2024 HSC Mathematics Extension 1
Question 14(d):
Vector motion
A particle is projected from the origin with initial speed at an angle to the horizontal. The position vector of the particle is (do NOT prove this).
Let be the distance of the particle from the origin at time , so .
Show that for the distance is increasing for all .
How to recognise this question
Question type: vector motion
- The command and notation point to vector motion.
- The requested response is a show that.
How to handle it: differentiate position to get velocity and acceleration, then translate the angle condition into a dot product.
Watch out: Do not select a formula until its domain, interval, sign and units match the question.
Step-by-step answer
- Work with to avoid the square root: .
- Expand: .
- Differentiate: .
- For , sign of matches the quadratic .
- This quadratic in (opening upwards) is positive for all if its discriminant is negative: .
- Need (since for launch angles in ).
- Thus implies for all , so is strictly increasing.
- NESA 14(d): 4 marks correct proof; 3 marks for explaining why the discriminant is negative; 2 marks for a correct derivative of or ; 1 mark for writing or .
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 4 total marks for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.
Try your hand at it again by answering these questions
Question 1
With and , show that if then is increasing for all .
Question 2
Starting from for , show that the quadratic factor is positive for all precisely when .
Question 3
Explain why showing for proves that the distance is increasing, then state the critical angle condition in terms of .