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Calculus, 2ADV C1 EQ-Bank 12

A block of ice is melting. The mass \(M\) kilograms of the ice block remaining at time \(t\) hours after it begins to melt is given by  \(M(t)=50(12-3t)^2, 0 \leqslant t \leqslant 4\).

  1. Find the rate of change of the ice block's mass at any time \(t\).   (1 mark)

  2. How long does it take for the ice block to completely melt?   (1 mark)

  3. At what time is the ice melting at a rate of 2100 kilograms per hour?   (2 marks)

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a.    \(\dfrac{dM}{dt}=-300(12-3t)\)

b.    \(4\ \text{hours}\)

c.    \(t=\dfrac{5}{3}\ \text{hours}\)

Show Worked Solution

a.    \(M(t)=50(12-3t)^2\)

\(\dfrac{dM}{dt}=50 \times 2 \times (-3) \times(12-3t)=-300(12-3t)\)
 

b.    \(\text{Find}\ t\ \text{when}\ \ M(t)=0:\)

\(50(12-3t)^2=0 \ \Rightarrow \ t=4\)

\(\text{Ice block is completely melted at} \ \ t=4 \ \ \text {hours}\)
 

c.    \(\text{Find}\ t \ \text{when}\ \ \dfrac{d M}{d t}=-2100:\)

\(-300(12-3t)\) \(=-2100\)
\(12-3t\) \(=7\)
\(-3t\) \(=-5\)
\(t\) \(=\dfrac{5}{3}\ \text{hours}\)

Calculus, 2ADV C1 EQ-Bank 11

An oil slick on the surface of water forms a circular shape. The radius \(r\) metres of the oil slick is increasing according to the formula  \(r(t)=3 \sqrt{t}\), where \(t\) is the time in minutes after the oil begins to spread,  \(t \geqslant 0\).

  1. Find the rate at which the radius is increasing at any time \(t\).   (1 mark)

  2. At what rate is the area of the oil slick increasing when \(t=16\) minutes?   (1 mark)

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a.    \(\dfrac{d r}{d t}=\dfrac{3}{2 \sqrt{t}}\)

b.    \(\dfrac{d r}{d t}=\dfrac{3}{8} \ \text{metres/min}\)

Show Worked Solution

a.    \(r(t)=3 \sqrt{t}\)

\(\dfrac{d r}{d t}=\dfrac{1}{2} \times 3 \times t^{-\tfrac{1}{2}}=\dfrac{3}{2 \sqrt{t}}\)
 

b.    \(\text{Find} \ \dfrac{d r}{d t} \ \text{when} \ \ t=16:\)

\(\dfrac{d r}{d t}=\dfrac{3}{2 \times \sqrt{16}}=\dfrac{3}{8} \ \text{metres/min}\)

Calculus, 2ADV C1 EQ-Bank 14

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 19

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

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

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

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