Differentiate parametric curves and integrate vector motion

Connect the existing prerequisites

Senior 7A and 7C establish spatial vectors and line/plane representations; 4A–5B establish scalar derivatives and integrals. The companion lessons reuse Unit 2 conic paths and constant relative motion. Here circular arguments are radians and every derivative names its variable

A parameterisation includes more than a Cartesian equation

A curve r(t)=(x(t), y(t)) or (x(t), y(t), z(t)) specifies a point for each permitted t. Eliminating t can lose the range, orientation, repeated traversal or excluded parameter values. If t is time, the coordinate functions additionally specify when the particle reaches each point

Derive the first parametric derivative

Where x′(t)≠0, the chain rule dy/dt=(dy/dx)(dx/dt) gives dy/dx=y′(t)/x′(t). This is a local Cartesian slope, not the derivative vector itself. Keep the nonzero denominator condition

Work a polynomial parameter slope

At t=1, the point is (3, 1). Differentiate both coordinates to find the tangent slope, then use the negative reciprocal for the normal slope

Give dy/dx at t=1 for the curve below

Use the same rule for circular coordinates

For x=1+3cos t, y=-2+2sin t, the slope is -2cos t/(3sin t) wherever sin t≠0. At t=π/4 it is -2/3. At t=π/2 it is 0; at t=0 the regular tangent is vertical because x′=0 but y′=2.

Give the slope at t=π/4 for the ellipse below

Differentiate a second time with respect to x

First form dy/dx as a function of t. Differentiate it with respect to t, then divide by x′ again. Thus d²y/dx²=(x′y″-y′x″)/(x′)³ where x′≠0. The ratio y″/x″ is generally wrong

Work a nonconstant second derivative

For x=t², y=t³ with t≠0, the first slope is 3t/2. Its t derivative is 3/2, so the second x derivative is 3/(4t). At t=2 the first slope is 3 and the second derivative 3/8. On each regular branch its sign describes concavity as a graph over x.

Give d²y/dx² at t=2 for x=t², y=t³

Regular horizontal and vertical tangent tests

If y′=0 and x′≠0, the tangent is horizontal. If x′=0 and y′≠0, the tangent is vertical. If both vanish, the parameterisation is singular at that value and those two tests do not decide the tangent. Investigate the local curve or a justified derivative ratio limit

A zero parameter velocity can still have a tangent

At t=0, the curve (t³, t³) has zero derivative vector but lies on y=x, so its tangent has slope 1. For (t², t³), the quotient 3t/2 tends to 0 from both sides, giving a horizontal tangent at the cusp. For (t³, t²), the reciprocal slope tends to 0, giving a vertical tangent

At the displayed cusp, which conclusion is justified?

Differentiate a vector component by component

For a differentiable position vector r(t), velocity is v(t)=r′(t) and acceleration is a(t)=r″(t). In orthonormal coordinates, speed is |v|=√(vx²+vy²+vz²), omitting the third component in a planar problem. Signed velocity components are not speeds

Work a three-dimensional motion

Let r(t)=(1+t+t²,-2+2t+t³, 3-t²) metres for t≥0 seconds. Then v=(1+2t, 2+3t²,-2t) m/s and a=(2, 6t,-2) m/s². At t=1 these are v=(3, 5,-2), a=(2, 6,-2); speed is √38 m/s.

Give the squared speed at t=1 for the position vector below, measured in metres

Give the y-component of acceleration at t=1 for the position vector below, measured in metres

Constant speed does not mean constant velocity

For r(t)=(2cos3t, 2sin3t, t), velocity is (-6sin3t, 6cos3t, 1), with constant speed √37. Acceleration is (-18cos3t,-18sin3t, 0), which is nonzero. The helix changes direction while keeping its speed. If the axial coefficient were 0, the path would be a circle

For the helix below, find the x-component of acceleration at t=0

Acceleration and change in speed are different

When |v|>0, d|v|/dt=(v·a)/|v|. An acceleration perpendicular to velocity changes direction without changing speed. At v=0 this quotient is undefined; analyse the speed function separately. Uniform circular motion has nonzero inward acceleration and constant speed

Integrate with a constant vector

If r′(t)=v(t), integrate each coordinate and include one constant per component. A position condition r(t0)=r0 determines all position constants. If acceleration is supplied, first integrate to velocity using its condition, then integrate to position using a separate condition

Work integration from a time-dependent acceleration

For a(t)=(2, 6t,-2), v(0)=(1, 2, 0), r(0)=(1,-2, 3), integration gives v=(1+2t, 2+3t²,-2t) and r=(1+t+t²,-2+2t+t³, 3-t²). At t=2, the position is (7, 10,-1) and velocity (5, 14,-4). Differentiate and check both initial conditions

Give the y-position at t=2 from the conditions below

Circular primitives also need initial constants

If v(t)=(6cos2t,-4sin2t, 1) and r(0)=(1,-1, 2), then r(t)=(1+3sin2t,-3+2cos2t, 2+t). The y constant is -3 because cosine equals 1 at 0. At t=π/4, the position is (4,-3, 2+π/4)

Find the constant in the y-coordinate below

The vector integral gives displacement

For a continuous velocity,∫ from a to b of v(t)dt=r(b)-r(a). Integrating speed instead gives distance travelled:∫ from a to b of|v(t)|dt for a<b. The norm of net displacement need not equal distance if the path curves or reverses. This distinction supplies the arc-length setup used in 5D

Work a reversal with zero net displacement

For r(t)=((t-1)², 0), 0≤t≤2, the particle goes from (1, 0) to (0, 0) and back. Endpoint displacement is 0, but speed is 2|t-1| and distance is 2. Split its speed integral at t=1, where the signed x velocity changes sign

Give the total distance travelled along the path below

Find crossing paths using independent parameters

A geometric crossing solves rP(t)=rQ(s), with both parameter domains checked. A collision solves rP(t)=rQ(t) using the same physical clock. The first calculation may give different arrival times, or a time outside one particle’s allowed domain

Do the particles in this diagram collide?

Do not silently extend a path into negative time

For P(t)=(t,(t-2)²) and Q(s)=(s+3, 0), with t, s≥0, the algebraic crossing needs t=2, s=-1. That point is absent from Q’s allowed path, so there is no allowed path crossing and no collision

A parameter change alters derivatives

If t=f(s), then dr/ds=(dr/dt)f′(s) and d²r/ds²=(d²r/dt²)[f′(s)]²+(dr/dt)f″(s). For an affine change t=αs+β, the last term vanishes. Negative α reverses traversal; speed with respect to s scales by |α|. Match the intervals before calling the traced sets equal

Work an affine reparameterisation

For r(t)=(t, t², t³), t=1-2s, at s=0 we have t=1. Thus dr/ds=-2(1, 2, 3)=(-2,-4,-6), while d²r/ds²=4(0, 2, 6)=(0, 8, 24). The second factor is 4, not -2.

Find the y-component of the second derivative with respect to s for the curve and parameter change below

Derive the parametric derivative rules, ungraded

Starting from the chain rule, derive dy/dx and d²y/dx² in terms of t derivatives. State the nonzero-denominator condition. Then explain why (t³, t³) and (t², t³) at 0 cannot be classified from 0/0 alone

Model: apply the chain and quotient rules in order

Where x′≠0, dy/dt=(dy/dx)x′ gives dy/dx=y′/x′. Applying (1/x′)d/dt again gives (x′y″-y′x″)/(x′)³. At t=0 both example quotients are 0/0. The first curve is the line y=x, with slope 1. For the second, the slope 3t/2 tends to 0, giving a horizontal tangent. Local geometry or limits supply information the bare quotient lacks

Prove the speed and vector-integral identities, ungraded

For a twice differentiable position vector, derive d|v|/dt=(v·a)/|v| where v≠0. Explain why integrating velocity gives displacement, while integrating its norm gives path length. Include a case with nonzero acceleration and constant speed

Model: differentiate the squared norm and apply the scalar theorem

Write |v|²=v·v. Differentiating gives 2|v|d|v|/dt=2v·a; division requires |v|>0. Apply the scalar fundamental theorem to each coordinate of ∫v dt to get r(b)-r(a). Local path increments have length approximately |v|dt, so their limiting sum is ∫|v|dt for a smooth path. For the helix,|v|²=37 while v·a=0 and a≠0. The reversal example shows why net displacement norm can be smaller than travelled length

Construct a full vector initial-value solution, ungraded

Solve a(t)=(2, 6t,-2), v(0)=(1, 2, 0), r(0)=(1,-2, 3). State position, velocity and acceleration units, check both conditions and derivatives, and describe the path and instantaneous direction at t=1. Do not treat a coordinate projection as the full spatial path

Model: preserve all six constants through the two integrations

Integrate the given acceleration componentwise. The initial velocity fixes its three constants, and a second integration followed by the initial position fixes the three position constants. The resulting spatial path has the position and tangent direction shown at t=1. Position is measured in metres, velocity in metres per second and acceleration in metres per second squared. The displayed derivatives and values at zero check the given acceleration and both initial vectors

Distinguish a curved-path crossing from collision, ungraded

For P(t)=(t,(t-2)²) and Q(s)=(s-d, 0), t, s≥0, classify the allowed path crossing and collision as the real constant d varies. Prove that all possible crossings have been considered

Model: use the unique zero before applying time conditions

Equal y-coordinates force (t-2)²=0, so t=2. Equal x-coordinates then give s=2+d. Hence the allowed paths cross exactly when d≥-2. A collision further requires s=t, so only d=0 gives a collision, at time 2. For -2≤d≠0 the paths cross at different allowed times; for d<-2 the algebraic point is not on Q’s allowed path. The squared term has a unique real zero, so no second crossing was missed

Justify a reparameterised path and its orientation, ungraded

For r(t)=(t, t², t³),-1≤t≤1, write the same path using t=1-2s. Find the corresponding s interval, orientation, first and second derivatives. Explain how the chain rule changes for nonlinear t(s)

Model: map the interval and keep both chain-rule terms

The correspondence t=1-2s maps 0≤s≤1 onto -1≤t≤1 with reversed orientation. The new derivative is -2(1, 2t, 3t²) and the second derivative 4(0, 2, 6t). It traces the same points because the affine map is a bijection of these intervals. For nonlinear t(s), differentiate the first chain-rule expression to obtain r″(t)(t′)²+r′(t)t″; the extra term cannot be dropped. Changes in traversal speed do not alter which points belong when the parameter mapping covers the same interval

Assessment boundary

Practice checks exact bounded derivatives, vector components, initial constants and crossing/collision conditions. Full proofs, path construction, unrestricted model analysis and independent technology use remain ungraded