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Calculus, 2ADV C1 EQ-Bank 5

A drone travels vertically from its launch pad.

It's height above ground, \(h\) metres, at time \(t\) minutes is modelled by

\(h(t)=-0.2 t^3+3 t^2+5 t\)  for  \(0 \leq t \leq 12\)

  1. Find the velocity of the drone at time \(t\) minutes.   (1 mark)

  2. Determine the exact time interval during which the drone is descending.   (2 marks)

Show Answers Only

a.    \(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)
 

b.    \(\dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)

Show Worked Solution

a.    \(h=-0.2 t^3+3 t^2+5 t\)

\(\text{Velocity of the drone}=\dfrac{d h}{d t}.\)

\(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)
 

b.    \(\text{Drone is descending when} \ \ \dfrac{dh}{dt}<0:\)

\(-0.6 t^2+6 t+5\) \(<0\)  
\(0.6 t^2-6 t-5\) \(>0\)  
\(6 t^2-60 t-50\) \(>0\)  

 
\(\text{Solve}\ \ 6 t^2-60 t-50=0:\)

\(t=\dfrac{60 \pm \sqrt{(-60)^2+4 \times 6 \times 50}}{2 \times 6}=\dfrac{60 \pm \sqrt{4800}}{12}=\dfrac{15 \pm 10 \sqrt{3}}{3}\)

 
\(\text{Since parabola is concave up:}\)

\(6 t^2-60 t-50>0\ \ \text{when}\ \ t>\dfrac{15+10 \sqrt{3}}{3} \quad\left( t=\dfrac{15-10 \sqrt{3}}{3}<0\right)\)

\(\therefore \text{Drone is descending for} \ \ \dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)

Calculus, 2ADV C1 EQ-Bank 5

Following a magpie plague in Raymond Terrace, a bird researcher estimated that the magpie population, \(M\), in hundreds, \(t\) months after 1st January, was given by  \(M=7+20t-3t^2\).

  1. Find the magpie population on 1st March.   (1 mark)
  2. At what rate was the population changing at this time?   (1 mark)
  3. In what month does the magpie population start to decrease?   (2 marks)
Show Answers Only

a.    \(\text{Population}\ =35 \times 100=3500\)

b.   \(M\ \text{is increasing at 800 per month}\)

c.   \(\text{Population starts to decrease in May.}\)

Show Worked Solution

a.    \(\text{Find}\ M\ \text{when}\ \ t=2:\)

\(M=7+20 \times 2-3 \times 2^2 = 35\)

\(\therefore \text{Population}\ =35 \times 100=3500\)
 

b.    \(M=7+20t-3t^2\)

\(\dfrac{dM}{dt}=20-6t\)

\(\text{Find}\ \dfrac{dM}{dt}\ \text{when}\ \ t=2: \)

\(\dfrac{dM}{dt}=20-6 \times 2 = 8\)

\(\therefore M\ \text{is increasing at 800 per month}\)
 

c.    \(\text{Find}\ t\ \text{when}\ \dfrac{dM}{dt}=0: \)

\(\dfrac{dM}{dt}=20-6t = 0\ \ \Rightarrow \ t= 3\ \dfrac{1}{3}\)

\(\dfrac{dM}{dt}<0\ \ \text{when}\ \ t>3\ \dfrac{1}{3} \)

\(\therefore\ \text{Population starts to decrease in May.}\)