Year 13 parametric equations test answers to download as PDF

These Year 13 parametric equations test answers give the worked solutions and the marks for every question of the parametric equations test. The test is marked out of 20, so you can work out your score.

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Answer to Question 1 (4 marks)
Method: to find a Cartesian equation, eliminate the parameter. If one equation is linear in \(t\), make \(t\) the subject and substitute. If the equations involve \(\sin t\) and \(\cos t\), isolate them and use \(\sin^2t+\cos^2t=1\) or a double angle formula. Mark scheme as shown in brackets.
1. Use the \(x\) coordinate to find \(t\), then check with the \(y\) coordinate. \(3t-1=8\) gives \(t=3\) (1); check \(y=9+2=11\) (1). So the point \((8,11)\) corresponds to \(t=3\).
2. Make \(t\) the subject of the linear equation and substitute into \(y\). \(t=\dfrac{x+1}3\) (1), so \(y=\dfrac{(x+1)^2}9+2\) (1). The curve is a parabola with vertex \((-1,2)\).
Answer to Question 2 (4 marks)
Method: to find a Cartesian equation, eliminate the parameter. If one equation is linear in \(t\), make \(t\) the subject and substitute. If the equations involve \(\sin t\) and \(\cos t\), isolate them and use \(\sin^2t+\cos^2t=1\) or a double angle formula. Mark scheme as shown in brackets.
\(\cos t=\dfrac{x-1}3\), \(\sin t=\dfrac{y-2}3\) (1). \((x-1)^2+(y-2)^2=9\) (2). Centre \((1,2)\), radius 3 (1).
Answer: the curve is the circle \((x-1)^2+(y-2)^2=9\), with centre \((1,2)\) and radius 3.
Answer to Question 3 (4 marks)
Method: to find a Cartesian equation, eliminate the parameter. If one equation is linear in \(t\), make \(t\) the subject and substitute. If the equations involve \(\sin t\) and \(\cos t\), isolate them and use \(\sin^2t+\cos^2t=1\) or a double angle formula. Mark scheme as shown in brackets.
1. Use the double angle formula that contains only \(\sin t\), then replace \(\sin t\) by \(x\). \(\cos2t=1-2\sin^2t\) (1) so \(y=1-2x^2\) (1).
2. \(x=\sin t\) lies between \(-1\) and 1; then \(x^2\) lies between 0 and 1. \(-1\le x\le1\) (1) and \(-1\le y\le1\) (1). So only the arch of the parabola between \(x=-1\) and \(x=1\) is drawn.
Answer to Question 4 (4 marks)
Method: to find where a parametric curve meets a line or an axis, substitute the expressions for \(x\) and \(y\) into the equation of the line, solve for \(t\), then substitute each value of \(t\) back to get the coordinates. Mark scheme as shown in brackets.
\(2t=t^2-8\) (1), so \(t^2-2t-8=0\) (1) and \((t-4)(t+2)=0\) (1). Points \((16,8)\) and \((4,-4)\) (1).
Check: \(8=16-8\) and \(-4=4-8\). Answer: the points are \((16,8)\) and \((4,-4)\).
Answer to Question 5 (4 marks)
Method: in a parametric model, \(t\) is usually time. Answer each question by deciding first whether you need a value of \(t\) (from a condition on \(x\) or \(y\)) or a position (by substituting \(t\)). Mark scheme as shown in brackets.
1. Greatest height: \(10-9.8t=0\) gives \(t=1.02\) s and \(y=2+\dfrac{100}{19.6}=7.10\) m (1). Landing: \(4.9t^2-10t-2=0\) gives \(t=2.224\) s (1), so \(x=26.7\) m (1).
2. No air resistance, ball treated as a particle (1).
Answer: the ball rises to about 7.10 m and lands about 26.7 m away.

Total : 20 marks

How to use these Year 13 parametric equations test answers

First, sit the test under timed conditions before you read anything here. Then mark each answer and give yourself the marks shown, including method marks. When you lose marks, read the worked solution and find the step you missed. After that, revise the parametric equations lesson and redo the worksheet.

Also, the Year 13 parametric equations test answers are available as a PDF, so a teacher can mark printed tests quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday.