Curves in the three-dimensional space, derivative and integral of vector functions

Curves in the space

Consider the following parametralized curve $$x=f(t),\quad y=g(t),\quad z=h(t),\quad t\in I.$$ The points $(x,y,z)=(f(t),g(t),z(t)),~t\in I$ make up the curve in the space that is also called the particle's path. It has an equivalent form $$\vec{r}(t)=\overrightarrow{OP}=f(t)\vec{i}+g(t)\vec{j}+h(t)\vec{k}.$$ The position vector $\vec{r}$ = $\overrightarrow{OP}$ of a particle moving in the space is a function of time.

Example: Graph the vector function $$\vec{r}(t) = (\cos t)\vec{i} + (\sin t)\vec{j} + t\vec{k}$$

Limits and continuity

Definition Let $\vec{r}(t) = \overrightarrow{OP}=f(t)\vec{i}+g(t)\vec{j}+h(t)\vec{k}$ be a vector function with domain $D$ and $\vec{L}$ a vector. We say that $\vec{r}$ has limit $\vec{L}$ as $t$ approaches $t_0$ and write $$\lim_{t\rightarrow t_0}\vec{r}(t)=\vec{L}$$ if, for every number $\epsilon>0$, there exists a corresponding number $\delta>0$ such that for all $t\in D$, $$|\vec{r}(t)-\vec{L}|<\epsilon,\mbox{ whenever } 0<|t-t_0|<\delta.$$

Definition A vector function $\vec{r}(t)$ is continuous at a point $t=t_0$ in its domain if $\lim_{t\rightarrow t_0}\vec{r}(t)=\vec{r}(t_0)$. The function is continuous if it is continuous over its domain.

Note When we consider the limit and continuity of a vector function, we consider each component function independently.

A vector function is continuous if and only if its component functions are continuous. Each component of the limit of a vector function comes from the limit of each component function.

Derivatives of vector functions

Definition The vector function $$\vec{r}(t)=f(t)\vec{i}+g(t)\vec{j}+h(t)\vec{k}$$ has a derivative (is differentiable) at $t$ if $f$, $g$, and $h$ have derivatives at $t$. The derivative is the vector function $$\vec{r}'(t)={d\vec{r}\over dt}=\lim_{\Delta t\rightarrow 0}{\vec{r}(t+\Delta t)-\vec{r}(t)\over \Delta t}={df\over dt}\vec{i}+{dg\over dt}\vec{j}+{dh\over dt}\vec{k}.$$

Motion

Definition If $\vec{r}$ is the position vector of a particle moving along a smooth curve in space, then $$\vec{v}(t) = {d\vec{r}\over dt}$$ is the particle's velocity vector, which is tangent to the curve. At any time $t$, the direction of $\vec{v}$ is the direction of motion, the magnitude of $\vec{v}$ is the particle's speed, and the derivative $\vec{a}={d\vec{v}\over dt}$ when it exists, is the particle's acceleration vector. In summary,

• Velocity is the derivative of position $\vec{v}(t) = {d\vec{r}\over dt}$

• Speed is the magnitude of velocity $|\vec{v}|$

• The unit vector ${\vec{v}\over |\vec{v}|}$ is the direction of motion at time $t$

• Acceleration is the derivative of velocity: $\vec{a}(t) = {d\vec{v}\over dt}$

Example: Find the velocity, speed, and acceleration of $\vec{r}(t)=2\cos t\vec{i}+2\sin t\vec{j}+5\cos^2t\vec{k}$

Differentiation rules

Let $\vec{u}$ and $\vec{v}$ be differentiable vector functions of $t$, $\vec{C}$ a constant vector, $c$ any scalar, and $f$ any differentiable scalar function.

• Constant Function Rule: ${d\over dt}\vec{C}=\vec{0}$

• Scalar Multiple Rules: ${d\over dt}[c\vec{u}(t)] = c{d\over dt}[\vec{u}(t)]$

• ${d\over dt}[f(t)\vec{u}(t)] = {d\over dt}[f(t)]\vec{u}(t) + f(t){d\over dt}[\vec{u}(t)]$

• Sum Rule: ${d\over dt}[\vec{u}(t)+\vec{v}(t)] = {d\over dt}[\vec{u}(t)] + {d\over dt}[\vec{v}(t)]$

• Difference Rule: ${d\over dt}[\vec{u}(t)-\vec{v}(t)] = {d\over dt}[\vec{u}(t)] - {d\over dt}[\vec{v}(t)]$

• Dot Product Rule: ${d\over dt}[\vec{u}(t)\cdot\vec{v}(t)] = {d\over dt}[\vec{u}(t)]\cdot\vec{v}(t) + \vec{u}(t)\cdot{d\over dt}[\vec{v}(t)]$

• Cross Product Rule: ${d\over dt}[\vec{u}(t)\times \vec{v}(t)] = {d\over dt}[\vec{u}(t)]\times\vec{v}(t) + \vec{u}(t)\times{d\over dt}[\vec{v}(t)]$

• Chain Rule: ${d\over dt}[\vec{u}(f(t))] = {d\over df(t)}[\vec{u}[f(t))]{d\over dt}[f(t)]$

Vector functions of constant length

If $\vec{r}(t)$ is a differentiable vector function of $t$ of constant length, i.e., $|\vec{r}(t)|=c$, then

Integrals of vector functions

Definition The indefinite integral of $\vec{r}$ with respect to $t$ is the set of all antiderivatives of $\vec{r}$, denoted as $\int \vec{r}(t)dt$. If $\vec{R}$ is any antiderivative of $\vec{r}$, then $$\int \vec{r}(t)dt=\vec{R}(t)+\vec{C}.$$

Example: $$\int((\cos t)\vec i + \vec j - 2t\vec k)dt = (\sin t)\vec i + t\vec j - t^2\vec k + \vec C.$$

Definition If the components of $\vec{r}(t)=f(t)\vec i+g(t)\vec j+h(t)\vec k$ are integrable over $[a,b]$, then so is $\vec r$, and the definite integral of $\vec r$ from $a$ to $b$ is $$\int_a^b\vec r(t)dt=\left(\int_a^bf(t)dt\right)\vec i+\left(\int_a^bg(t)dt\right)\vec j+\left(\int_a^bh(t)dt\right)\vec k.$$

Fundamental theorem of calculus for vector functions

If $${d\vec{R}(t)\over dt} = \vec{r}(t),$$ we have $$\int_a^b\vec r(t)dt= \left.\vec{R}(t)\right]_{a}^b=\vec{R}(b)-\vec{R}(a).$$

Example: Suppose we do not know the path of a hang glider, but only its acceleration vector $\vec a(t) = -(3 \cos t)\vec i - (3 \sin t)\vec j + 2\vec k$. We also know that the glider departed from the point $(4, 0, 0)$ with velocity $\vec v(0) = 3\vec j$ at time $t = 0$. Find the glider’s position as a function $\vec r(t)$ of $t$.

Solution: $$\vec v(t)=\int\vec a(t)dt= -(3 \sin t)\vec i + (3 \cos t)\vec j + 2t\vec k + \vec {C_1}$$ Since $\vec v(0) = 3\vec j$, we know $\ \vec {C_1}=\vec 0$ and $\vec v(t)= -(3 \sin t)\vec i + (3 \cos t)\vec j + 2t\vec k$. $$\vec r(t)=\int\vec v(t)dt= (3 \cos t)\vec i + (3 \sin t)\vec j + t^2\vec k + \vec {C_2}$$ Since $\vec r(0) = (4,0,0)$, we know $\ \vec {C_2}=(1,0,0)$ and $\vec r(t)= (3 \cos t+1)\vec i + (3 \sin t)\vec j + t^2\vec k$.

Ideal projectile motion

Height, flight time, and range for ideal projectile motion

For ideal projectile motion, when an object is launched from the origin over a horizontal surface with an initial speed $v_0$ and launch angle $\alpha$:

Since $\vec{v}(t)=-gt\vec{j}+\vec{C}$ and the velocity at $t=0$ is $$\vec{v}(0)=v_0\cos\alpha\vec{i} + v_0\sin\alpha\vec{j}.$$ So we have $$\vec{v}(t)=-gt\vec{j}+v_0\cos\alpha\vec{i} + v_0\sin\alpha\vec{j}.$$ So the position vector is $\vec{r}(t)=-{1\over 2}gt^2\vec{j}+v_0\cos\alpha t\vec{i} +v_0\sin\alpha t\vec{j}$ since the starting point is 0.

• Maximum height: $v_0^2\sin^2\alpha\over2g$

The height is $-{1\over 2}gt^2+v_0\sin\alpha t$, which has the maximum value when $t=v_0\sin\alpha/g$. So the maximum height is ${v_0^2\sin^2\alpha\over 2g}$

• Flight time: $2v_0\sin\alpha\over g$

When $t=2v_0\sin\alpha/g$, the height becomes $0$.

• Range: $2v_0^2\sin\alpha\cos\alpha\over g$

The range happens when $t=2v_0\sin\alpha/g$, so the total range is $${2v_0^2\sin\alpha\cos\alpha\over g}$$

Projectile motion with wind gusts

A baseball is hit when it is 1m above the ground. It leaves the bat with an initial speed of $50$ m/s, making an angle of 20° with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of $-2.5\vec{i}$ (m/s) to the ball's initial velocity ($2.5$ m/s = $9$ km/h).

(a) Find a vector equation (position vector) for the baseball path.

Since $\vec{v}(t)=-gt\vec{j}+\vec{C}$ and the velocity at $t=0$ is $$\vec{v}(0)=(50\cos20^o-2.5)\vec{i} + 50\sin20^o\vec{j}.$$ So we have $$\vec{v}(t)=-gt\vec{j}+(50\cos20^o-2.5)\vec{i} + 50\sin20^o\vec{j}.$$ So the position vector is $\vec{r}(t)=-{1\over 2}gt^2\vec{j}+(50\cos20^o-2.5) t\vec{i} +(50\sin20^ot+1)\vec{j}$ since the starting point is (0,1).

(b) How high does the baseball go, and when does it reach maximum height?

The height is $(50\sin20^ot+1)-{1\over 2}gt^2$, which reaches the maximum when $t=50\sin20^o/g$. So the maximum height is $2500\sin^220^o/g+1$.

(c) Assuming that the ball is not caught, find its range and flight time.

The total time the ball travels is when $(50\sin20^ot+1)-{1\over 2}gt^2=0$. Then we can also find the range based on the total time.