SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, 2ADV EQ-Bank 13

A quantity of radioactive material decays according to the equation

\(\dfrac{d M}{d t}=-k M\),

where \(M\) is the mass of the material in kilograms, \(t\) is the time in years and \(k\) is a constant.

  1. Verify that  \(M=A e^{-k t}\)  is a solution to this equation, where \(A\) is a constant.   (1 mark)

  2. The time for half of the material to decay is 300 years. The initial amount of material is 20 kg .
  3. Find the amount of material remaining after 1000 years.   (3 marks)

Show Answers Only

a.    \(M=A e^{-k t}\)

\(\dfrac{d M}{d t}=-k A e^{-k t}=-k M \ \left( \text{since} \ M=A e^{-k t}\right)\)
 

b.    \(1.98\text{ kg}\)

Show Worked Solution

a.    \(M=A e^{-k t}\)

\(\dfrac{d M}{d t}=-k A e^{-k t}=-k M \ \left( \text{since} \ M=A e^{-k t}\right)\)
 

b.    \(\text{When} \ t=0, M=20:\)

\(20=A e^{-k(0)}\ \ \Rightarrow\ \ A=20\)

\(M=20 e^{-k t}\)
 

\(\text{When }\ t=300, M=10:\)

\(10\) \(=20 e^{-300 k}\)
\(\dfrac{1}{2}\) \(=e^{-300 k}\)
\(-300 k\) \(=\ln \frac{1}{2}\)
\(-300k\) \(=- \ln2\)
\(k\) \(=\dfrac{\ln 2}{300}\)

 
\(\text{When } \ t=1000:\)

\(M=20 e^{-1000 k}=20 e^{-\tfrac{10}{3} \ln 2}=1.984 \ldots\) 

\(\therefore \ \text{The amount remaining after 1000 years} \approx 1.98\text{ kg}\)

Calculus, 2ADV C1 EQ-Bank 16

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)

Show Answers Only

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)

Show Answers Only

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 20

A cylindrical water tank is being filled. The volume \(V\) litres of water in the tank at time \(t\) minutes after filling begins is given by  \(V(t)=500 \sqrt{(2 t+1)}, t \geqslant 0\).

  1. At what rate is water entering the tank at any time \(t\) ?   (1 mark)

  2. Find the rate at which the tank is being filled when \(t=12\) minutes.   (1 mark)

  3. At what time is the water flowing into the tank at a rate of 125 litres per minute?   (2 marks)

Show Answers Only

a.    \(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{2t+1}} \)

b.    \(100\ \text{L/min} \)

c.    \(\dfrac{15}{2}\ \text{mins}\)

Show Worked Solution

a.    \(V(t)=500(2t+1)^{\frac{1}{2}}\)

\(\dfrac{dV}{dt}=\dfrac{1}{2} \times 2 \times 500(2t+1)^{-\frac{1}{2}} = \dfrac{500}{\sqrt{2t+1}} \)
 

b.    \(\text{When}\ \ t=12:\)

\(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{25}} = 100\ \text{L/min} \)
 

c.    \(\text{Find}\ t\ \text{when}\ \dfrac{dV}{dt}=125:\)

\(125\) \(=\dfrac{500}{\sqrt{2t+1}}\)  
\(\sqrt{2t+1}\) \(=4\)  
\(2t+1\) \(=16\)  
\(t\) \(=\dfrac{15}{2}\ \text{mins}\)  

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 18

The volume of water in a tank, \(V\) litres, at time \(t\) minutes is given by:

\(V(t)=2 t^3-15 t^2+24 t+50\)  for  \(0 \leqslant t \leqslant 6\)

  1. Find an expression for the rate at which water is flowing at time \(t\).   (1 mark)

  2. Calculate the rate of flow at  \(t=4\)  minutes and interpret this value.   (1 mark)

  3. Deduce when the water level in the tank is increasing.   (2 marks)

Show Answers Only

a.    \(\dfrac{dV}{d t}=6 t^2-30 t+24\)
 

b.    \(\text{At} \ \ t=4:\ \ \dfrac{dV}{d t}=0 \ \text{litres/minute}\)

\(\text{Interpretation: At \(\ t=4 \ \), water has stopped flowing either into or}\)

\(\text{out of the tank.}\)
 

c.    \(\text{If water level is increasing} \ \Rightarrow \ \dfrac{dV}{d t}>0:\)

\(\text {Find \(t\) when}\ \dfrac{dV}{d t}>0:\)

\(6 t^2-30t+24\) \(\gt 0\)  
\(6\left(t^2-5 t+4\right)\) \(\gt 0\)  
\((t-4)(t-1)\) \(\gt 0\)  

 

\(\therefore \ \text{Water level increases for}\ \ t \in [0,1) \cup (4, 6]\)

Show Worked Solution

a.    \(V=2 t^3-15 t^2+24 t+50\)

\(\dfrac{dV}{d t}=6 t^2-30 t+24\)
 

b.    \(\text{At} \ \ t=4:\)

\(\dfrac{dV}{d t}=6 \times 4^2-30 \times 4+24=0 \ \text{litres/minute}\)

\(\text{Interpretation: At \(\ t=4 \ \), water has stopped flowing either into or}\)

\(\text{out of the tank.}\)
 

c.    \(\text{If water level is increasing} \ \Rightarrow \ \dfrac{dV}{d t}>0:\)

\(\text {Find \(t\) when}\ \dfrac{dV}{d t}>0:\)

\(6 t^2-30t+24\) \(\gt 0\)  
\(6\left(t^2-5 t+4\right)\) \(\gt 0\)  
\((t-4)(t-1)\) \(\gt 0\)  

 

\(\therefore \ \text{Water level increases for}\ \ t \in [0,1) \cup (4, 6]\)

Calculus, 2ADV C1 EQ-Bank 15

The displacement \(x\) metres from the origin at time, \(t\) seconds, of a particle travelling in a straight line is given by

\(x=t^3-9 t^2+9 t, \quad t \geqslant 0\)

  1. Find the time(s) when the particle is at the origin.   (2 marks)
  2. On the graph below, sketch the displacement, \(x\) metres, with respect to time \(t\).   (2 marks)
     
       

  3. Find the velocity of the particle when  \(t=2\).   (2 marks)

Show Answers Only

a.  \(\text{Particle at origin when}\ \ t=0, t=3.\)

b.
       
 

c.   \(\dot{x}=-15\  \text{m s}^{-1}\)

Show Worked Solution

a.     \(x\) \(=t^3-9 t^2+9 t\)
    \(=t\left(t^2-9 t+9\right)\)
    \(=t(t-3)^2\)

 
\(\text{Particle at origin when}\ \ t=0, t=3.\)

 
b.
       
 

c.    \(x=t^3-9 t^2+9 t\)

\(\dot{x}= \dfrac{dx}{dt} = 3 t^2-18 t+9\)

\(\text {When } t=2:\)

\(\dot{x}=3 \times 2^2-18 \times 2+9=-15\  \text{m s}^{-1}\)

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

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

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

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

Calculus, 2ADV C1 EQ-Bank 4 MC

The displacement of a particle is given by  \(x=3t^{3}-6t^{2}-15\) . The acceleration is zero at:

  1. \(t=\dfrac{2}{3}\)
  2. \(t=\dfrac{4}{3}\)
  3. \(t=\dfrac{5}{2}\)
  4. \(\text{never}\)
Show Answers Only

\(A\)

Show Worked Solution

\(x=3t^{3}-6t^{2}-15\)

\(v=9t^{2}-12t\)

\(a=18t-12\)

\(\text{Find}\ t\ \text{when}\ \ a=0:\)

\(18t-12=0\ \ \Rightarrow\ \ t=\dfrac{2}{3} \)

\(\Rightarrow A\)

Calculus, 2ADV C1 EQ-Bank 17

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = 2t^3-t^2-3t + 11`  when  `t >= 0`

  1.  Calculate the velocity when  `t = 2`.   (1 mark)

  2.  When is the particle stationary?   (2 marks)

Show Answers Only

 a.    `17\ text(ms)^(−1)`

 b.    `(1 + sqrt19)/6`

Show Worked Solution

a.   `x =2t^3-t^2-3t + 11` 

`v = (dx)/(dt) = 6t^2-2t-3`

 
`text(When)\ t = 2:`

`v= 6 xx 2^2-2 · 2-3= 17\ text(ms)^(−1)`
 

b.   `text(Particle is stationary when)\ \ v = 0`

`6t^2-2t-3=0`

`:. t` `= (2 ±sqrt((−2)^2-4 · 6 · (−3)))/12`
  `= (2 ± sqrt76)/12`
  `= (1 ± sqrt19)/6`
  `= (1 + sqrt19)/6 qquad(t >= 0)`

Calculus, 2ADV C1 EQ-Bank 23

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = sqrt(20t-2t^2)`  metres per second

Find the acceleration of the particle when  `t = 4`.

Express your answer as an exact value in its simplest form.   (3 marks)

Show Answers Only

`sqrt3/6\ \ text(ms)^(−2)`

Show Worked Solution

`v = sqrt(20t-2t^2)`

`a` `= (dv)/(dt)`
  `= 1/2 · (20t-2t^2)^(−1/2) · (20-4t)`

 
`text(When)\ \ t = 4:`

`a` `= 1/2(20 · 4-2 · 4^2)^(−1/2)(20-16)`
  `= 2/(sqrt48)`
  `= 2/(4sqrt3) xx sqrt3/sqrt3`
  `= sqrt3/6\ \ text(ms)^(−2)`

Calculus, 2ADV C1 2008 HSC 6b

The graph shows the velocity of a particle,  `v`  metres per second, as a function of time,  `t`  seconds.

  1. What is the initial velocity of the particle?   (1 mark)

  2. When is the velocity of the particle equal to zero?   (1 mark)

  3. When is the acceleration of the particle equal to zero?   (1 mark)

Show Answers Only
  1. `20\ text(m/s)`
  2. `t=10\ text(seconds)`
  3. `t=6\ text(seconds)`
Show Worked Solution

i.    `text(Find)\ v\ text(when)  t=0:`

`v=20\ \ text(m/s)`

ii.    `text(Particle comes to rest at)\  t=10\ text{seconds  (from graph)}`

iii.  `text(Acceleration is zero when)\ t=6\ text{seconds  (from graph)}`

Calculus, 2ADV C1 2018 HSC 12d

The displacement of a particle moving along the `x`-axis is given by

`x = t^3/3-2t^2 + 3t,`

where `x` is the displacement from the origin in metres and `t` is the time in seconds, for `t >= 0`.

  1. What is the initial velocity of the particle?   (1 mark)

  2. At which times is the particle stationary?   (2 marks)

  3. Find the position of the particle when the acceleration is zero.   (2 marks)

Show Answers Only

i.    `3\ text(ms)^(-1)`

ii.   `t = 1 or 3\ text(seconds)`

iii.  `2/3\ text(m)`

Show Worked Solution

i.    `x = t^3/3-2t^2 + 3t`

`v = (dx)/(dt) = t^2-4t + 3`
 

`text(Find)\ v\ text(when)\ \ t = 0:`

`v= 0-0 + 3= 3\ text(ms)^(-1)`
 

ii.  `text(Particle is stationary when)\ \ v = 0`

`t^2-4t + 3 = 0`

`(t-3) (t-1) = 0`

`t = 1 or 3\ text(seconds)`
 

iii.  `a = (dv)/(dt) = 2t-4`
 

`text(Find)\ t\ text(when)\ \ a = 0:`

`2t-4= 0\ \ =>\ \ t=2`

`x(2)= 2^3/3-2(2^2) + 3(2)= 8/3-8 + 6= 2/3`

Calculus, 2ADV C1 2006 HSC 8a

A particle is moving in a straight line. Its displacement, `x` metres, from the origin, `O`, at time `t` seconds, where  `t ≥ 0`, is given by  `x = 1-7/(t + 4)`.

  1. Find the initial displacement of the particle.   (1 mark)

  2. Find the velocity of the particle as it passes through the origin.   (3 marks)

  3. Show that the acceleration of the particle is always negative.   (1 mark)

  4. Sketch the graph of the displacement of the particle as a function of time.   (2 marks)

Show Answers Only

i.    `text(–3/4 m)`

ii.   `1/7\ text(ms)^-1`

iii.  `text(Proof)\ \ text{(See Worked Solutions)}`

iv.   

Show Worked Solution

i.    `x = 1-7/(t + 4)`

`text(When)\ \ t = 0:`

`x= 1-7/4= -3/4\ \text(m)`

`:.\ text(Initial displacement is)\ 3/4\ text(metres to the left of the origin.)`

 

ii.  `x = 1-7/(t+4) = 1-7(t + 4)^-1`

`dot x` `= (-1)  -7(t + 4)^-2 xx d/(dt)(t + 4)`
  `= 7 (t + 4)^-2 xx 1`
  `= 7/(t + 4)^2`

 
`text(Find)\ t\ text(when)\ x = 0:`

`0` `= 1-7/(t + 4)`
`7/(t + 4)` `= 1`
`7` `= (t + 4)`
`t` `= 3`

 

`text(When)\ t = 3:`

`dot x= 7/(3 + 4)^2= 1/7\ text(ms)^-1`

`:.\ text(The velocity of the particle through the origin is)\ 1/7\ text(ms)^-1.`

 

iii.  `dot x` `= 7(t + 4)^-2`
`ddot x` `= (d dot x)/(dt) = -14 (t +4)^-3`

 
`text(Given)\ t >= 0:`

`=>  (t + 4)^-3 >= 0`

`=> -14 (t + 4)^-3 <= 0`

`:. ddot x\ text(is always negative.)`
 

iv.  2UA HSC 2006 8a

Calculus, 2ADV C1 2014 HSC 13c

The displacement of a particle moving along the `x`-axis is given by

 `x = t-1/(1 + t)`,

where `x` is the displacement from the origin in metres, `t` is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.    (2 marks)

  2. What value does the velocity approach as `t` increases indefinitely?    (1 mark)

Show Answers Only

i.    `text(Proof)\ \ text{(See Worked Solutions)}`

ii.   `1`

Show Worked Solution

i.    `x= t-1/(1 + t)= t-(1 + t)^(-1)`

`dot x= 1-(-1) (1 + t)^(-2)= 1 + 1/((1 + t)^2)`

`ddot x= -2(1 + t)^(-3)= – 2/((1 + t)^3)`

`text(S)text(ince)\ \ t >= 0\ \ =>\ \ -2/((1 + t)^3) < 0`

`:.\ text(Acceleration is always negative.)`
 

ii.   `text(Velocity)\ (dot x) = 1 + 1/((1 + t)^2)`

`text(As)\ t -> oo,\ 1/((1 + t)^2) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 1`

Calculus, 2ADV C1 2014 HSC 9 MC

The graph shows the displacement  `x`  of a particle moving along a straight line as a function of time  `t`.

2014 9 mc

 Which statement describes the motion of the particle at the point  `P`? 

  1. The velocity is negative and the acceleration is positive.
  2. The velocity is negative and the acceleration is negative.
  3. The velocity is positive and the acceleration is positive.
  4. The velocity is positive and the acceleration is negative.
Show Answers Only

`A`

Show Worked Solution

`text(At)\ P,\ text(the particle is moving back towards)\ O`

`text{after hitting a max (positive) displacement}`

`:.\ text(Velocity is negative.)`

`text(Its displacement hits a minimum just after)\ P`

`text(and increases again.)`

`:.\ text{Acceleration is working against (negative) velocity}`

`text(and must be positive.)`

`=>  A`