Parametric Equations, Polar Coordinates, and Vector-Valued Functions

A parameter specifies x and y together. Position is a vector, while speed is the magnitude of velocity. Polar coordinates specify a directed radius and an angle; a negative radius places the point opposite the stated direction.

Parametric equations x=x(t), y=y(t) describe position as a parameter changes. Eliminating the parameter may reveal the geometric curve, but can lose direction, speed, and domain restrictions. Two parametrizations can trace the same curve differently.

When dx/dt is nonzero, dy/dx = (dy/dt)(dx/dt). A horizontal tangent requires dy/dt = 0 with dx/dt ≠ 0; a vertical tangent requires dx/dt = 0 with dy/dt ≠ 0. If both vanish, investigate further rather than declaring both tangent types.

Differentiate dy/dx with respect to t and then divide by dx/dt to obtain d²ydx² where the formula is valid. Dividing y″(t) by x″(t) is generally wrong because it does not differentiate slope with respect to x.

Parametric arc length integrates √[(x′)²+(y′)²] over the parameter interval. The curve may be traced more than once, so the integral counts traveled path length with multiplicity. Choose an interval appropriate to the requested single traversal.

A position vector has velocity equal to its componentwise derivative and acceleration equal to the velocity derivative. Speed is the magnitude of velocity, √[(x′)²+(y′)²], not the sum of components. Integrate each velocity component and use initial position to recover location.

Displacement is a vector difference; distance traveled is the integral of speed. A particle can return to its starting point with nonzero traveled distance. Speeding up depends on the alignment of velocity and acceleration, captured by the sign of their dot product when velocity is nonzero.

For r=r(θ), write x=r cosθ and y=r sinθ, then differentiate with respect to θ. The slope is (r′sinθ+r cosθ)(r′cosθ−r sinθ) where the denominator is nonzero. Radius sign and angle determine location, so r′ alone is not the Cartesian slope.

Zeros of the numerator or denominator can locate horizontal or vertical tangent candidates, but check the companion expression and the actual point. At the pole, both coordinates vanish and multiple branches can meet; inspect the limiting geometry if the direct expression is indeterminate.

A thin sector has area approximately ½r²Δθ, leading to area ½∫r²dθ over an interval tracing the intended region. Negative radius does not produce negative sector area because radius is squared, but repeated tracing can double count.

For area between curves over shared angular bounds, subtract squared inner radius from squared outer radius when those radial descriptions correctly represent the region. Intersections and changes of outer boundary can require splitting. Sketch or sample points before trusting a single integral.

If x and y depend on t and dx/dt≠0, dy/dx=(dy/dt)(dx/dt). A horizontal tangent is suggested by dy/dt=0 with dx/dt≠0; a vertical tangent by dx/dt=0 with dy/dt≠0. If both vanish, neither simple test resolves the tangent. Examine a limit or the local curve.

For the second derivative with respect to x, first differentiate dy/dx with respect to t and then divide by dx/dt again. Differentiating only with respect to t gives the wrong variable. Retain the parameter until the requested quantity is expressed correctly. A curve can pass through the same point at different parameter values with different tangent directions.

For position ⟨x(t),y(t)⟩, velocity is ⟨x′(t),y′(t)⟩ and acceleration is ⟨x″(t),y″(t)⟩. Speed is √((x′)²+(y′)²), a nonnegative scalar. The integral of speed gives distance traveled. Integrating the velocity components gives displacement, which can have a much smaller magnitude when a path bends or loops.

To recover position from velocity, integrate each component and use the initial position separately. Do not use one arbitrary constant for both components. A particle’s speed is not obtained by adding velocity components, and acceleration magnitude is not generally the derivative of speed. Directional changes can produce acceleration even at constant speed.

For r=f(θ), Cartesian coordinates are x=r cosθ and y=r sinθ. Differentiate both with respect to θ and take their quotient for dy/dx. The resulting numerator and denominator include both r and drdθ. The derivative drdθ alone measures how signed radius changes, not the slope of the curve in the xy-plane.

Polar area uses one-half the integral of r² over an angular interval. Determine which interval traces the intended region once; symmetry can help but must be checked. For area between polar curves, compare squared radii along the same rays and split intervals when the outer boundary changes. Negative r-values still contribute through r², but tracing and overlap require geometric attention.

For x=x(t) and y=y(t), dy/dx=(dy/dt)(dx/dt) when dx/dt is nonzero. A horizontal tangent requires dy/dt=0 with nonzero dx/dt; a vertical tangent requires the corresponding opposite condition. If both vanish, further analysis is necessary.

The second derivative with respect to x is not simply y″(t)x″(t). Differentiate dy/dx with respect to t, then divide by dx/dt. This extra conversion accounts for changing x along the parameterized path.

A curve can pass through the same location at several parameter values with different tangent directions. Keep the parameter interval and direction of travel when interpreting a sketch. A Cartesian equation alone can hide this timing information.

Position has components; velocity and acceleration differentiate each component. Speed is the magnitude of velocity, √[(x′)²+(y′)²], and distance integrates speed. Displacement is the difference between final and initial position vectors, not the accumulated path length.

Recover position by integrating velocity components and using initial coordinates. Constants must be applied to both components. An acceleration vector need not point along the velocity vector; it can change direction as well as speed.

For arc length, integrate the magnitude of the derivative over the correct parameter interval. If the curve is traced multiple times, the integral counts repeated travel. Determine whether the question asks for distance traveled or length of the geometric curve traced once.

For r=r(θ), convert to x=r cos θ and y=r sin θ when finding a tangent slope. Differentiate both with respect to θ and form dy/dx. The derivative drdθ alone measures radial change, not Cartesian slope.

Polar area uses one-half the integral of r² with respect to θ, derived from small sectors. Bounds should trace the intended region once. Negative r does not make the squared sector contribution negative, but it changes which direction represents the point, so a sketch and tracing analysis remain important.

For area between polar curves on a suitable common angular interval, subtract inner radius squared from outer radius squared. Intersections can involve the pole or different coordinate descriptions of the same point. Algebraic equality at the same angle is useful but not always the whole geometric story.

Before integrating a polar region, identify how the curve moves as θ changes. Record where r=0, where r changes sign, and which angles reach extreme radial values. A loop can be traced over less than a full 2π interval, so integrating over an automatic full turn can double-count area.

For r=2cos θ, using −π2≤θ≤π2 traces the circle once with nonnegative radius. The equation corresponds to a circle centered at (1,0) with radius 1. The sector-area integral over that interval gives π. This geometric check confirms the bounds and factor of one-half.

When two curves intersect at the pole, their θ-values there need not match. A graph or tracing table can reveal regions missed by simply solving r₁(θ)=r₂(θ). Describe which curve is outer on each chosen interval.

For r(t)=〈x(t),y(t)〉, integrate acceleration component by component to recover velocity, then use initial velocity. Integrate again for position with a second initial condition. One constant for the entire vector is shorthand for separate component constants, not a single number forced into both coordinates.

Speed is √(vx²+vy²). An object can have zero x-velocity while still moving vertically. A particle is at rest only when all velocity components are zero simultaneously. Likewise, a zero component of acceleration does not mean the whole acceleration vector vanishes.

The rate of change of speed depends on the component of acceleration along velocity. Curving motion can have acceleration perpendicular to velocity and constant speed. This connects calculus-based vector motion with the physical distinction between changing magnitude and changing direction.