We have looked at applying differentiation to rates of change, optimisation and properties of graphs. Many applied questions may require us to draw from different sections of our knowledge within the one question.
Worked examples
Example 1
The function $f(x)=ax^2+bx+c$f(x)=ax2+bx+c passes through the point $\left(6,13\right)$(6,13), has a minimum turning point at $x=3$x=3 and the equation of the tangent to the graph at $x=1$x=1 is $y=-8x+11$y=−8x+11. Find the values of $a$a, $b$b and $c$c and state the equation of the function.
Think: We have three unknowns and we have been given three pieces of information. We can use the information to create equations in terms of the unknowns which we can then solve simultaneously. We have two pieces of information about the derivative, let's find the derivative and then use the information to create our first two equations.
Do: Find the derivative $f'\left(x\right)$f′(x):
$f'\left(x\right)=2ax+b$f′(x)=2ax+b
Using the fact that there is a minimum at $x=3$x=3, we know that $f'\left(3\right)=0$f′(3)=0:
| $f'\left(3\right)$f′(3) | $=$= | $0$0 |
|
| $6a+b$6a+b | $=$= | $0$0 |
Equation $1$1 |
Using the fact that the tangent at $x=1$x=1 has a gradient of $-8$−8, we know that $f'\left(1\right)=-8$f′(1)=−8:
| $f'\left(1\right)$f′(1) | $=$= | $-8$−8 |
|
| $2a+b$2a+b | $=$= | $-8$−8 |
Equation $2$2 |
Solving equation $1$1 and equation $2$2 simultaneously:
| $6a+b-\left(2a+b\right)$6a+b−(2a+b) | $=$= | $0-\left(-8\right)$0−(−8) |
Equation $1$1 - Equation $2$2 |
| $4a$4a | $=$= | $8$8 |
|
| $a$a | $=$= | $2$2 |
|
Substituting $a=2$a=2 into equation $2$2:
| $2(2)+b$2(2)+b | $=$= | $-8$−8 |
|
| $4+b$4+b | $=$= | $-8$−8 |
Take $4$4 from both sides |
| $b$b | $=$= | $-12$−12 |
|
Lastly, let's find $c$c using the final bit of information that the function passes through $\left(6,13\right)$(6,13), that is $f\left(6\right)=13$f(6)=13:
| $ax^2+bx+c$ax2+bx+c | $=$= | $f(x)$f(x) |
substitute $x=6$x=6, $f\left(x\right)=13$f(x)=13, $a=2$a=2 and $b=-12$b=−12 |
| $2(6)^2-12(6)+c$2(6)2−12(6)+c | $=$= | $13$13 |
|
| $72-72+c$72−72+c | $=$= | $13$13 |
|
| $\therefore\ c$∴ c | $=$= | $13$13 |
|
Hence, $a=2$a=2, $b=-12$b=−12, $c=13$c=13 and our function has the equation $f(x)=2x^2-12x+13$f(x)=2x2−12x+13.
Example 2
A colony of ants establishes a new nest and the number of ants in the nest after $t$t months can be roughly modelled by the function: $A(t)=\frac{6400}{1+159e^{-0.5t}}$A(t)=64001+159e−0.5t
a) How many ants established the colony?
Think: The initial population will be at $t=0$t=0.
Do:
| $A(0)$A(0) | $=$= | $\frac{6400}{1+159e^{-0.5\times0}}$64001+159e−0.5×0 |
| $=$= | $\frac{6400}{1+159\times1}$64001+159×1 | |
| $=$= | $\frac{6400}{160}$6400160 | |
| $=$= | $40$40 |
Thus, $40$40 ants established ants the colony.
b) What is the average growth rate of the colony in the first year?
Think: The average rate of change is given by $\frac{\text{Change in population}}{\text{Change in time}}$Change in populationChange in time. Recall the time is in months, so here we want the average rate of change from $t=0$t=0 to $t=12$t=12.
Do:
| Average rate of change | $=$= | $\frac{A\left(12\right)-A\left(0\right)}{12-0}$A(12)−A(0)12−0 |
| $\approx$≈ | $\frac{4591-40}{12}$4591−4012 | |
| $=$= | $379.2$379.2 ants per year (to $1$1 decimal place) |
c) The model suggests there is a limit to the colony size. What is the limit?
Think: As $t$t get very large the term $e^{-0.5t}$e−0.5t approaches zero.
Do:
| Population limit | $=$= | $\lim_{t\rightarrow\infty}\frac{6400}{1+159e^{-0.5t}}$limt→∞64001+159e−0.5t |
| $=$= | $\frac{6400}{1+0}$64001+0 | |
| $=$= | $6400$6400 |
The population limit of the colony according to the model is $6400$6400 ants. The could also be confirmed using technology to graph the function.
d) Find the rate of change of the colony population $A'\left(t\right)$A′(t).
Think: We consider the numerator and the denominator and make the appropriate designations for $u$u and $v$v in order to use the quotient rule.
Do:
| Let | $u=6400$u=6400 | then | $u'=0$u′=0 |
| and | $v=1+159e^{-0.5t}$v=1+159e−0.5t | then | $v'=-0.5\times159e^{-0.5t}$v′=−0.5×159e−0.5t |
| $A'\left(t\right)$A′(t) | $=$= | $\frac{vu'-uv'}{v^2}$vu′−uv′v2 |
| $=$= | $\frac{0-6400\left(-0.5\times159e^{-0.5t}\right)}{\left(1+159e^{-0.5t}\right)^2}$0−6400(−0.5×159e−0.5t)(1+159e−0.5t)2 | |
| $=$= | $\frac{508800e^{-0.5t}}{\left(1+159e^{-0.5t}\right)^2}$508800e−0.5t(1+159e−0.5t)2 |
e) Using technology find the coordinates of the point of inflection in the graph of $A(t)$A(t).
Think: The point of inflection in the graph of $A(t)$A(t) is where the graph changes from concave-up (increasing gradient) to concave down (decreasing gradient). This will be a maximum turning point in the graph of $A'(t)$A′(t).
Do: Obtain a graph of $A'(t)$A′(t) and find the value of $t$t for which the the rate of change is at a maximum.
The maximum rate of change occurs at approximately $t=10.138$t=10.138, substituting this value into the function $A(t)$A(t) we find the coordinates of the point of inflection of $A(t)$A(t) are approximately $\left(10.138,3200\right)$(10.138,3200).
f) Sketch a graph of the population over the first $2$2 years displaying all key features.
Think: The key features to display in our graph are:
- The horizontal axis needs to be shown for the interval $0\le t\le24$0≤t≤24
- The limit for the population is $6400$6400, this can be shown as a dotted line the curve approaches
- The point of inflection and initial population should be labelled
- Axes should be labelled
- The general shape can be found using technology and plotting a few additional points may assist in obtaining an accurate sketch
Do:
Practice questions
Question 1
The function $f\left(x\right)=ax^3+bx^2+9x+4$f(x)=ax3+bx2+9x+4 has a horizontal point of inflection at $x=1$x=1.
Use information about $f'\left(x\right)$f′(x) to write an equation involving $a$a and $b$b.
Give your answer such that all terms are on one side and the coefficient of $a$a is positive.
Use information about $f''\left(x\right)$f′′(x) to write an equation for $b$b in terms of $a$a.
Hence solve for the value of $a$a.
Hence solve for $b$b.
Question 2
The number of cars ($N$N) parked in the drive through section of a fast food restaurant $t$t hours after midnight is given by $N=\frac{18t}{2+t^2}$N=18t2+t2.
Evaluate the number of cars parked in the drive through at 12:30 am.
Determine the rate of change of the number of cars parked in the drive through at time $t$t.
Express your answer in factorised form.
Determine the average rate of change of the number of cars parked in the drive through in the second hour.
Solve for the time $t$t when the number of cars in the drive through is decreasing most rapidly.
Express your answer in exact form.
Question 3
A population, $P$P in thousands of insects, was observed to follow the model $P\left(t\right)=1.5\sin\left(\frac{\pi t}{6}\right)+5$P(t)=1.5sin(πt6)+5 for $0\le t\le12$0≤t≤12 where $t$t is the time in months after January $1$1st in a given year.
Find $P'\left(t\right)$P′(t).
At what value of $t$t is the population at its highest?
At what possible value of $t$t is the population increasing at the fastest rate?
How would we observe the population increasing at its fastest rate in the graphs of $P\left(t\right)$P(t) and its derivative $P'\left(t\right)$P′(t)?
A maximum turning point in $P\left(t\right)$P(t) and a point of inflection in $P'\left(t\right)$P′(t).
AA maximum turning point $P'\left(t\right)$P′(t) and a point of inflection in $P\left(t\right)$P(t).
BA minimum turning point in $P'\left(t\right)$P′(t) and a point of inflection in $P\left(t\right)$P(t).
CA minimum turning point in $P\left(t\right)$P(t) and a point of inflection in $P'\left(t\right)$P′(t).
DWhat is the fastest rate of decrease in the population over the year?
Give your answer in $1000$1000 of insects per month.
What is the average rate of change in population between the population's peak and consecutive low?
Give your answer in $1000$1000 insects per month.