\vec{r}'(t) = \langle -3\sin t, 3\cos t \rangle, \quad t \in [0, 2\pi]

["# Understanding the Vector Derivative: \vec{r}'(t) = \langle -3\sin t, 3\cos t \rangle \ ext{ for } t \in [0, 2\pi]", "When exploring motion and velocity in calculus, one of the most essential concepts is the derivative of a position vector — commonly written as \vec{r}'(t). This vector derivative represents the instantaneous velocity of an object moving along a path parametrized by time ( t ). In this article, we analyze the specific vector function:", "[\n\vec{r}'(t) = \langle -3\sin t, 3\cos t \rangle, \quad t \in [0, 2\pi]\n]", "We’ll break down its geometric meaning, interpret its physical significance, and discuss how it traces the motion of a particle over one full cycle.", "---", "## What Is \vec{r}'(t) and Why Does It Matter?", "For a position vector (\vec{r}(t) = \langle x(t), y(t) \rangle), the derivative:\n[\n\vec{r}'(t) = \langle x'(t), y'(t) \rangle\n]\nrepresents the velocity vector at time ( t ). Its magnitude gives speed, and its direction indicates the instantaneous direction of motion.", "Given:\n[\nx(t) = -3\sin t, \quad y(t) = 3\cos t\n]\nThen:\n[\nx'(t) = \frac{d}{dt}(-3\sin t) = -3\cos t\n]\n[\ny'(t) = \frac{d}{dt}(3\cos t) = -3\sin t\n]\nSo:\n[\n\vec{r}'(t) = \langle -3\cos t, -3\sin t \rangle\n]", "---", "## Geometric Interpretation: The Path and Velocity Direction", "We recognize (x(t) = -3\sin t), (y(t) = 3\cos t) as parametric equations describing a circle with radius 3, centered at the origin.", "To see this more clearly, observe:\n[\nx^2 + y^2 = (-3\sin t)^2 + (3\cos t)^2 = 9\sin^2 t + 9\cos^2 t = 9(\sin^2 t + \cos^2 t) = 9\n]\nThus, the particle moves along the circle (x^2 + y^2 = 9), counterclockwise (assuming standard parametrization), though note the negative signs affect direction — stay tuned!", "Now consider the velocity vector:\n[\n\vec{r}'(t) = \langle -3\cos t, -3\sin t \rangle\n]", "This vector points toward the center (inward radial direction), scaled by 3. That means the particle’s velocity is always directed toward the origin — a key insight.", "---", "## Analyzing Speed and Direction", "The speed (magnitude of velocity) is:\n[\n|\vec{r}'(t)| = \sqrt{(-3\cos t)^2 + (-3\sin t)^2} = \sqrt{9\cos^2 t + 9\sin^2 t} = \sqrt{9(\sin^2 t + \cos^2 t)} = \sqrt{9} = 3\n]", "Speed is constant and equal to 3 units per time. This makes sense — motion along a circle of radius 3 at constant angular speed yields uniform speed.", "The direction of (\vec{r}'(t)) reveals the instantaneous tangent to the circle:\n- At (t = 0): ( \vec{r}'(0) = \langle -3, 0 \rangle ) → points left (west)\n- At (t = \frac{\pi}{2}): ( \vec{r}'(\pi/2) = \langle 0, -3 \rangle ) → points down (south)\n- At (t = \pi): ( \vec{r}'(\pi) = \langle 3, 0 \rangle ) → points right (east)\n- At (t = \frac{3\pi}{2}): ( \vec{r}'(3\pi/2) = \langle 0, 3 \rangle ) → points up (north)", "Thus, the velocity vector rotates with the particle — continuously turning, maintaining constant speed, and always pointing toward the center.", "---", "## Physical Interpretation: Circular Motion", "This parametric velocity describes uniform circular motion around the origin:", "- Position moves along a circle of radius 3.\n- Velocity is tangential but inversely related to radius — in circular kinematics, speed ( v = r\omega ). Here ( r = 3 ), and from integration, ( \omega = 3 ) radians per unit time (since angular speed is rate of change of angle and ( \ heta(t) ) satisfies ( x = 3\sin\ heta ), ( y = 3\cos\ heta ), yielding ( \ heta = t + \frac{\pi}{2} ), so ( \omega = 1 ) rad/unit time? Wait — correction: derivative of ( \ heta = t + \pi/2 \Rightarrow \omega = 1 )). But our speed is 3 — so radius would need to be 1 for ( v = r\omega ). But radius is 3 and speed is 3 ⇒ ( \omega = 1 ) rad/unit time. Still consistent.", "Wait: wait — let’s double-check angle. If ( x = -3\sin t = 3\cos(t + \pi/2) ), and ( y = 3\cos t = 3\sin(t + \pi/2 - \pi/2) ), but actually:\n[\nx = -3\sin t = 3\cos\left(t + \frac{\pi}{2}\right), \quad y = 3\cos t = 3\sin\left(t + \frac{\pi}{2}\right) \ ext{? No — } \sin(t + \pi/2) = \cos t, \cos(t + \pi/2) = -\sin t, so\n]\n[\nx = -3\sin t = 3 \cos\left(t + \frac{\pi}{2}\right), \quad y = 3\cos t = 3 \sin\left(t + \frac{\pi}{2}\right)\n]\nThus, ( (x(t), y(t)) = 3\left( \cos\left(t + \frac{\pi}{2}\right), \sin\left(t + \frac{\pi}{2}\right) \right) ), so the particle moves counterclockwise around the circle with radius 3, completing one full revolution as ( t ) goes from 0 to ( 2\pi ).", "But velocity is:\n[\n\vec{r}'(t) = \langle -3\cos t, -3\sin t \rangle = -3 \langle \cos t, \sin t \rangle\n]\nThis is a unit vector inward scaled by 3? Wait — ( \langle \cos t, \sin t \rangle ) is the tangent vector (direction of motion), and multiplying by (-3) flips it to radial inward. But in circular motion, velocity should be tangential, not radial. Contradiction?", "Ah — here’s the subtlety: if the position is ( (x, y) = (-3\sin t, 3\cos t) ), then:", "Differentiate:\n[\n\vec{r}'(t) = \langle -3\cos t, -3\sin t \rangle\n]", "But compute the tangential velocity:\nOn a circle ( x = R\sin\ heta ), ( y = R\cos\ heta ), standard parametrization, velocity is ( \langle R\omega\cos\ heta, -R\omega\sin\ heta \rangle ), which is tangential.", "Compare:\n[\nx(t) = -3\sin t = 3 \cos(t + \pi/2), \quad y(t) = 3\cos t = 3\sin(t + \pi/2)\n]\nSo it’s equivalent to counterclockwise motion at angular speed 1, with angular position ( \ heta(t) = t + \pi/2 ).", "Then tangential velocity is:\n[\n\vec{v}(t) = R\omega \langle -\sin\ heta, \cos\ heta \rangle = 3 \cdot 1 \cdot \langle -\sin\ heta, \cos\ heta \rangle = \langle -3\sin(t + \pi/2), 3\cos(t + \pi/2) \rangle\n]\nBut ( \sin(t + \pi/2) = \cos t ), ( \cos(t + \pi/2) = -\sin t ), so:\n[\n\vec{v}(t) = \langle -3\cos t, -3\sin t \rangle\n]\nSame as computed — thus, although the position vector rotates as if counterclockwise with angular speed 1, the velocity is indeed tangential and inward-radial, but in circular motion, inward radial and tangential velocity are perpendicular — and here both components are nonzero, confirming smooth turning.", "Their dot product:\n[\n\vec{r}'(t) \cdot \vec{v}(t) = (-3\cos t)(-3\cos t) + (-3\sin t)(-3\sin t) = 9\cos^2 t + 9\sin^2 t = 9 <br/>\ne 0\n]\nWait — they are not perpendicular? But in circular motion, radial and tangential vectors must be perpendicular. Contradiction?", "Ah — error: earlier we said velocity is radial inward, but that’s incorrect for circular motion.", "Let’s reevaluate:\nGiven circular motion:\n[\nx(t) = R\cos(t + \alpha), \quad y(t) = R\sin(t + \alpha)\n]\nThen:\n[\nx' = -R\sin(t + \alpha), \quad y' = R\cos(t + \alpha)\n]\nSo velocity is ( \langle -R\sin\ heta, R\cos\ heta \rangle ), where ( \ heta = t + \alpha ) — this vector has magnitude ( R ), and is always tangent, since dot product with radial ( \langle \cos\ heta, \sin\ heta \rangle ) is:\n[\n(-R\sin\ heta)(\cos\ heta) + (R\cos\ heta)(\sin\ heta) = 0\n]\nBut in our case:\n[\nx(t) = -3\sin t = 3 \cos(t + \pi/2), \quad y(t) = 3\cos t = 3\sin(t + \pi/2)\n]\nSo ( \ heta = t + \pi/2 ), then:\n[\nx' = -3\cos t = 3\sin(t + \pi/2), \quad y' = -3\sin t = 3\cos(t + \pi/2)\n]\nSo velocity:\n[\n\vec{r}'(t) = \langle 3\sin(t + \pi/2), 3\cos(t + \pi/2) \rangle = 3 \langle \sin(t + \pi/2), \cos(t + \pi/2) \rangle\n]\nNow dot product with radius:\n[\n\vec{r}'(t) \cdot \langle \cos(t + \pi/2), \sin(t + \pi/2) \rangle = 3\left[ \sin\ heta \cos\ heta + \cos\ heta \sin\ heta \right] = 3(2\sin\ heta\cos\ heta) <br/>\ne 0\n]\nStill not zero — confused.", "Better: define unit tangent vector.", "For ((x(t), y(t)) = (-3\sin t, 3\cos t)), compute derivative:\n[\n\vec{r}'(t) = (-3\cos t, -3\sin t)\n]\nNow compute tangent vector: direction of motion. At (t=0): velocity (\langle -3, 0 \rangle) → left. The circle centered at origin: at (0,3), moving downward — so clockwise? Wait:\n- (t=0): ((0, 3))\n- (t = \pi/2): ((-3, 0))\n- (t = \pi): ((0, -3))\n- (t = 3\pi/2): ((3, 0))\nSo from (0,3) → (-3,0) → (0,-3) → (3,0) → back — this is clockwise motion.", "Thus, velocity vector (\langle -3\cos t, -3\sin t \rangle) at (t=0): ((-3, 0)) — west, correct for clockwise.", "magnitude: 3, direction: southwest — tangential to circle, oriented clockwise.", "So yes, velocity is always tangent and clockwise, with speed 3 along the circle.", "---", "## Key Takeaways", "- (\vec{r}'(t) = \langle -3\cos t, -3\sin t \rangle) describes uniform circular motion of radius 3.\n- The particle moves clockwise around the origin.\n- Speed is constant: ( \boxed{3} ).\n- Direction of velocity is tangent to the circle, always pointing opposite to the radius (inward radial), confirming centripripetal motion.\n- This derivative is fundamental in kinematics, dynamics, and vector calculus — used to define angular velocity, centripetal acceleration, and more.", "---", "## Using \vec{r}'(t) to Study Curves in \mathbb{R}^2", "For parametric curves:\n[\n\vec{r}(t) = \langle x(t), y(t) \rangle, \quad t \in [a,b]\n]\n[\n\vec{r}'(t) = \langle x'(t), y'(t) \rangle\n]\ntells us the instantaneous direction, speed, and curvature tendencies. For circular motion, the magnitude gives circular speed, and the orthogonal unit vector gives the correct tangent.", "---", "## Conclusion", "The vector function (\vec{r}'(t) = \langle -3\sin t, 3\cos t \rangle) captures the instantaneous velocity of a particle executing uniform clockwise circular motion of radius 3 centered at the origin, repeated once as (t) traverses ([0, 2\pi]). It combines speed, direction, and centripetal orientation, offering deep insight into rotational dynamics through calculus.", "Whether analyzing physical systems, designing motion paths, or studying optics and mechanics, understanding such derivatives is foundational.", "---", "Keywords: vector derivative, \vec{r}'(t), \vec{r}'(t) = \langle -3\sin t, 3\cos t \rangle, circular motion, velocity vector, parametrized curve, tangential velocity, uniform circular motion, (t \in [0, 2\pi])", "Meta description: Explore the vector derivative \vec{r}'(t) = \langle -3\sin t, 3\cos t \rangle for (t \in [0, 2\pi]), revealing instantaneous velocity, constant speed (3), clockwise circular motion, and its geometric meaning in 2D space."]