Multivariable Calculus 13.2 Textbook Notes

Status: Baby

Integrals of Vector Functions; Projectile Motion

When $r(t)=〈f(t),g(t),h(t)〉$ and the (indefinite or definite) integrals of $f(t),g(t),h(t)$ are defined, we can define the integral of $r(t)$ accordingly:

$\int r(t)dt=<\int f(t)dt,\int g(t)dt,\int h(t)dt>\text{ or }{\int }_a^{b}r(t)dt=<{\int }_a^{b}f(t)dt,{\int }_a^{b}g(t)dt,{\int }_a^{b}h(t)dt>$

If we apply this definition to $\frac{dr(t)}{dt}$ in place of $r(t)$, we see that ${\int }_a^b\frac{dr(t)}{dt}dt={\int }_a^b{r}^{\prime }(t)dt=r(b)-r(a)$ holds and it represents the net displacement of $r(t)$ from $t=a$ to $t=b$, while ${\int }_a^b||\frac{dr(t)}{dt}||dt={\int }_a^b||r^{\prime }(t)||dt$ represents the total distance the particle traveled following the
curve $r(t)$ from $t=a$ to $t=b$.

In the study of ideal projectile motion, one studies a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ satisfying $r\prime \prime (t)=〈0,-g,0〉,r\prime (0)=v_0\text{ and }r(0)=r_0$ given,
where $g$ is the gravitational constant (with a value of 9.8 $\frac{m}{s^2}$). One can immediately reduce this into the one-variable calculus problems:
$x\prime \prime (t)=0,x(0)=x_0,x\prime (0)=v_1;y\prime \prime (0)=-g,y(0)=y_0,y\prime (0)=v_2;z\prime \prime (t)=0,z(0)=z_0,z\prime (0)=v_3$

Reading Questions/Quizzes

1. Find solutions of the above and reconcile this with the discussion as given in the textbook, especially the relation between v0, α in the textbook and v1, v2, v3 here (Note that in the textbook discussion, z-component is not introduced and j is used to represent the vertical direction, which corresponds to setting v3 = 0 so z(t) = z0 for all t here.)
2. Compute ${\int }_0^{\pi }[(-\sin t)i+(\cos t)j+k]dt$
   1. $[(\cos t)i+(\sin t)j+tk{]}_0^{\pi }=(-1-1)i+(0-0)j+(\pi -0)k=-2i+\pi k$