AS 91578 · 2023 paper · trig, quotients, the helicopter
Question 2(e) Merit to Excellence
Practises Related rates
A quotient with cos, a cot gradient, a stationary point on an exponential quotient, inflection on x2 ln x, and a related-rates chase where the rate given is an angle.
A police helicopter flies above a straight horizontal motorway chasing a speeding car, at a constant 72 m s⁻¹ and a constant height of 400 metres. When the direct distance from the helicopter to the car is 2500 metres, the angle of depression θ between the horizontal and the line of sight is increasing at 0.002 rad s⁻¹.
Calculate the speed of the car at this instant.
| helicopter 72 m s⁻¹ | |
| height 400 m | |
| line of sight 2500 m | |
| dθ/dt | 0.002 rad s⁻¹ |
|---|
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Let x be the horizontal distance from the helicopter to the car. How are x and θ linked?
tanθ = 400/x, so x = 400 cotθ. -
2 Differentiate with respect to θ, then chain in dθ/dt.
dx/dθ = −400cosec2θ = (−400)/(sin2θ)
dx/dt = dx/dθ × dθ/dt = (−400)/(sin2θ) × 0.002 = (−0.8)/(sin2θ) -
3 At this moment the line of sight is 2500 m long. What is sin θ?
sinθ = 400/2500 = 0.16, so sin2θ = 0.0256 and dx/dt = (−0.8)/0.0256 = −31.25. -
4 What does dx/dt = −31.25 mean? How does it give the car’s speed?
The gap is shrinking at 31.25 m s⁻¹. The helicopter chases at 72 m s⁻¹ in the same direction, so the car is 31.25 m s⁻¹ slower than it:
vcar = 72 − 31.25 = 40.75
What the marker wanted
- Achieved dx/dθ found.
- Merit An expression for dx/dt.
- Excellence dx/dt = −31.25 from correct derivatives, and the car’s speed (E8). dx/dt = −31.25 without the speed, or a correct solution with one minor error, is E7.
The schedule’s Excellence cell gives the car’s speed as 40.76 m s⁻¹, but its own working, 72 − 31.25, gives 40.75 m s⁻¹, and no rounding along the way reaches 40.76.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.