CLASSICAL-MECHANICS · Interactive Practice | Unit 2 · Video 2

Why Unit Vectors Don't Always Point the Same Way

IKey Formulas

Formula Name Description
x=rcos⁡θ,y=rsin⁡θx = r\cos\theta,\quad y = r\sin\theta Polar → Cartesian Build x,yx,y from radius and angle
r=x2+y2,θ=tan⁡−1(y/x)r = \sqrt{x^2 + y^2},\quad \theta = \tan^{-1}(y/x) Cartesian → Polar Recover radius and angle
r^=cos⁡θ ı^+sin⁡θ ȷ^\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath} Radial unit vector Points away from the zz-axis
θ^=−sin⁡θ ı^+cos⁡θ ȷ^\hat{\theta} = -\sin\theta\,\hat{\imath} + \cos\theta\,\hat{\jmath} Tangential unit vector Points counterclockwise, tangent to the circle

IIInteractive Visualizations

Visualization 1 — Polar coordinates to a point

Where does the pair (r,θ)(r,\theta) land in the xyxy-plane, and what right triangle links them?

Visualization 2 — Unit vectors rotate with position

Do r^\hat{r} and θ^\hat{\theta} point the same way at every position, the way ı^,ȷ^\hat{\imath},\hat{\jmath} do?

💡 The Cartesian unit vectors ı^,ȷ^,k^\hat{\imath},\hat{\jmath},\hat{k} point identically at every point in space; only the polar frame turns with position.

Visualization 3 — Building r^\hat{r} and θ^\hat{\theta} from ı^,ȷ^\hat{\imath},\hat{\jmath}

The rotating polar frame is just the fixed Cartesian axes resolved into cos⁡θ\cos\theta and sin⁡θ\sin\theta components.

IIIQuiz Questions

Question 1

A point has polar coordinates r=2r = 2, θ=60°\theta = 60°. What are its Cartesian coordinates (x,y)(x, y)?

(Recall: cos⁡60°=12\cos 60° = \tfrac{1}{2}, sin⁡60°=32\sin 60° = \tfrac{\sqrt{3}}{2}.)

✅ Correct! x=2cos⁡60°=1x = 2\cos 60° = 1 and y=2sin⁡60°=3y = 2\sin 60° = \sqrt{3}.

❌ Not quite. You swapped sine and cosine — xx uses cosine, yy uses sine.

❌ Not quite. Multiply rr by the cosine (for xx) and sine (for yy) of the angle.

Show solution

Solution:

Use x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta:

x=2cos⁡60°=2⋅12=1x = 2\cos 60° = 2 \cdot \tfrac{1}{2} = 1 y=2sin⁡60°=2⋅32=3y = 2\sin 60° = 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}

So (x,y)=(1,3)(x, y) = (1, \sqrt{3}).

Question 2

True or False: In cylindrical coordinates, the radial unit vector at point P is always equal to the radial unit vector at point S, just like in Cartesian coordinates where ı^P=ı^S\hat{\imath}_P = \hat{\imath}_S everywhere.

✅ Correct! Polar unit vectors rotate with position, so generally r^P≠r^S\hat{r}_P \ne \hat{r}_S.

❌ Not quite. Re-watch Visualization 2 — r^\hat{r} points a different way at different angles.

Show solution

Solution:

This is False. The crucial difference between Cartesian and cylindrical coordinates is that the polar unit vectors depend on position.

  • r^\hat{r} points radially outward from the zz-axis — that direction is different at different angular positions.
  • θ^\hat{\theta} points counterclockwise, tangent to the circle — also different at different positions.

So in general r^P≠r^S\hat{r}_P \ne \hat{r}_S and θ^P≠θ^S\hat{\theta}_P \ne \hat{\theta}_S. Only k^\hat{k} stays the same everywhere (it matches Cartesian).

Question 3

At an angular position θ=90°\theta = 90°, what is the radial unit vector r^\hat{r} expressed in Cartesian components?

(Recall: r^=cos⁡θ ı^+sin⁡θ ȷ^\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath}, with cos⁡90°=0\cos 90° = 0, sin⁡90°=1\sin 90° = 1.)

✅ Correct! At θ=90°\theta=90°, r^=0 ı^+1 ȷ^=ȷ^\hat{r} = 0\,\hat{\imath} + 1\,\hat{\jmath} = \hat{\jmath}.

❌ Not quite. That would be θ=0°\theta = 0°. At 90°90°, cosine is 0 and sine is 1.

❌ Not quite. Plug cos⁡90°=0\cos 90° = 0 and sin⁡90°=1\sin 90° = 1 into r^=cos⁡θ ı^+sin⁡θ ȷ^\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath}.

Show solution

Solution:

r^=cos⁡θ ı^+sin⁡θ ȷ^\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath}

At θ=90°\theta = 90°:

r^=cos⁡90° ı^+sin⁡90° ȷ^=0⋅ı^+1⋅ȷ^=ȷ^\hat{r} = \cos 90°\,\hat{\imath} + \sin 90°\,\hat{\jmath} = 0\cdot\hat{\imath} + 1\cdot\hat{\jmath} = \hat{\jmath}

This makes geometric sense: at 90°90° the point is straight up on the yy-axis, so "radially outward" means pointing in the +ȷ^+\hat{\jmath} direction.

Question 4

Why are r^\hat{r} and θ^\hat{\theta} guaranteed to be a valid orthonormal pair (each of length 1 and perpendicular to each other) at every angle θ\theta?

Given r^=cos⁡θ ı^+sin⁡θ ȷ^\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath} and θ^=−sin⁡θ ı^+cos⁡θ ȷ^\hat{\theta} = -\sin\theta\,\hat{\imath} + \cos\theta\,\hat{\jmath}.

✅ Correct! The Pythagorean identity gives unit length, and the dot product is 0 for all θ.

❌ Not quite. Orthonormality comes from cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 and r^⋅θ^=0\hat{r}\cdot\hat{\theta} = 0.

Show solution

Solution:

Unit length: ∣r^∣2=cos⁡2θ+sin⁡2θ=1,∣θ^∣2=(−sin⁡θ)2+cos⁡2θ=1|\hat{r}|^2 = \cos^2\theta + \sin^2\theta = 1, \qquad |\hat{\theta}|^2 = (-\sin\theta)^2 + \cos^2\theta = 1

Perpendicular (dot product vanishes): r^⋅θ^=(cos⁡θ)(−sin⁡θ)+(sin⁡θ)(cos⁡θ)=0\hat{r}\cdot\hat{\theta} = (\cos\theta)(-\sin\theta) + (\sin\theta)(\cos\theta) = 0

Because the Pythagorean identity holds for every θ\theta, the pair stays orthonormal no matter where the point is — even though both vectors rotate with position.

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