Question

Calc III Mini-Tutorial: Integrals of Vectors + Projectile Motion

Answer 1

\({\bf{Integration:}}\) basically follows the same principles as scalar functions

\(\int\limits_{}^{}r(t)dt = R(t)+c\)
for vector function r(t), indefinite integral
\(\int\limits_{a}^{b}r(t)dt=(\int\limits_{a}^{b}f(t)dt )i + (\int\limits_{a}^{b}g(t)dt )j + (\int\limits_{a}^{b}h(t)dt )k\)
for definite integrals

special note: often times you will be given the second derivative of a function (such as acceleration) and asked to find the original function, given initial conditions. in that case each integration will produce a C-value that must be taken into consideration before taking the next integral.

Answer 2

\({\bf{Projectile~Motion:}}\)
vector v0 = (|v0|cos(theta))i + (|v0|sin(theta))j
where v0 is the initial velocity given to the object and theta is the initial angle the object is thrown at

after the object is launched the only force acting upon it is gravity (welltechnicallyairresistancetoobutthatsreallycomplicatedtoaccountfor)

so d^2r/dt^2 = -gj

assuming initial position = r0 and initial velocity = v0, taking the integral of this gives us

dr/dt = (-gt)j + v0
and r(t) = -(1/2)gt^2 + v0t + r(0)

you can also take this expression: vector v0 = (|v0|cos(theta))i + (|v0|sin(theta))j

and plug this in for v0 to get

-(1/2)gt^2 + (t|v0|cos(theta))i + (t|v0|sin(theta))j + r(0)

so if we wanted to launch, say, a human baby at 1000 m/s at an angle of 89 degrees, we can plug in 1000 for v0 and 89degrees for theta, pick a time and find r(t) at that time

Answer 3

we can re-write y as a function of x instead of t

x = vcos(theta) * t
t = x/(vcos(theta))

if y = -(1/2)gt^2 + (t|v0|sin(theta))j

then y = -(1/2)g(x/(vcos(theta))^2 + xtan(theta), remembering that sin(theta)/cos(theta) = tan(theta), and that t|v0| = x

Answer 4

\({\bf{Supplementary~Equations:}}\) all of these equations can be derived from the basic kinematics equations

max height = (v0 sin(theta))^2/(2g)
flight time: t = (2v0sin(theta))/g
range = (v0)^2/g * sin(2theta)

starting with flight time, you can just set max height equal to 0 and solve for t (make sure to consider nontrivial t)

0 = v_0 sin(theta) t - (1/2)gt^2
dividing both sides by t, you can do this as long as you are taking the nontrivial value of t rather than t = 0

v_0sin(theta) = (1/2)gt

t = (2v0sin(theta))/g

for range simply plug that t value into the x-velocity equation

r = v0 cos(theta) * t = [v0 cos(theta) * t](2v0sin(theta))/g
remembering the identity 2(cos(theta) sin(theta) = sin(2theta)
r = (v0)^2/g * sin(2theta)

for the max height equation start w/ the kinematic equation

vf^2 = v_0^2 - 2gh

set vf = 0, solve for h, then let v_0 be the vertical component v_0 sin(theta)

Answer 5

Source material is section 13.2 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.