Defining and Differentiating Vector-Valued Functions
Introduced| English | Meaning |
|---|---|
| vector-valued function/ˈvektə ˈvæljuːd ˈfʌŋkʃn/ | A function whose output is a vector rather than a single scalar. |
| position vector/pəˈzɪʃn ˈvektə/ | position vector |
| velocity/vəˈlɒsɪti/ | The rate of change of displacement. |
| acceleration/əkˌseləˈreɪʃn/ | The rate of change of velocity. |
| speed/spiːd/ | The distance travelled per unit time. |
Track where a robot is and how it moves
- A robot's location and its direction of travel are different facts. A vector-valued function 向量值函数 packages its coordinates as $\mathbf r(t)=\langle x(t),y(t)\rangle$.
- This is the position vector 位置向量 from the origin to the point. Its derivative describes how that position changes, rather than where the point is.
Differentiate each component
- If both coordinate functions are differentiable, $\mathbf r'(t)=\langle x'(t),y'(t)\rangle$. Apply the ordinary derivative rules separately; do not add or multiply the components together first.
- Differentiate the functions before substituting a parameter value. Substituting first produces a fixed vector and loses the information about how the position changes nearby.
To differentiate $\mathbf{r}(t)=\langle x(t),y(t)\rangle$, you...
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$\mathbf{r}'=\langle x',y'\rangle$.
To find velocity at a chosen parameter, first replace the position function by its fixed value and differentiate that constant vector.
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Differentiate the coordinate functions first, then evaluate the derivative at the chosen parameter.
Work the whole derivative chain
- The velocity 速度 is $\mathbf v=\mathbf r'$, and the acceleration 加速度 is $\mathbf a=\mathbf v'=\mathbf r''$. For $\mathbf r(t)=\langle t^2,\sin t\rangle$, these are $\mathbf v(t)=\langle2t,\cos t\rangle$ and $\mathbf a(t)=\langle2,-\sin t\rangle$.
- At $t=0$, position is $\langle0,0\rangle$, velocity is $\langle0,1\rangle$, and acceleration is $\langle2,0\rangle$. Being at the origin does not mean the point is stationary.
For $\mathbf{r}(t)=\langle t^2,\sin t\rangle$, the velocity $\mathbf{v}(t)=$
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Differentiate each component.
The acceleration vector is...
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Differentiate position twice: $\mathbf{a}=\mathbf{r}''$.
For $\mathbf r(t)=\langle t^2,\sin t\rangle$, find the first component of $\mathbf a(0)$.
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The second derivative of $t^2$ is 2.
For that same function, find the second component of $\mathbf a(0)$.
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The second derivative of $\sin t$ is $-\sin t$, giving 0 at t=0.
Speed is a scalar magnitude
- The speed 速率 is $|\mathbf v(t)|=\sqrt{(x'(t))^2+(y'(t))^2}$, a nonnegative scalar. In the example, $|\mathbf v(0)|=\sqrt{0^2+1^2}=1$.
- Keep the vector until you need its magnitude. If $\mathbf v=\langle3,-4\rangle$, speed is $5$, not $3+(-4)=-1$. Taking a magnitude before differentiating also loses direction information.
Position and its vector derivatives
For r(t)=⟨t²,sin t⟩ at t=0, compare position⟨0,0⟩, velocity⟨0,1⟩, acceleration⟨2,0⟩ and speed 1. The derivative arrows are shown from a common origin for comparison.
For $\mathbf{v}(0)=\langle 0,1\rangle$, find the speed $|\mathbf{v}(0)|$.
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$\sqrt{0^2+1^2}=1$.
Speed is a vector, the same as velocity.
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Speed is the scalar magnitude $|\mathbf{v}|$.
Speed equals the ____ of the velocity vector.
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$|\mathbf{v}|=\sqrt{(x')^2+(y')^2}$.
If $\mathbf v=\langle3,-4\rangle$, what is the speed?
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$\sqrt{3^2+(-4)^2}=5$, not the component sum-1.
Work a nontrivial polynomial example
- For $\mathbf r(t)=\langle t^2,t^3-t\rangle$, $\mathbf v(t)=\langle2t,3t^2-1\rangle$ and $\mathbf a(t)=\langle2,6t\rangle$.
- At $t=1$, the point is $(1,0)$, velocity is $\langle2,2\rangle$, acceleration is $\langle2,6\rangle$, and speed is $2\sqrt2$. Evaluate every requested quantity at the same parameter value.
For $\mathbf r(t)=\langle t^2,t^3-t\rangle$, find the second velocity component at t=1.
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$3t^2-1=2$ at t=1.
For that same function, find the second acceleration component at t=1.
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The second derivative of $t^3-t$ is $6t$, giving 6.
For that same function, find the speed at t=1.
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$\mathbf v(1)=\langle2,2\rangle$, so speed is $\sqrt8=2\sqrt2$.
Read direction from velocity components
- For the polynomial example at $t=-1$, $\mathbf v(-1)=\langle-2,2\rangle$: the point moves left and up. Its position is $(1,0)$, which does not by itself tell you that direction.
- A zero component does not mean a zero vector. The original example has $\mathbf v(0)=\langle0,1\rangle$, so its motion at that instant is vertical with positive speed.
What direction is described by velocity $\langle-2,2\rangle$?
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The first component is negative and the second is positive.
A point in the right half-plane must have a positive first velocity component.
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Position does not determine velocity. The polynomial example at t=-1 has x=1 but first velocity component-2.
A nonzero velocity is tangent to the path
- At a regular point where $\mathbf v\ne\mathbf0$, the velocity gives the tangent direction. If $v_x\ne0$, the tangent slope is $v_y/v_x$; a nonzero vertical velocity gives a vertical tangent.
- For $\mathbf r(t)=\langle t^2,t^3-t\rangle$ at $t=1$, the tangent slope is $2/2=1$. The position vector $\langle1,0\rangle$ is not the tangent vector $\langle2,2\rangle$.
At a point with nonzero velocity $\langle2,2\rangle$, what is the tangent slope?
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$v_y/v_x=2/2=1$.
A zero velocity needs a separate direction check
- For $\mathbf r(t)=\langle t^3,t^3\rangle$, $\mathbf v(0)=\langle0,0\rangle$ and speed is $0$. The zero vector supplies no direction at that instant; you cannot divide it by its magnitude to create a unit direction.
- The geometric path is still the straight line $y=x$, with a tangent line at the origin. Zero velocity alone does not show that no tangent exists, or that the point stays at rest at other times.
What can the zero velocity vector $\langle0,0\rangle$ tell you about a tangent direction at that instant?
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It has magnitude 0 and no direction; the geometric path needs separate analysis.
If $\mathbf r(t_0)=\langle0,0\rangle$, then $\mathbf v(t_0)=\langle0,0\rangle$.
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For $\mathbf r(t)=\langle t^2,\sin t\rangle$, position at 0 is zero but velocity is $\langle0,1\rangle$.
Constant velocity gives zero acceleration
- For $\mathbf r(t)=\langle1+2t,3-t\rangle$, velocity is the constant vector $\langle2,-1\rangle$, acceleration is $\langle0,0\rangle$, and speed is $\sqrt5$.
- A zero acceleration vector means the velocity is not changing at that instant. It does not mean the point's position or velocity must be zero.
For $\mathbf r(t)=\langle1+2t,3-t\rangle$, what is the magnitude of acceleration?
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Both velocity components are constant, so the acceleration vector is zero.
Constant speed can coexist with acceleration
- For $\mathbf r(t)=\langle\cos t,\sin t\rangle$, velocity is $\langle-\sin t,\cos t\rangle$ and acceleration is $\langle-\cos t,-\sin t\rangle$. Speed is constantly $1$.
- Acceleration is nonzero because velocity changes direction around the circle. Acceleration is the derivative of the velocity vector, not simply the derivative of its scalar magnitude.
For $\mathbf r(t)=\langle\cos t,\sin t\rangle$, what is the speed?
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$\sqrt{\sin^2t+\cos^2t}=1$.
Constant scalar speed always implies a zero acceleration vector.
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On the unit circle the speed is 1 but the velocity changes direction, giving nonzero acceleration.
Try fresh vector functions
- For $\mathbf r(t)=\langle3t^2+1,t^3-2t\rangle$, find position, velocity, acceleration and speed at $t=1$. State the direction of travel using the velocity components.
- For $\mathbf r(t)=\langle2\cos t,2\sin t\rangle$, find the same quantities at $t=\pi/2$. Then explain the velocity and tangent distinction at $t=0$ for $\mathbf r(t)=\langle t^3,2t^3\rangle$.
For $\mathbf r(t)=\langle3t^2+1,t^3-2t\rangle$, find position, velocity, acceleration and speed at t=1. State the direction of travel.
Compare your component derivatives, vectors and scalar speed with the model.
For $\mathbf r(t)=\langle2\cos t,2\sin t\rangle$, find position, velocity, acceleration and speed at $t=\pi/2$. Explain why nonzero acceleration is compatible with constant speed.
Compare your component derivatives, vectors and scalar speed with the model.
Keep the requested quantity clear
- Differentiate position once for velocity and twice for acceleration. Substitute the requested parameter after differentiating. Use the velocity magnitude only when speed is requested.
- Separate position, direction and speed. A nonzero velocity gives a tangent direction; a zero velocity needs further geometric analysis. A change in velocity can involve direction even when speed stays constant.
For $\mathbf r(t)=\langle t^3,2t^3\rangle$ at t=0, find velocity, acceleration and speed. Does zero velocity rule out a geometric tangent?
Compare your component derivatives, vectors and scalar speed with the model.