SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, EXT1 C1 2004 HSC 3c

A ferry wharf consists of a floating pontoon linked to a jetty by a 4 metre long walkway. Let  `h`  metres be the difference in height between the top of the pontoon and the top of the jetty and let  `x`  metres be the horizontal distance between the pontoon and the jetty.
 

2004 3c
 

  1. Find an expression for  `x`  in terms of  `h`.  (1 mark)

When the top of the pontoon is 1 metre lower than the top of the jetty, the tide is rising at a rate of 0.3 metres per hour.

  1. At what rate is the pontoon moving away from the jetty?  (3 marks)

Show Answers Only

a.    `x=sqrt(16 − h^2)`

b.    `0.077\ text(m/hr)`

Show Worked Solution

a.    `text(Using Pythagoras,)`

`x^2 + h^2` `= 4^2`
`x^2` `= 16 − h^2`
`x` `= sqrt(16 − h^2)`

 

b.    `text(Find)\ \ (dx)/(dt)\ \ text(when)\ \ h = 1`

`(dx)/(dt)` `= (dx)/(dh)·(dh)/(dt)`
`x` `= (16 − h^2)^(1/2)`
`(dx)/(dh)` `= 1/2 xx (16 − h^2)^(−1/2) xx d/(dh)(16 − h^2)`
  `= 1/2 (16 − h^2)^(−1/2) xx −2h`
  `= (−h)/(sqrt(16 − h^2))`

 

`text(When)\ \ h = 1, (dh)/(dt)= −0.3\ text(m/hr)`

`text{(}h\ \ text{decreases when the tide is rising)}`

`(dx)/(dt)` `= (−h)/(sqrt(16 − h^2)) xx −0.3`
  `= (−1)/sqrt(16 − 1^2) xx −0.3`
  `= 0.3/sqrt15`
  `= 0.0774…`
  `= 0.077\ \ \ text{metres per hr (to 2 d.p.)}`

 

`:.\ text(When)\ \ h = 1,\ text(the pontoon is moving away)`

`text(at 0.077 metres per hr.)`

Calculus, EXT1 C1 2014 HSC 13b

One end of a rope is attached to a truck and the other end to a weight. The rope passes over a small wheel located at a vertical distance of  40 m above the point where the rope is attached to the truck.

The distance from the truck to the small wheel is `L` m, and the horizontal distance between them is `x` m. The rope makes an angle `theta` with the horizontal at the point where it is attached to the truck.

The truck moves to the right at a constant speed of 3 metres per second, as shown in the diagram.
 

  1. Using Pythagoras’ Theorem, or otherwise, show that  `(dL)/(dx) = cos theta`.   (2 marks)

  2.  Show that  `(dL)/(dt) = 3 cos theta`.   (1 mark)

Show Answers Only

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

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

Show Worked Solution
a.      

`text(Show)\ \ (dL)/(dx) = cos theta`

`text(Using Pythagoras:)`

`L^2` `=40^2 + x^2`
`L` `=(40^2 + x^2)^(1/2)`
`(dL)/(dx)` `=1/2 * 2x * (40^2 + x^2)^(-1/2)`
  `=x/ sqrt((40^2 + x^2))`
  `=x/L`
  `=cos theta\ \ \ text(… as required.)`

 

b.    `text(Show)\ \ (dL)/(dt) = 3 cos theta`

`(dL)/(dt)` `= (dL)/(dx) * (dx)/(dt)`
  `= cos theta xx 3`
  `= 3 cos theta\ \ \ text(… as required)`

Calculus, EXT1 C1 2010 HSC 2d

A radio transmitter `M` is situated 6 km from a straight road. The closest point on the road to the transmitter is `S`.

A car is travelling away from `S` along the road at a speed of 100 km h−1.  The distance from the car to `S` is `x` km and from the car to `M` is `r` km.
 

2010 2d

Find an expression in terms of `x` for `(dr)/(dt)`, where `t` is time in hours.   (3 marks) 

Show Answers Only

 `(100x)/sqrt(x^2 + 36)\ \ text(km/hr)`

Show Worked Solution

`text(Using Pythagoras:)`

`r^2` `= x^2 + 6^2`
`r` `= sqrt(x^2 + 36),\ r > 0`
`(dr)/(dt)` `= (dx)/(dt) * (dr)/(dx)\ \ \ \ …\ (1)`
`(dx)/(dt)` `= 100\ \ \ text{(given)}`
`(dr)/(dx)` `= 1/2 (x^2 + 36)^(-1/2) xx d/(dx) (x^2 + 36)= x/sqrt(x^2 + 36)`

 
`text{Substituting into (1):}`

`:.\ (dr)/(dt)= (100x)/sqrt(x^2 + 36)\ \ text(km/hr)`

Calculus, EXT1 C1 2012 HSC 14c

A plane `P` takes off from a point `B`. It flies due north at a constant angle `alpha` to the horizontal. An observer is located at `A`, 1 km from `B`, at a bearing 060° from `B`.

Let `u` km be the distance from `B` to the plane and let `r` km be the distance from the observer to the plane. The point `G` is on the ground directly below the plane.
 

2012 14c

  1. Show that  `r = sqrt(1 + u^2 -u cos alpha)`.   (3 marks)

  2. The plane is travelling at a constant speed of 360 km/h.
  3. At what rate, in terms of `alpha`, is the distance of the plane from the observer changing 5 minutes after take-off?   (2 marks)

Show Answers Only

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

b.    `(180 (60-cos alpha))/sqrt(901-30 cos alpha)\ \ text(km/hr)`

Show Worked Solution

a.    `text(Show)\ \ r = sqrt(1 + u^2-u cos alpha)`

♦ Mean mark (a) 42%
IMPORTANT: Look for opportunities to use the identity `sin^2 alpha+cos^2 alpha=1` to clean up trig calculations.

`text(In)\ Delta PGB:`

`cos alpha= (BG)/u\ \ =>\ \ BG= u cos alpha`

`sin alpha= (PG)/(u)\ \ =>\ \ PG= u sin alpha`
 

`text(In)\ Delta PGA\ text(using Pythagoras):`

`AG^2= r^2-PG^2= r^2-u^2 sin^2 alpha`
 

`text(Using cosine rule in)\ Delta ABG:`

`AG^2` `= BG^2 + AB^2-2 xx BG xx AB xx cos 60^@`
`r^2-u^2 sin^2 alpha` `= u^2 cos^2 alpha + 1-2 (u cos alpha) xx 1 xx 1/2`
`r^2` `= u^2 cos^2 alpha + u^2 sin^2 alpha + 1-u cos alpha`
  `= u^2 (cos^2 alpha + sin^2 alpha) + 1-u cos alpha`
  `= u^2 + 1-u cos alpha`
`r` `= sqrt(u^2 + 1-u cos alpha)\ \ text(… as required)`

 

b.    `text(Find)\ (dr)/(dt)\ text(when)\ t =5:`

♦♦♦ Mean mark (b) 14%

`(dr)/(dt) = (dr)/(du) xx (du)/(dt)`

`r` `= (u^2 + 1-u cos alpha)^(1/2)`
`(dr)/(du)` `= 1/2 (u^2 + 1-u cos alpha)^(-1/2) xx (2u-cos alpha)`
  `= (2u-cos alpha)/(2 sqrt(u^2 + 1-u cos alpha))`

 
`(du)/(dt)= 360\ text{km/hr   (plane’s speed)}`

 
`text(After 5 mins:)`

`u` `= 5/60 xx 360 = 30\ text(km)`
`(dr)/(dt)` `= ((2 xx 30)-cos alpha)/( 2 sqrt(30^2 + 1-30 cos alpha)) xx 360`
  `= (180 (60-cos alpha))/sqrt(901-30 cos alpha)\ \ text(km/hr)`