C$s(t)=t42t36t2+9t+C

gives

$s(0)=0^4-2\cdot0^3-6\cdot0^2+9\cdot0+C$s(0)=042·036·02+9·0+C

which then simplifies down to

$1=C$1=C

Thus the particular $s(t)$s(t) function satisfying all the conditions laid out in the question is

$s(t)=t^4-2t^3-6t^2+9t+1$s(t)=t42t36t2+9t+1

This $s(t)$s(t) is a function born from a rate of change function.   From here we can find the displacement at any point in the given domain. 

(a) What is the displacement at $t=5$t=5? This is a substitution exercise. 

$s(5)=5^4-2\cdot5^3-6\cdot5^2+9\cdot5+1=625-250-150+45+1=271$s(5)=542·536·52+9·5+1=625250150+45+1=271cm

(b) What is the displacement at $t=10$t=10? Same thing!

$s(10)=10^4-2\cdot10^3-6\cdot10^2+9\cdot10+1=10000-2000-600+91=7491$s(10)=1042·1036·102+9·10+1=100002000600+91=7491cm

 

Example 2

The rate of change of volume (in litres) of water in a bathtub is modeled using the volume function $\frac{dv}{dt}=3t^2-12t+9$dvdt=3t212t+9 for $0\le t\le3$0t3minutes.  Find the maximum volume of water the bathtub has during this period of time. At $t=0$t=0, the bathtub is empty. 

Think:  By finding the antiderivative of $\frac{dv}{dt}$dvdt, we will get the function for volume $v(t)$v(t).  

$\int3t^2-12t+9dt=\frac{3t^3}{3}-\frac{12t^2}{2}+9t+C$3t212t+9dt=3t3312t22+9t+C

$v(t)=t^3-6t^2+9tC$v(t)=t36t2+9tC

To find the value of $C$C, we need to use our initial condition which is that at the beginning the bathtub is empty, which means that at $t=0,v=0$t=0,v=0.

So the function then becomes $v(t)=t^3-6t^2+9t$v(t)=t36t2+9t

This is the function from the rate of change function.  This function can now be used to answer any question about the volume of the bath over the time from $0$0 to $3$3 minutes.  

Specifically in this case we are asked to find the maximum volume of the bathtub.  Which means we are looking for the maximum point of $v(t)$v(t) on the domain $0\le t\le3$0t3

We know that the maximum will occur where the derivative is $0$0.  And we already have the derivative function, $\frac{dv}{dt}=3t^2-12t+9$dvdt=3t212t+9   By solving this for $\frac{dv}{dt}=0$dvdt=0 we get $t=1$t=1 and $t=3$t=3.  By looking at the graph of the function we can see that at $t=1$t=1, this is our maximum point. 

To find the maximum volume at this point we need only substitute in $t=1$t=1, into $v(t)=t^3-6t^2+9t$v(t)=t36t2+9t

$v(1)=1-6+9=4$v(1)=16+9=4.  So the maximum volume is $4$4L. 

 

Worked Examples

Question 1

A rate of change function is given by $\frac{dx}{dt}=\sqrt{t}+t^5$dxdt=t+t5 :

  1. Determine $x\left(t\right)$x(t). Use $C$C as the constant of integration.

Question 2

A rate of change function is given by $\frac{dy}{dx}=\left(4x+\frac{1}{x}\right)\left(5x-\frac{2}{x}\right)$dydx=(4x+1x)(5x2x):

  1. Expand and simplify the expression for $\frac{dy}{dx}$dydx

  2. Hence determine $y$y. Use $C$C as the constant of integration.

Question 3

The marginal revenue, in dollars, at the "Ah, Fudge!" chocolate factory, is given by $R'\left(x\right)=400-\frac{4}{5}x$R(x)=40045x per additional unit produced.

  1. Given that total revenue is equal to the price per unit sold times the number of units sold, find a function, $R\left(x\right)$R(x), for the total revenue.

  2. Find the total revenue, to the nearest dollar, from producing and selling $138$138 units.

Outcomes

M7-10

Apply differentiation and anti-differentiation techniques to polynomials

91262

Apply calculus methods in solving problems

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