Q4 - Calculus for vector motion

Q4 - Calculus for vector motion

Differentiate and integrate two-dimensional motion component by component, use vector conditions to determine constants, and interpret changing speeds and directions. You will distinguish acceleration from the rate of change of speed, and use a common time when solving a direction condition.

Differentiating vector position

Can a particle accelerate while its speed is momentarily not changing? Yes: acceleration changes velocity, which includes direction. Begin by finding that velocity correctly from a position vector, then we will measure how its magnitude changes.

You already know

Along a line, v=dr/dtv=dr/dt and a=dv/dta=dv/dt. Differentiate each component of a vector separately. The magnitude of xi+yjx\mathbf i+y\mathbf j, for perpendicular unit vectors, is x2+y2\sqrt{x^2+y^2}.

Let r=x(t)i+y(t)j\mathbf r=x(t)\mathbf i+y(t)\mathbf j, where i\mathbf i and j\mathbf j are fixed perpendicular unit vectors. For example, differentiating t2it^2\mathbf i gives 2ti2t\mathbf i: the component changes but the chosen axis does not.

Vector time derivatives

v=r˙=dxdti+dydtja=r¨=d2xdt2i+d2ydt2j\begin{aligned} \mathbf v&=\dot{\mathbf r}=\frac{dx}{dt}\mathbf i+\frac{dy}{dt}\mathbf j \\ \mathbf a&=\ddot{\mathbf r}=\frac{d^2x}{dt^2}\mathbf i+\frac{d^2y}{dt^2}\mathbf j \end{aligned}

These relationships do not require constant acceleration. Not in the formula booklet: learn it.

Position is measured in metres, velocity in m s−1\mathrm{m\,s^{-1}} and acceleration in m s−2\mathrm{m\,s^{-2}}. Column-vector notation gives the same equations. The third Cartesian unit vector k\mathbf k is not needed here. Velocity is tangent to the path; position points from the origin to the particle, which is a different direction in general.

Worked example 1: a curved path

Question. A particle has r=(t2+1)i+(t3−3t)j\mathbf r=(t^2+1)\mathbf i+(t^3-3t)\mathbf j metres for 0≤t≤20\le t\le2, with tt in seconds. Find velocity and acceleration, then describe the vectors at t=1t=1.

Plan. Differentiate each position component twice, then substitute the same time in each vector.

v=2ti+(3t2−3)jfirst derivativesa=2i+6tjsecond derivativesr(1)=2i−2jpositionv(1)=2ihorizontal velocitya(1)=2i+6jacceleration\begin{aligned} \mathbf v&=2t\mathbf i+(3t^2-3)\mathbf j && \text{first derivatives} \\ \mathbf a&=2\mathbf i+6t\mathbf j && \text{second derivatives} \\ \mathbf r(1)&=2\mathbf i-2\mathbf j && \text{position} \\ \mathbf v(1)&=2\mathbf i && \text{horizontal velocity} \\ \mathbf a(1)&=2\mathbf i+6\mathbf j && \text{acceleration} \end{aligned}

Why differentiate before substituting? Substituting t=1t=1 first would turn the position into a constant vector and erase its rate of change.

Answer. At 11 second the particle is at (2,−2)(2,-2), moving horizontally right at 2 m s−12\,\mathrm{m\,s^{-1}}, with acceleration 2i+6j m s−22\mathbf i+6\mathbf j\,\mathrm{m\,s^{-2}}.

Check. The vertical position has derivative 3t2−3=03t^2-3=0 at 11, agreeing with the horizontal tangent below.

The diagram shows why position direction and motion direction must be kept separate.

Exact parametric path with its position at one second and horizontal velocity direction

Worked example 2 changes the component functions to exponential and trigonometric ones. The componentwise method stays the same; the scalar calculus rules change.

Worked example 2: nonpolynomial components

Question. A particle has r=2eti+sin⁡tj\mathbf r=2e^t\mathbf i+\sin t\mathbf j metres, with tt in seconds and the argument in radians. Find velocity and acceleration, and their values at t=0t=0.

Plan. Apply the appropriate scalar derivative to each component.

v=2eti+cos⁡tjdifferentiate positiona=2eti−sin⁡tjdifferentiate velocityv(0)=2i+je0=1, cos⁡0=1a(0)=2isin⁡0=0\begin{aligned} \mathbf v&=2e^t\mathbf i+\cos t\mathbf j && \text{differentiate position} \\ \mathbf a&=2e^t\mathbf i-\sin t\mathbf j && \text{differentiate velocity} \\ \mathbf v(0)&=2\mathbf i+\mathbf j && e^0=1,\ \cos0=1 \\ \mathbf a(0)&=2\mathbf i && \sin0=0 \end{aligned}

Why keep the unit vectors fixed? They describe the chosen Cartesian axes, not directions that turn with the particle.

Answer. The vector functions are those above; initial velocity is 2i+j m s−12\mathbf i+\mathbf j\,\mathrm{m\,s^{-1}} and initial acceleration is 2i m s−22\mathbf i\,\mathrm{m\,s^{-2}}.

Check. Differentiating the vertical component twice takes sin⁡t\sin t to cos⁡t\cos t, then −sin⁡t-\sin t, as in scalar calculus.

The elementary power, sine, cosine and exponential derivatives are recall results; they are not printed in the formula booklet. Include chain-rule factors for arguments such as 2t2t.

Watch out

Wrong: speed=d∣r∣/dt\text{speed}=d|\mathbf r|/dt. That measures how fast the particle's distance from the origin changes, which is a different quantity. Right: differentiate r\mathbf r to obtain v\mathbf v, then calculate ∣v∣|\mathbf v|.

The next model changes the exponential rate and trigonometric frequency. Both components must retain their chain-rule factors.

Integrating with vector initial conditions

Each integrated component has its own constant, and a vector condition supplies one equation for each component.

Integrating 4ti+6j4t\mathbf i+6\mathbf j gives (2t2+Cx)i+(6t+Cy)j(2t^2+C_x)\mathbf i+(6t+C_y)\mathbf j. The constant is a vector, Cxi+CyjC_x\mathbf i+C_y\mathbf j; neither component is automatically zero. Integrating again introduces a separate position constant vector.

Vector integration

Integrate each component with respect to the common time:

v=∫a dtr=∫v dt\mathbf v=\int\mathbf a\,dt \qquad \mathbf r=\int\mathbf v\,dt

Displacement over an interval is

r(t2)−r(t1)=∫t1t2v dt\mathbf r(t_2)-\mathbf r(t_1)=\int_{t_1}^{t_2}\mathbf v\,dt

Not in the formula booklet: learn it.

Use a velocity condition to find the velocity constants and a position condition to find the position constants. Each condition must be applied at the stated time, which need not be zero. Elementary antiderivatives are recalled from Pure Mathematics.

Worked example 3: two vector conditions

Question. A particle has a=4ti+6j m s−2\mathbf a=4t\mathbf i+6\mathbf j\,\mathrm{m\,s^{-2}}, v(0)=i−2j m s−1\mathbf v(0)=\mathbf i-2\mathbf j\,\mathrm{m\,s^{-1}} and r(0)=3i+j\mathbf r(0)=3\mathbf i+\mathbf j metres. Find velocity and position.

Plan. Integrate twice, applying the matching vector condition after each integration.

v=(2t2+Cx)i+(6t+Cy)jintegrate accelerationCx=1,Cy=−2initial velocityr=(23t3+t+Dx)i+(3t2−2t+Dy)jintegrate complete velocityDx=3,Dy=1initial position\begin{aligned} \mathbf v&=(2t^2+C_x)\mathbf i+(6t+C_y)\mathbf j && \text{integrate acceleration} \\ C_x&=1,&C_y&=-2 && \text{initial velocity} \\ \mathbf r&=(\tfrac23t^3+t+D_x)\mathbf i \\ &\quad +(3t^2-2t+D_y)\mathbf j && \text{integrate complete velocity} \\ D_x&=3,&D_y&=1 && \text{initial position} \end{aligned}

Why are four constants needed? There are two component integrations for velocity and two more for position, each constrained by its own data.

Answer. v=(2t2+1)i+(6t−2)j m s−1\mathbf v=(2t^2+1)\mathbf i+(6t-2)\mathbf j\,\mathrm{m\,s^{-1}} and r=(23t3+t+3)i+(3t2−2t+1)j\mathbf r=(\frac23t^3+t+3)\mathbf i+(3t^2-2t+1)\mathbf j metres.

Check. Differentiating position twice recovers 4ti+6j4t\mathbf i+6\mathbf j; substituting zero recovers both given vectors.

Worked example 4 changes only the time at which the same velocity and position vectors are known. The constants must change to satisfy these later observations.

Worked example 4: conditions at a later time

Question. Suppose a=4ti+6j m s−2\mathbf a=4t\mathbf i+6\mathbf j\,\mathrm{m\,s^{-2}}, but now v(1)=i−2j m s−1\mathbf v(1)=\mathbf i-2\mathbf j\,\mathrm{m\,s^{-1}} and r(1)=3i+j\mathbf r(1)=3\mathbf i+\mathbf j metres. Find velocity and position.

Plan. Use the same antiderivatives, substituting t=1t=1 when determining each constant.

v=(2t2+Cx)i+(6t+Cy)j1=2+Cx,−2=6+Cycondition at 1Cx=−1,Cy=−8r=(23t3−t+Dx)i+(3t2−8t+Dy)j3=23−1+Dx,1=3−8+Dyposition at 1Dx=103,Dy=6\begin{aligned} \mathbf v&=(2t^2+C_x)\mathbf i+(6t+C_y)\mathbf j \\ 1&=2+C_x,&-2&=6+C_y && \text{condition at }1 \\ C_x&=-1,&C_y&=-8 \\ \mathbf r&=(\tfrac23t^3-t+D_x)\mathbf i+(3t^2-8t+D_y)\mathbf j \\ 3&=\tfrac23-1+D_x,&1&=3-8+D_y && \text{position at }1 \\ D_x&=\tfrac{10}{3},&D_y&=6 \end{aligned}

Why not copy the constants from the previous example? The same observed vectors at a different time describe a different motion.

Answer. v=(2t2−1)i+(6t−8)j m s−1\mathbf v=(2t^2-1)\mathbf i+(6t-8)\mathbf j\,\mathrm{m\,s^{-1}} and r=(23t3−t+103)i+(3t2−8t+6)j\mathbf r=(\frac23t^3-t+\frac{10}{3})\mathbf i+(3t^2-8t+6)\mathbf j metres.

Check. At t=1t=1, these give components (1,−2)(1,-2) and (3,1)(3,1), exactly the stated conditions.

Definite component integrals provide an alternative: start from the known vector at its known time and integrate the change from there. They produce the same constants implicitly.

Watch out

Wrong: for a=2j\mathbf a=2\mathbf j and v(1)=0\mathbf v(1)=\mathbf0, write v=2tj\mathbf v=2t\mathbf j, omitting the constant. This gives v(1)=2j\mathbf v(1)=2\mathbf j, contradicting the condition. Right: v=(2t−2)j\mathbf v=(2t-2)\mathbf j. A zero observed component still supplies a condition.

In the next task the condition is again given after the clock starts. Find both component constants at that time.

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