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.


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).

  • A๐‘ฅ=5๐‘ก,๐‘ฆ=9๐‘กcoshsinh
  • B๐‘ฅ=โˆ’5๐‘ก,๐‘ฆ=โˆ’9๐‘กcoshsinh
  • C๐‘ฅ=25๐‘ก,๐‘ฆ=81๐‘กcoshsinh
  • D๐‘ฅ=โˆ’5๐‘ก,๐‘ฆ=9๐‘กcoshsinh
  • E๐‘ฅ=5๐‘ก,๐‘ฆ=โˆ’9๐‘กcoshsinh


The first figure shows the graphs of cos2๐œ‹๐‘ก and sin2๐œ‹๐‘ก that 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)


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.

  • A๐‘ฅ=4(๐‘ก+1),๐‘ฆ=3(๐‘ก+1)
  • B๐‘ฅ=๐‘ก+4,๐‘ฆ=๐‘ก+3
  • C๐‘ฅ=4๐‘ก+1,๐‘ฆ=3๐‘ก+1
  • D๐‘ฅ=3๐‘ก+1,๐‘ฆ=4๐‘ก+1
  • E๐‘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.

  • A๐‘“(๐‘ก)=4๐‘ก+1,๐‘”(๐‘ก)=3๐‘ก+1
  • B๐‘“(๐‘ก)=๐‘ก+4,๐‘”(๐‘ก)=๐‘ก+3
  • C๐‘“(๐‘ก)=3๐‘ก+1,๐‘”(๐‘ก)=4๐‘ก+1
  • D๐‘“(๐‘ก)=๐‘ก+45,๐‘”(๐‘ก)=๐‘ก+35
  • E๐‘“(๐‘ก)=45๐‘ก+1,๐‘”(๐‘ก)=35๐‘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‘ ))?

  • A๐‘ โˆ’1
  • B๐‘ +1
  • C๐‘ 
  • D๐‘ +2
  • E๐‘ 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.

  • A๐‘ฅ=1213๐‘ก+1,๐‘ฆ=513๐‘ก+1 on 5โ‰ค๐‘กโ‰ค13
  • B๐‘ฅ=1213๐‘ก+1,๐‘ฆ=513๐‘ก+1 on 0โ‰ค๐‘กโ‰ค13
  • C๐‘ฅ=513๐‘ก+1,๐‘ฆ=1213๐‘ก+1 on 0โ‰ค๐‘กโ‰ค13
  • D๐‘ฅ=๐‘ก+1213,๐‘ฆ=513๐‘ก on 0โ‰ค๐‘กโ‰ค13
  • E๐‘ฅ=513๐‘ก+1,๐‘ฆ=1213๐‘ก+1 on 5โ‰ค๐‘กโ‰ค13


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‘.

  • A2๐‘ฅ+3๐‘ฆ=7
  • B3๐‘ฅ+2๐‘ฆ=7
  • C3๐‘ฅ+2๐‘ฆ=13
  • D3๐‘ฅโˆ’2๐‘ฆ=7
  • E3๐‘ฅโˆ’2๐‘ฆ=โˆ’1

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

  • A๐‘ฅ=1, at ๐‘ก=โˆ’1โˆš2
  • B๐‘ฅ=1, at ๐‘ก=0
  • C๐‘ฅ=0.5, at ๐‘ก=โˆ’1โˆš2
  • D๐‘ฅ=0.5, at ๐‘ก=0
  • E๐‘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?

  • A๐‘ฅ=๏€ฟ๐‘กโˆ’1โˆš2๏‹+1, ๐‘ฆ=๏€ฟ๐‘กโˆ’1โˆš2๏‹+1, the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • B๐‘ฅ=2๏€ฟ๐‘ก+1โˆš2๏‹โˆ’1, ๐‘ฆ=โˆ’3๏€ฟ๐‘ก+1โˆš2๏‹โˆ’1, the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • C๐‘ฅ=2๏€ฟ๐‘ก+1โˆš2๏‹+1, ๐‘ฆ=โˆ’3๏€ฟ๐‘ก+1โˆš2๏‹+1, the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • D๐‘ฅ=๏€ฟ๐‘ก+1โˆš2๏‹+1, ๐‘ฆ=๏€ฟ๐‘ก+1โˆš2๏‹+1, the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2
  • E๐‘ฅ=2๏€ฟ๐‘กโˆ’1โˆš2๏‹+1๏Šจ, ๐‘ฆ=โˆ’3๏€ฟ๐‘กโˆ’1โˆš2๏‹+2๏Šจ, the interval 0โ‰ค๐‘กโ‰คโˆš2+1โˆš2


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

  • A๐‘ฅโˆ’๐‘ฆ=9๏Šจ๏Šจ
  • B๐‘ฆ=3๐‘ฅsin
  • C๐‘ฅ+๐‘ฆ=3๏Šจ๏Šจ
  • D๐‘ฅ+๐‘ฆ=9๏Šจ๏Šจ
  • E๐‘ฅ=3๐‘ฆcos


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

  • A๐‘ฅ=๐‘ก,๐‘ฆ=25๐‘กsincos๏Šจ๏Šจ
  • B๐‘ฅ=5๐‘ก,๐‘ฆ=5๐‘กcossin
  • C๐‘ฅ=5๐‘ก,๐‘ฆ=5๐‘กcossin๏Šจ๏Šจ
  • D๐‘ฅ=๐‘ก,๐‘ฆ=๐‘กsincos
  • E๐‘ฅ=25๐‘ก,๐‘ฆ=25๐‘กsincos


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

  • A๐‘ฅ=๐‘ฆ+2๏Šจ
  • B๐‘ฅ=๐‘ฆ+13+2
  • C๐‘ฅ=๏€ผ๐‘ฆ+13๏ˆ+2๏Šจ
  • D๐‘ฅ=3๐‘ฆโˆ’1
  • E๐‘ฅ=๏€ผ๐‘ฆ+13๏ˆ๏Šจ


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

  • A๐‘ฅ=3๐‘ก,๐‘ฆ=3๐‘กcoshsinh
  • B๐‘ฅ=9๐‘ก,๐‘ฆ=9๐‘กcoshsinh
  • C๐‘ฅ=3๐‘ก,๐‘ฆ=3๐‘กcossin
  • D๐‘ฅ=9๐‘ก,๐‘ฆ=9๐‘กsincos
  • E๐‘ฅ=3๐‘ก,๐‘ฆ=3๐‘กsinhcosh


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

  • A๐‘ฅ=1โˆ’๐‘ก,๐‘ฆ=5๐‘กโˆ’1
  • B๐‘ฅ=๐‘ก+1,๐‘ฆ=5๐‘กโˆ’1
  • C๐‘ฅ=5๐‘ก+1,๐‘ฆ=1โˆ’๐‘ก
  • D๐‘ฅ=5๐‘ก+1,๐‘ฆ=1+๐‘ก
  • E๐‘ฅ=5๐‘กโˆ’1,๐‘ฆ=๐‘ก+1


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

  • A๐‘ฆ=๐‘ฅโˆ’92
  • B๐‘ฆ=2๐‘ฅ+1
  • C๐‘ฆ=๐‘ฅ+92
  • D๐‘ฆ=๐‘ฅโˆ’4
  • E๐‘ฆ=๐‘ฅโˆ’12


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

  • A๐‘ฅ=3(๐‘ก)+3๐‘ฆ=3(๐‘ก)+5cosandsin
  • B๐‘ฅ=3(๐‘ก)โˆ’3๐‘ฆ=3(๐‘ก)โˆ’5cosandsin
  • C๐‘ฅ=3(๐‘ก)๐‘ฆ=3(๐‘ก)cosandsin
  • D๐‘ฅ=9(๐‘ก)+3๐‘ฆ=9(๐‘ก)+5cosandsin
  • E๐‘ฅ=9(๐‘ก)โˆ’3๐‘ฆ=9(๐‘ก)โˆ’5cosandsin


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

  • A๐‘ฆ=12๐‘’๏Ž ๏Žก๏—
  • B๐‘ฆ=12๐‘’๏Šจ๏—
  • C๐‘ฆ=38๐‘’๏Šจ๏—
  • D๐‘ฆ=6๐‘’๏Šจ๏—
  • E๐‘ฆ=3๐‘’๏Šจ๏—


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

  • A๐‘ฆ=5๐‘ฅ+4โˆš๐‘ฅ๏Šจ
  • B๐‘ฆ=5๐‘ฅ+5๐‘ฅ๏Šฎ๏Šจ
  • C๐‘ฆ=5๐‘ฅ+4๐‘ฅ๏Šช
  • D๐‘ฆ=5๐‘ฅโˆ’4๐‘ฅ๏Šฎ๏Šจ
  • E๐‘ฆ=5๐‘ฅ+4๐‘ฅ๏Šฎ๏Šจ


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

  • A๐‘ฆ=(๐‘ฅ)sin
  • B๐‘ฅ+๐‘ฆ=1
  • C๐‘ฅ=(๐‘ฆ)cos
  • D๐‘ฅ+๐‘ฆ=1๏Šจ๏Šจ
  • E๐‘ฅโˆ’๐‘ฆ=1๏Šจ๏Šจ


Convert the rectangular equation (๐‘ฅ+2)โˆ’(๐‘ฆ+5)=16๏Šจ๏Šจ to parametric form.

  • A๐‘ฅ=4(๐‘ก)โˆ’2๐‘ฆ=4(๐‘ก)โˆ’5coshandsinh
  • B๐‘ฅ=4(๐‘ก)๐‘ฆ=4(๐‘ก)coshandsinh
  • C๐‘ฅ=16(๐‘ก)โˆ’2๐‘ฆ=16(๐‘ก)โˆ’5coshandsinh
  • D๐‘ฅ=4(๐‘ก)โˆ’2๐‘ฆ=4(๐‘ก)โˆ’5cosandsin
  • E๐‘ฅ=4(๐‘ก)+2๐‘ฆ=4(๐‘ก)+5cosandsin

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