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Parametric equations in modellingAQA A-Level Maths: Flashcards

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Why use parametric equations in modelling?

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Why use parametric equations in modelling?
They give position (x,y)(x,y) and the time tt together.
How do you find the time of flight of a projectile?
Set y=0y=0 and take the non-zero root.
How do you find the range?
Substitute the time of flight into xx.
How do you find the greatest height?
Use symmetry (tt halfway) or dydt=0\frac{dy}{dt}=0, then find yy.
Model for a wheel: x=rsin⁡ωtx=r\sin\omega t, y=h−rcos⁡ωty=h-r\cos\omega t. Period?
2πω\dfrac{2\pi}{\omega}.
Greatest and least yy for y=h−rcos⁡ωty=h-r\cos\omega t?
h+rh+r and h−rh-r.
Convert x=80cos⁡θx=80\cos\theta, y=50sin⁡θy=50\sin\theta to Cartesian form.
x26400+y22500=1\dfrac{x^2}{6400}+\dfrac{y^2}{2500}=1.
When do two objects collide?
When they are at the same point at the same time.
Paths cross at a point. Does this prove a collision?
No; the objects must reach it at the same time.
Name a limitation of a projectile model.
It ignores air resistance (and spin and the object's size).
Distance between (25,10)(25,10) and (30,8)(30,8)?
29≈5.4\sqrt{29}\approx5.4.
What should you state before using a model?
Meaning of variables, units, and the valid range of tt.
Ball: y=16t−5t2y=16t-5t^2. Time of flight?
3.23.2 s.

Exam questions on Parametric equations in modelling

  1. A ball is kicked from level ground. At time tt seconds after the kick, its horizontal distance from the starting point is x=12tx=12t metres and its height is y=16t−5t2y=16t-5t^2 metres. The model applies until the ball lands.
    Find the greatest height reached by the ball.2 marks
  2. A Ferris wheel has its centre 1212 m above the ground. The position of a passenger pod is modelled by x=10sin⁡(0.2t)x=10\sin(0.2t), y=12−10cos⁡(0.2t)y=12-10\cos(0.2t), where xx is the horizontal distance in metres from the vertical line through the centre, yy is the height above the ground in metres and tt is the time in seconds after the pod passes its lowest point. The angle 0.2t0.2t is in radians.
    State the greatest height of the pod above the ground, and justify your answer.2 marks
  3. A runner follows an oval track. Relative to the centre of the track, the runner's position at time tt seconds is x=80cos⁡θx=80\cos\theta, y=50sin⁡θy=50\sin\theta, where θ=0.1t\theta=0.1t radians and xx and yy are in metres.
    Show that the runner's path has the Cartesian equation x26400+y22500=1\dfrac{x^2}{6400}+\dfrac{y^2}{2500}=1.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).