Worksheet: Conversion between Parametric and Rectangular Equations

In this worksheet, we will practice converting from the parametric form of an equation to its equivalent rectangular form and vice versa.

Q1:

Use the fact that coshsinh๏Šจ๏Šจ๐‘ฅโˆ’๐‘ฅ=1 to find a parametrization of the part of the hyperbola ๐‘ฅ25โˆ’๐‘ฆ81=1๏Šจ๏Šจ that contains the point (โˆ’5,0).

  • A ๐‘ฅ = 5 ๐‘ก , ๐‘ฆ = 9 ๐‘ก c o s h s i n h
  • B ๐‘ฅ = โˆ’ 5 ๐‘ก , ๐‘ฆ = โˆ’ 9 ๐‘ก c o s h s i n h
  • C ๐‘ฅ = 2 5 ๐‘ก , ๐‘ฆ = 8 1 ๐‘ก c o s h s i n h
  • D ๐‘ฅ = โˆ’ 5 ๐‘ก , ๐‘ฆ = 9 ๐‘ก c o s h s i n h
  • E ๐‘ฅ = 5 ๐‘ก , ๐‘ฆ = โˆ’ 9 ๐‘ก c o s h s i n h

Q2:

The first figure shows the graphs of cos2๐œ‹๐‘ก and sin2๐œ‹๐‘ก, which parameterize the unit circle for 0โ‰ค๐‘กโ‰ค1. What do the two functions graphed in the second figure parameterize?

  • Athe unit circle
  • Bthe square on (0.5,0.5), (0.5,โˆ’0.5), (โˆ’0.5,โˆ’0.5), (โˆ’0.5,0.5)
  • Cthe square on (0,0), (0,1), (1,1), (0,1)
  • Dthe square on (1,0), (0,1), (โˆ’1,0), (0,โˆ’1)

Q3:

Consider the points ๐ด=(1,1) and ๐ต=(5,4).

What is the length of ๐ด๐ต?

Find a parameterization of the segment ๐ด๐ต over 0โ‰ค๐‘กโ‰ค1.

  • A ๐‘ฅ = 4 ( ๐‘ก + 1 ) , ๐‘ฆ = 3 ( ๐‘ก + 1 )
  • B ๐‘ฅ = ๐‘ก + 4 , ๐‘ฆ = ๐‘ก + 3
  • C ๐‘ฅ = 4 ๐‘ก + 1 , ๐‘ฆ = 3 ๐‘ก + 1
  • D ๐‘ฅ = 3 ๐‘ก + 1 , ๐‘ฆ = 4 ๐‘ก + 1
  • E ๐‘ฅ = 3 ( ๐‘ก + 1 ) , ๐‘ฆ = 4 ( ๐‘ก + 1 )

Find ๐‘“ and ๐‘” so that ๐‘ฅ=๐‘“(๐‘ก), ๐‘ฆ=๐‘”(๐‘ก) parameterizes ๐ด๐ต over 0โ‰ค๐‘กโ‰ค5.

  • A ๐‘“ ( ๐‘ก ) = 4 ๐‘ก + 1 , ๐‘” ( ๐‘ก ) = 3 ๐‘ก + 1
  • B ๐‘“ ( ๐‘ก ) = ๐‘ก + 4 , ๐‘” ( ๐‘ก ) = ๐‘ก + 3
  • C ๐‘“ ( ๐‘ก ) = 3 ๐‘ก + 1 , ๐‘” ( ๐‘ก ) = 4 ๐‘ก + 1
  • D ๐‘“ ( ๐‘ก ) = ๐‘ก + 4 5 , ๐‘” ( ๐‘ก ) = ๐‘ก + 3 5
  • E ๐‘“ ( ๐‘ก ) = 4 5 ๐‘ก + 1 , ๐‘” ( ๐‘ก ) = 3 5 ๐‘ก + 1

Using the functions above for 0โ‰ค๐‘ โ‰ค5, what is the distance between the point (1,1) and the point (๐‘“(๐‘ ),๐‘”(๐‘ ))?

  • A ๐‘  โˆ’ 1
  • B ๐‘  + 1
  • C ๐‘ 
  • D ๐‘  + 2
  • E ๐‘  โˆ’ 2

The parameterization of ๐ด๐ต above is an example of an {arc-length parameterization} of a plane curve. Find an arc-length parameterization ๐‘ฅ=๐‘“(๐‘ก), ๐‘ฆ=๐‘”(๐‘ก) of ๐ด๐ถ, where ๐ถ=(13,6) and the parameter starts at ๐‘ก=0. Give the interval used.

  • A ๐‘ฅ = 1 2 1 3 ๐‘ก + 1 , ๐‘ฆ = 5 1 3 ๐‘ก + 1 on 5โ‰ค๐‘กโ‰ค13
  • B ๐‘ฅ = 1 2 1 3 ๐‘ก + 1 , ๐‘ฆ = 5 1 3 ๐‘ก + 1 on 0โ‰ค๐‘กโ‰ค13
  • C ๐‘ฅ = 5 1 3 ๐‘ก + 1 , ๐‘ฆ = 1 2 1 3 ๐‘ก + 1 on 0โ‰ค๐‘กโ‰ค13
  • D ๐‘ฅ = ๐‘ก + 1 2 1 3 , ๐‘ฆ = 5 1 3 ๐‘ก on 0โ‰ค๐‘กโ‰ค13
  • E ๐‘ฅ = 5 1 3 ๐‘ก + 1 , ๐‘ฆ = 1 2 1 3 ๐‘ก + 1 on 5โ‰ค๐‘กโ‰ค13

Q4:

A particle moves along the curve given by the parametric equations ๐‘ฅ=2๐‘ก+1๏Šจ, ๐‘ฆ=โˆ’3๐‘ก+2๏Šจ with โˆ’1โˆš2โ‰ค๐‘กโ‰ค1.

At which point is the particle located when ๐‘ก=1โˆš2? Give your answer to one decimal place, if necessary.

  • A ( 2 , 1 )
  • B ( 2 , 0 . 5 )
  • C ( 1 . 5 , 0 . 5 )
  • D ( 1 , 2 )
  • E ( 0 . 5 , 2 )

At which points is the particle located when ๐‘ก=0 and ๐‘ก=1? Give your answers to one decimal place, if necessary.

  • A ( 2 , โˆ’ 3 ) and (1,3)
  • B ( 1 , 2 ) and (3,1)
  • C ( 2 , 1 ) and (โˆ’1,3)
  • D ( 1 , 2 ) and (3,โˆ’1)
  • E ( 1 , 2 ) and (3,5)

Find an equation for the line that the particle moves along in the form ๐‘Ž๐‘ฅ+๐‘๐‘ฆ=๐‘.

  • A 2 ๐‘ฅ + 3 ๐‘ฆ = 7
  • B 3 ๐‘ฅ + 2 ๐‘ฆ = 7
  • C 3 ๐‘ฅ + 2 ๐‘ฆ = 1 3
  • D 3 ๐‘ฅ โˆ’ 2 ๐‘ฆ = 7
  • E 3 ๐‘ฅ โˆ’ 2 ๐‘ฆ = โˆ’ 1

What is the smallest value of ๐‘ฅ during the particleโ€™s motion? When is it reached?

  • A ๐‘ฅ = 1 , at ๐‘ก=โˆ’1โˆš2
  • B ๐‘ฅ = 1 , at ๐‘ก=0
  • C ๐‘ฅ = 0 . 5 , at ๐‘ก=โˆ’1โˆš2
  • D ๐‘ฅ = 0 . 5 , at ๐‘ก=0
  • E ๐‘ฅ = โˆ’ 1 , at ๐‘ก=1

Describe the motion from ๐‘ก=โˆ’1โˆš2 to ๐‘ก=1 in terms of position on the line.

  • AThe particle starts at (0.5,2), goes left and upward to (2,1), and then goes back right and downward to (โˆ’1,3).
  • BThe particle starts at (2,0.5), goes left and upward to (1,2), and then goes back right and downward to (3,โˆ’1).
  • CThe particle starts at (2,0.5), goes right and downward to (1,2), and then goes back left and upward to (3,โˆ’1).
  • DThe particle starts at (1,2), goes left and upward to (2,1), and then goes back right and downward to (3,โˆ’1).
  • EThe particle starts at (0.5,2), goes right and downward to (2,1), and then goes back left and upward to (โˆ’1,3).

Give the parameters ๐‘ฅ=๐‘“(๐‘ก), ๐‘ฆ=๐‘”(๐‘ก) that describe the same motion but on an interval starting at ๐‘ก=0 instead of โˆ’1โˆš2. In what interval does ๐‘ก lie?

  • A ๐‘ฅ = ๏€ฟ ๐‘ก โˆ’ 1 โˆš 2 ๏‹ + 1 , ๐‘ฆ = ๏€ฟ ๐‘ก โˆ’ 1 โˆš 2 ๏‹ + 1 , the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • B ๐‘ฅ = 2 ๏€ฟ ๐‘ก + 1 โˆš 2 ๏‹ โˆ’ 1 , ๐‘ฆ = โˆ’ 3 ๏€ฟ ๐‘ก + 1 โˆš 2 ๏‹ โˆ’ 1 , the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • C ๐‘ฅ = 2 ๏€ฟ ๐‘ก + 1 โˆš 2 ๏‹ + 1 , ๐‘ฆ = โˆ’ 3 ๏€ฟ ๐‘ก + 1 โˆš 2 ๏‹ + 1 , the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • D ๐‘ฅ = ๏€ฟ ๐‘ก + 1 โˆš 2 ๏‹ + 1 , ๐‘ฆ = ๏€ฟ ๐‘ก + 1 โˆš 2 ๏‹ + 1 , the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • E ๐‘ฅ = 2 ๏€ฟ ๐‘ก โˆ’ 1 โˆš 2 ๏‹ + 1 ๏Šจ , ๐‘ฆ = โˆ’ 3 ๏€ฟ ๐‘ก โˆ’ 1 โˆš 2 ๏‹ + 2 ๏Šจ , the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2

Q5:

Convert the parametric equations ๐‘ฅ=3๐‘กcos and ๐‘ฆ=3๐‘กsin to rectangular form.

  • A ๐‘ฅ โˆ’ ๐‘ฆ = 9 ๏Šจ ๏Šจ
  • B ๐‘ฆ = 3 ๐‘ฅ s i n
  • C ๐‘ฅ + ๐‘ฆ = 3 ๏Šจ ๏Šจ
  • D ๐‘ฅ + ๐‘ฆ = 9 ๏Šจ ๏Šจ
  • E ๐‘ฅ = 3 ๐‘ฆ c o s

Q6:

Convert the rectangular equation ๐‘ฅ+๐‘ฆ=25๏Šจ๏Šจ to parametric form.

  • A ๐‘ฅ = ๐‘ก , ๐‘ฆ = 2 5 ๐‘ก s i n c o s ๏Šจ ๏Šจ
  • B ๐‘ฅ = 5 ๐‘ก , ๐‘ฆ = 5 ๐‘ก c o s s i n
  • C ๐‘ฅ = 5 ๐‘ก , ๐‘ฆ = 5 ๐‘ก c o s s i n ๏Šจ ๏Šจ
  • D ๐‘ฅ = ๐‘ก , ๐‘ฆ = ๐‘ก s i n c o s
  • E ๐‘ฅ = 2 5 ๐‘ก , ๐‘ฆ = 2 5 ๐‘ก s i n c o s

Q7:

Convert the parametric equations ๐‘ฅ=๐‘ก+2๏Šจ and ๐‘ฆ=3๐‘กโˆ’1 to rectangular form.

  • A ๐‘ฅ = ๐‘ฆ + 2 ๏Šจ
  • B ๐‘ฅ = ๐‘ฆ + 1 3 + 2
  • C ๐‘ฅ = ๏€ผ ๐‘ฆ + 1 3 ๏ˆ + 2 ๏Šจ
  • D ๐‘ฅ = 3 ๐‘ฆ โˆ’ 1
  • E ๐‘ฅ = ๏€ผ ๐‘ฆ + 1 3 ๏ˆ ๏Šจ

Q8:

Convert the rectangular equation ๐‘ฅโˆ’๐‘ฆ=9๏Šจ๏Šจ to parametric form.

  • A ๐‘ฅ = 3 ๐‘ก , ๐‘ฆ = 3 ๐‘ก c o s h s i n h
  • B ๐‘ฅ = 9 ๐‘ก , ๐‘ฆ = 9 ๐‘ก c o s h s i n h
  • C ๐‘ฅ = 3 ๐‘ก , ๐‘ฆ = 3 ๐‘ก c o s s i n
  • D ๐‘ฅ = 9 ๐‘ก , ๐‘ฆ = 9 ๐‘ก s i n c o s
  • E ๐‘ฅ = 3 ๐‘ก , ๐‘ฆ = 3 ๐‘ก s i n h c o s h

Q9:

Consider the points ๐ด=(โˆ’1,1) and ๐ต=(4,2). Parameterize the segment ๐ด๐ต, where 0โ‰ค๐‘กโ‰ค1.

  • A ๐‘ฅ = 1 โˆ’ ๐‘ก , ๐‘ฆ = 5 ๐‘ก โˆ’ 1
  • B ๐‘ฅ = ๐‘ก + 1 , ๐‘ฆ = 5 ๐‘ก โˆ’ 1
  • C ๐‘ฅ = 5 ๐‘ก + 1 , ๐‘ฆ = 1 โˆ’ ๐‘ก
  • D ๐‘ฅ = 5 ๐‘ก + 1 , ๐‘ฆ = 1 + ๐‘ก
  • E ๐‘ฅ = 5 ๐‘ก โˆ’ 1 , ๐‘ฆ = ๐‘ก + 1

Q10:

Convert the parametric equations ๐‘ฅ=2๐‘ก+1 and ๐‘ฆ=๐‘กโˆ’4 to the rectangular form.

  • A ๐‘ฆ = ๐‘ฅ โˆ’ 9 2
  • B ๐‘ฆ = 2 ๐‘ฅ + 1
  • C ๐‘ฆ = ๐‘ฅ + 9 2
  • D ๐‘ฆ = ๐‘ฅ โˆ’ 4
  • E ๐‘ฆ = ๐‘ฅ โˆ’ 1 2

Q11:

Convert the rectangular equation (๐‘ฅ+3)+(๐‘ฆ+5)=9๏Šจ๏Šจ to parametric form.

  • A ๐‘ฅ = 3 ( ๐‘ก ) + 3 ๐‘ฆ = 3 ( ๐‘ก ) + 5 c o s a n d s i n
  • B ๐‘ฅ = 3 ( ๐‘ก ) โˆ’ 3 ๐‘ฆ = 3 ( ๐‘ก ) โˆ’ 5 c o s a n d s i n
  • C ๐‘ฅ = 3 ( ๐‘ก ) ๐‘ฆ = 3 ( ๐‘ก ) c o s a n d s i n
  • D ๐‘ฅ = 9 ( ๐‘ก ) + 3 ๐‘ฆ = 9 ( ๐‘ก ) + 5 c o s a n d s i n
  • E ๐‘ฅ = 9 ( ๐‘ก ) โˆ’ 3 ๐‘ฆ = 9 ( ๐‘ก ) โˆ’ 5 c o s a n d s i n

Q12:

Convert the parametric equations ๐‘ฅ=๏€ผ12๐‘ก๏ˆln and ๐‘ฆ=3๐‘ก๏Šจ to rectangular form.

  • A ๐‘ฆ = 1 2 ๐‘’ ๏Ž  ๏Žก ๏—
  • B ๐‘ฆ = 1 2 ๐‘’ ๏Šจ ๏—
  • C ๐‘ฆ = 3 8 ๐‘’ ๏Šจ ๏—
  • D ๐‘ฆ = 6 ๐‘’ ๏Šจ ๏—
  • E ๐‘ฆ = 3 ๐‘’ ๏Šจ ๏—

Q13:

Convert the parametric equations ๐‘ฅ=โˆš๐‘ก and ๐‘ฆ=5๐‘ก+4๐‘ก๏Šช to rectangular form.

  • A ๐‘ฆ = 5 ๐‘ฅ + 4 โˆš ๐‘ฅ ๏Šจ
  • B ๐‘ฆ = 5 ๐‘ฅ + 5 ๐‘ฅ ๏Šฎ ๏Šจ
  • C ๐‘ฆ = 5 ๐‘ฅ + 4 ๐‘ฅ ๏Šช
  • D ๐‘ฆ = 5 ๐‘ฅ โˆ’ 4 ๐‘ฅ ๏Šฎ ๏Šจ
  • E ๐‘ฆ = 5 ๐‘ฅ + 4 ๐‘ฅ ๏Šฎ ๏Šจ

Q14:

Convert the parametric equations ๐‘ฅ=(๐‘ก)cos and ๐‘ฆ=(๐‘ก)sin to rectangular form.

  • A ๐‘ฆ = ( ๐‘ฅ ) s i n
  • B ๐‘ฅ + ๐‘ฆ = 1
  • C ๐‘ฅ = ( ๐‘ฆ ) c o s
  • D ๐‘ฅ + ๐‘ฆ = 1 ๏Šจ ๏Šจ
  • E ๐‘ฅ โˆ’ ๐‘ฆ = 1 ๏Šจ ๏Šจ

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