Mathbench

Geometry: vectors and parametric curves

Vectors — combining them, the vector between two points and its length — lines given by a vector, parametric curves and their slope, and Lissajous figures. Set for OMPT-D.

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Questions

Sheet length
1

Give the vector AB⃗\vec{AB} from A(−6,1)A(-6, 1) to B(3,6).B(3, 6)\mathclose{.}

Answer:

2

A Lissajous figure is traced by x=sin⁡(3t), y=cos⁡(t).x = \sin(3 t),\ y = \cos(t)\mathclose{.} Give the point of the figure at t=π3.t = \frac{\pi}{3}\mathclose{.}

Answer:

3

Given a⃗=(−1,5)\vec{a} = (-1, 5) and b⃗=(1,−1),\vec{b} = (1, -1)\mathclose{,} give 2a⃗−b⃗.2\vec{a} - \vec{b}\mathclose{.}

Answer:

4

A Lissajous figure is traced by x=sin⁡(2t), y=cos⁡(3t).x = \sin(2 t),\ y = \cos(3 t)\mathclose{.} Give the slope dydx\dfrac{dy}{dx} at t=π6.t = \frac{\pi}{6}\mathclose{.}

Answer:

5

Give the exact length of the vector (−5,−2).(-5, -2)\mathclose{.}

Answer:

6

A line is given by (x,y)=(0,−1)+t(4,1).(x, y) = (0, -1) + t(4, 1)\mathclose{.} Give the point on it with t=2.t = 2\mathclose{.}

Answer:

7

A point moves along the curve x=t2−4t, y=t3+5t.x = t^{2} - 4 t,\ y = t^{3} + 5 t\mathclose{.} Give its speed — the length of the velocity vector — at t=1.t = 1\mathclose{.}

Answer:

8

Write the line (x,y)=(5,−3)+t(3,2)(x, y) = (5, -3) + t(3, 2) in the form y=mx+c.y = mx + c\mathclose{.}

Answer:

9

A Lissajous figure is traced by x=sin⁡(3t), y=cos⁡(2t).x = \sin(3 t),\ y = \cos(2 t)\mathclose{.} Give the point of the figure at t=π4.t = \frac{\pi}{4}\mathclose{.}

Answer:

10

Give the vector AB⃗\vec{AB} from A(3,−5)A(3, -5) to B(0,1).B(0, 1)\mathclose{.}

Answer:

11

Given a⃗=(4,6)\vec{a} = (4, 6) and b⃗=(−1,5),\vec{b} = (-1, 5)\mathclose{,} give 3a⃗−b⃗.3\vec{a} - \vec{b}\mathclose{.}

Answer:

12

A Lissajous figure is traced by x=sin⁡(2t), y=cos⁡(t).x = \sin(2 t),\ y = \cos(t)\mathclose{.} Give the slope dydx\dfrac{dy}{dx} at t=π6.t = \frac{\pi}{6}\mathclose{.}

Answer:

13

Give the exact length of the vector (−1,−6).(-1, -6)\mathclose{.}

Answer:

14

A curve is given by x=t2−3t, y=t3+5t.x = t^{2} - 3 t,\ y = t^{3} + 5 t\mathclose{.} Give the slope dydx\dfrac{dy}{dx} at t=−2.t = -2\mathclose{.}

Answer:

15

A line is given by (x,y)=(−1,−5)+t(−6,−1).(x, y) = (-1, -5) + t(-6, -1)\mathclose{.} Give the point on it with t=3.t = 3\mathclose{.}

Answer:

16

Write the line (x,y)=(3,2)+t(−3,1)(x, y) = (3, 2) + t(-3, 1) in the form y=mx+c.y = mx + c\mathclose{.}

Answer:

17

A Lissajous figure is traced by x=sin⁡(t), y=cos⁡(2t).x = \sin(t),\ y = \cos(2 t)\mathclose{.} Give the point of the figure at t=π3.t = \frac{\pi}{3}\mathclose{.}

Answer:

18

Give the vector AB⃗\vec{AB} from A(−2,0)A(-2, 0) to B(−6,−1).B(-6, -1)\mathclose{.}

Answer:

19

A Lissajous figure is traced by x=sin⁡(3t), y=cos⁡(t).x = \sin(3 t),\ y = \cos(t)\mathclose{.} Give the slope dydx\dfrac{dy}{dx} at t=π3.t = \frac{\pi}{3}\mathclose{.}

Answer:

20

Given a⃗=(5,−1)\vec{a} = (5, -1) and b⃗=(1,1),\vec{b} = (1, 1)\mathclose{,} give 3a⃗−b⃗.3\vec{a} - \vec{b}\mathclose{.}

Answer:

21

Give the exact length of the vector (−5,−12).(-5, -12)\mathclose{.}

Answer:

22

A curve is given by x=t2−t, y=t3−3t.x = t^{2} - t,\ y = t^{3} - 3 t\mathclose{.} Give the slope dydx\dfrac{dy}{dx} at t=2.t = 2\mathclose{.}

Answer:

23

A line is given by (x,y)=(1,4)+t(3,−5).(x, y) = (1, 4) + t(3, -5)\mathclose{.} Give the point on it with t=−1.t = -1\mathclose{.}

Answer:

24

A Lissajous figure is traced by x=sin⁡(t), y=cos⁡(3t).x = \sin(t),\ y = \cos(3 t)\mathclose{.} Give the point of the figure at t=π6.t = \frac{\pi}{6}\mathclose{.}

Answer:

25

Write the line (x,y)=(5,−1)+t(3,−6)(x, y) = (5, -1) + t(3, -6) in the form y=mx+c.y = mx + c\mathclose{.}

Answer:

26

Give the vector AB⃗\vec{AB} from A(−6,5)A(-6, 5) to B(2,−5).B(2, -5)\mathclose{.}

Answer:

27

A Lissajous figure is traced by x=sin⁡(3t), y=cos⁡(t).x = \sin(3 t),\ y = \cos(t)\mathclose{.} Give the slope dydx\dfrac{dy}{dx} at t=π4.t = \frac{\pi}{4}\mathclose{.}

Answer:

28

Given a⃗=(3,−2)\vec{a} = (3, -2) and b⃗=(−1,−1),\vec{b} = (-1, -1)\mathclose{,} give 2a⃗−b⃗.2\vec{a} - \vec{b}\mathclose{.}

Answer:

29

A point moves along the curve x=t2−3t, y=t3−t.x = t^{2} - 3 t,\ y = t^{3} - t\mathclose{.} Give its speed — the length of the velocity vector — at t=3.t = 3\mathclose{.}

Answer:

30

Give the exact length of the vector (−1,−2).(-1, -2)\mathclose{.}

Answer:

Answers

Every answer below was re-derived independently before this page was built.

1

Answer: (9,5)(9, 5)

End minus start: AB⃗=B−A=(3−(−6),6−1)=(9,5).\vec{AB} = B - A = (3 - (-6), 6 - 1) = (9, 5)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Subtracted the wrong way round, A−BA - B → (−9,−5)(-9, -5)
  • Added the coordinates → (−3,7)(-3, 7)
2

Answer: (0,12)(0, \frac{1}{2})

Put t=π3t = \frac{\pi}{3} into both: x=sin⁡(π)=0x = \sin\left(\pi\right) = 0 and y=cos⁡(π3)=12.y = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Swapped xx and yy → (12,0)(\frac{1}{2}, 0)
  • Left out the factors inside: used sin⁡(t)\sin(t) and cos⁡(t)\cos(t) → (32,12)(\frac{\sqrt{3}}{2}, \frac{1}{2})
3

Answer: (−3,11)(-3, 11)

Component by component: 2⋅(−1,5)−(1,−1)=(−2−1,10−(−1))=(−3,11).2 \cdot (-1, 5) - (1, -1) = (-2 - 1, 10 - (-1)) = (-3, 11)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied only the first component by 22 → (−3,6)(-3, 6)
  • Added b⃗\vec{b} instead of subtracting it → (−1,9)(-1, 9)
4

Answer: −3-3

dydx=dy/dtdx/dt=−3sin⁡(3t)2cos⁡(2t),\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-3\sin(3 t)}{2\cos(2 t)}\mathclose{,} which at t=π6t = \frac{\pi}{6} is −3.-3\mathclose{.}

Common mistakes, and the answer each one gives:

  • Divided dx/dtdx/dt by dy/dt,dy/dt\mathclose{,} upside down → −13- \frac{1}{3}
  • Left out the chain-rule factors from inside → −2-2
5

Answer: 29\sqrt{29}

By Pythagoras: (−5)2+(−2)2=29.\sqrt{(-5)^2 + (-2)^2} = \sqrt{29}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the components → −7-7
  • Forgot the square root → 2929
6

Answer: (8,1)(8, 1)

Start at (0,−1)(0, -1) and add the direction (4,1)(4, 1) 22 times: (0+2⋅4,−1+2⋅1)=(8,1).(0 + 2 \cdot 4, -1 + 2 \cdot 1) = (8, 1)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the direction vector once instead of 22 times → (4,0)(4, 0)
  • Multiplied the starting point by tt as well → (8,0)(8, 0)
7

Answer: 2172 \sqrt{17}

The velocity vector is (dxdt,dydt)=(2t−4,3t2+5),\left(\dfrac{dx}{dt}, \dfrac{dy}{dt}\right) = (2 t - 4, 3 t^{2} + 5)\mathclose{,} which at t=1t = 1 is (−2,8).(-2, 8)\mathclose{.} Its length is (−2)2+82=217.\sqrt{(-2)^2 + 8^2} = 2 \sqrt{17}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the two components' sizes → 1010
  • Forgot the square root → 6868
8

Answer: y=2x3−193y = \frac{2 x}{3} - \frac{19}{3}

The direction (3,2)(3, 2) goes 22 up for every 33 across, so the slope is m=23.m = \dfrac{2}{3}\mathclose{.} The line passes through (5,−3):(5, -3)\mathclose{:} y=2x3−193.y = \frac{2 x}{3} - \frac{19}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Turned the slope upside down, acrossup\frac{\text{across}}{\text{up}} → y=3x2−212y = \frac{3 x}{2} - \frac{21}{2}
  • Used the direction vector as a point on the line → y=2x3y = \frac{2 x}{3}
9

Answer: (22,0)(\frac{\sqrt{2}}{2}, 0)

Put t=π4t = \frac{\pi}{4} into both: x=sin⁡(3π4)=22x = \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} and y=cos⁡(π2)=0.y = \cos\left(\frac{\pi}{2}\right) = 0\mathclose{.}

Common mistakes, and the answer each one gives:

  • Swapped xx and yy → (0,22)(0, \frac{\sqrt{2}}{2})
  • Left out the factors inside: used sin⁡(t)\sin(t) and cos⁡(t)\cos(t) → (22,22)(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})
10

Answer: (−3,6)(-3, 6)

End minus start: AB⃗=B−A=(0−3,1−(−5))=(−3,6).\vec{AB} = B - A = (0 - 3, 1 - (-5)) = (-3, 6)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Subtracted the wrong way round, A−BA - B → (3,−6)(3, -6)
  • Added the coordinates → (3,−4)(3, -4)
11

Answer: (13,13)(13, 13)

Component by component: 3⋅(4,6)−(−1,5)=(12−(−1),18−5)=(13,13).3 \cdot (4, 6) - (-1, 5) = (12 - (-1), 18 - 5) = (13, 13)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied only the first component by 33 → (13,1)(13, 1)
  • Added b⃗\vec{b} instead of subtracting it → (11,23)(11, 23)
12

Answer: −12- \frac{1}{2}

dydx=dy/dtdx/dt=−sin⁡(t)2cos⁡(2t),\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-\sin(t)}{2\cos(2 t)}\mathclose{,} which at t=π6t = \frac{\pi}{6} is −12.- \frac{1}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Divided dx/dtdx/dt by dy/dt,dy/dt\mathclose{,} upside down → −2-2
  • Left out the chain-rule factors from inside → −1-1
13

Answer: 37\sqrt{37}

By Pythagoras: (−1)2+(−6)2=37.\sqrt{(-1)^2 + (-6)^2} = \sqrt{37}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the components → −7-7
  • Forgot the square root → 3737
14

Answer: −177-\dfrac{17}{7}

dydx=dy/dtdx/dt=3t2+52t−3.\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3 t^{2} + 5}{2 t - 3}\mathclose{.} At t=−2t = -2 that is 17−7=−177.\dfrac{17}{-7} = -\dfrac{17}{7}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Divided dx/dtdx/dt by dy/dt,dy/dt\mathclose{,} upside down → −717- \frac{7}{17}
  • Gave dy/dtdy/dt alone → 1717
15

Answer: (−19,−8)(-19, -8)

Start at (−1,−5)(-1, -5) and add the direction (−6,−1)(-6, -1) 33 times: (−1+3⋅(−6),−5+3⋅(−1))=(−19,−8).(-1 + 3 \cdot (-6), -5 + 3 \cdot (-1)) = (-19, -8)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the direction vector once instead of 33 times → (−7,−6)(-7, -6)
  • Multiplied the starting point by tt as well → (−21,−18)(-21, -18)
16

Answer: y=−x3+3y = - \frac{x}{3} + 3

The direction (−3,1)(-3, 1) goes 11 up for every −3-3 across, so the slope is m=−13.m = -\dfrac{1}{3}\mathclose{.} The line passes through (3,2):(3, 2)\mathclose{:} y=−x3+3.y = - \frac{x}{3} + 3\mathclose{.}

Common mistakes, and the answer each one gives:

  • Turned the slope upside down, acrossup\frac{\text{across}}{\text{up}} → y=−3x+11y = - 3 x + 11
  • Used the direction vector as a point on the line → y=−x3y = - \frac{x}{3}
17

Answer: (32,−12)(\frac{\sqrt{3}}{2}, - \frac{1}{2})

Put t=π3t = \frac{\pi}{3} into both: x=sin⁡(π3)=32x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} and y=cos⁡(2π3)=−12.y = \cos\left(\frac{2\pi}{3}\right) = - \frac{1}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Swapped xx and yy → (−12,32)(- \frac{1}{2}, \frac{\sqrt{3}}{2})
  • Left out the factors inside: used sin⁡(t)\sin(t) and cos⁡(t)\cos(t) → (32,12)(\frac{\sqrt{3}}{2}, \frac{1}{2})
18

Answer: (−4,−1)(-4, -1)

End minus start: AB⃗=B−A=(−6−(−2),−1−0)=(−4,−1).\vec{AB} = B - A = (-6 - (-2), -1 - 0) = (-4, -1)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Subtracted the wrong way round, A−BA - B → (4,1)(4, 1)
  • Added the coordinates → (−8,−1)(-8, -1)
19

Answer: 36\frac{\sqrt{3}}{6}

dydx=dy/dtdx/dt=−sin⁡(t)3cos⁡(3t),\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-\sin(t)}{3\cos(3 t)}\mathclose{,} which at t=π3t = \frac{\pi}{3} is −32−3=36.\dfrac{- \frac{\sqrt{3}}{2}}{-3} = \frac{\sqrt{3}}{6}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Divided dx/dtdx/dt by dy/dt,dy/dt\mathclose{,} upside down → 232 \sqrt{3}
  • Left out the chain-rule factors from inside → 32\frac{\sqrt{3}}{2}
20

Answer: (14,−4)(14, -4)

Component by component: 3⋅(5,−1)−(1,1)=(15−1,−3−1)=(14,−4).3 \cdot (5, -1) - (1, 1) = (15 - 1, -3 - 1) = (14, -4)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied only the first component by 33 → (14,−2)(14, -2)
  • Added b⃗\vec{b} instead of subtracting it → (16,−2)(16, -2)
21

Answer: 1313

By Pythagoras: (−5)2+(−12)2=169=13.\sqrt{(-5)^2 + (-12)^2} = \sqrt{169} = 13\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the components → −17-17
  • Forgot the square root → 169169
22

Answer: 33

dydx=dy/dtdx/dt=3t2−32t−1.\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3 t^{2} - 3}{2 t - 1}\mathclose{.} At t=2t = 2 that is 93=3.\dfrac{9}{3} = 3\mathclose{.}

Common mistakes, and the answer each one gives:

  • Divided dx/dtdx/dt by dy/dt,dy/dt\mathclose{,} upside down → 13\frac{1}{3}
  • Gave dy/dtdy/dt alone → 99
23

Answer: (−2,9)(-2, 9)

Start at (1,4)(1, 4) and add the direction (3,−5)(3, -5) −1-1 times: (1+(−1)⋅3,4+(−1)⋅(−5))=(−2,9).(1 + (-1) \cdot 3, 4 + (-1) \cdot (-5)) = (-2, 9)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the direction vector once instead of −1-1 times → (4,−1)(4, -1)
  • Multiplied the starting point by tt as well → (−4,1)(-4, 1)
24

Answer: (12,0)(\frac{1}{2}, 0)

Put t=π6t = \frac{\pi}{6} into both: x=sin⁡(π6)=12x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} and y=cos⁡(π2)=0.y = \cos\left(\frac{\pi}{2}\right) = 0\mathclose{.}

Common mistakes, and the answer each one gives:

  • Swapped xx and yy → (0,12)(0, \frac{1}{2})
  • Left out the factors inside: used sin⁡(t)\sin(t) and cos⁡(t)\cos(t) → (12,32)(\frac{1}{2}, \frac{\sqrt{3}}{2})
25

Answer: y=−2x+9y = - 2 x + 9

The direction (3,−6)(3, -6) goes −6-6 up for every 33 across, so the slope is m=−2.m = -2\mathclose{.} The line passes through (5,−1):(5, -1)\mathclose{:} y=−2x+9.y = - 2 x + 9\mathclose{.}

Common mistakes, and the answer each one gives:

  • Turned the slope upside down, acrossup\frac{\text{across}}{\text{up}} → y=−x2+32y = - \frac{x}{2} + \frac{3}{2}
  • Used the direction vector as a point on the line → y=−2xy = - 2 x
26

Answer: (8,−10)(8, -10)

End minus start: AB⃗=B−A=(2−(−6),−5−5)=(8,−10).\vec{AB} = B - A = (2 - (-6), -5 - 5) = (8, -10)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Subtracted the wrong way round, A−BA - B → (−8,10)(-8, 10)
  • Added the coordinates → (−4,0)(-4, 0)
27

Answer: 13\frac{1}{3}

dydx=dy/dtdx/dt=−sin⁡(t)3cos⁡(3t),\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-\sin(t)}{3\cos(3 t)}\mathclose{,} which at t=π4t = \frac{\pi}{4} is −22−322=13.\dfrac{- \frac{\sqrt{2}}{2}}{- \frac{3 \sqrt{2}}{2}} = \frac{1}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Divided dx/dtdx/dt by dy/dt,dy/dt\mathclose{,} upside down → 33
  • Left out the chain-rule factors from inside → 11
28

Answer: (7,−3)(7, -3)

Component by component: 2⋅(3,−2)−(−1,−1)=(6−(−1),−4−(−1))=(7,−3).2 \cdot (3, -2) - (-1, -1) = (6 - (-1), -4 - (-1)) = (7, -3)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied only the first component by 22 → (7,−1)(7, -1)
  • Added b⃗\vec{b} instead of subtracting it → (5,−5)(5, -5)
29

Answer: 685\sqrt{685}

The velocity vector is (dxdt,dydt)=(2t−3,3t2−1),\left(\dfrac{dx}{dt}, \dfrac{dy}{dt}\right) = (2 t - 3, 3 t^{2} - 1)\mathclose{,} which at t=3t = 3 is (3,26).(3, 26)\mathclose{.} Its length is 32+262=685.\sqrt{3^2 + 26^2} = \sqrt{685}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the two components' sizes → 2929
  • Forgot the square root → 685685
30

Answer: 5\sqrt{5}

By Pythagoras: (−1)2+(−2)2=5.\sqrt{(-1)^2 + (-2)^2} = \sqrt{5}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the components → −3-3
  • Forgot the square root → 55