Notes
Motion
- $v(t) = s'(t)$, $a(t) = v'(t) = s''(t)$. Speed is $|v|$.
- At rest when $v = 0$. The object speeds up when $v$ and $a$ have the same sign and slows down when they have opposite signs.
- Net displacement is $s(b) - s(a)$. Total distance requires splitting at direction changes.
Related rates
- Link the changing quantities with an equation, differentiate with respect to time using the chain rule, and substitute the instant's values after differentiating.
- Common equations: $A = \pi r^2$, $\displaystyle V = \frac{4}{3}\pi r^3$, $V = s^3$, and $x^2 + y^2 = L^2$.
Linear approximation
- $L(x) = f(a) + f'(a)(x - a)$ approximates $f$ near $a$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$v = 3t^2 - 3$, so $v(2) = 12 - 3 = 9$.
$v = 4t^3 - 2$ and $a = 12t^2$. So $a(2) = 48$.
The signs are opposite, so the speed is decreasing.
$\displaystyle \frac{dC}{dt} = 2\pi\frac{dr}{dt} = 6\pi$, for every radius.
$L(x) = 8 + 12(x - 2)$, so $L(2.1) = 9.2$.
$\displaystyle \frac{dV}{dt} = 3s^2\frac{ds}{dt} = 3(25)(0.2) = 15$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A particle moves along a line with position $s(t) = t^3 - 9t^2 + 24t$ meters, for $t \geq 0$ seconds.
(a) Find the velocity and acceleration at $t = 1$.
Show the answer and scoring guide
$v(t) = 3t^2 - 18t + 24$ and $a(t) = 6t - 18$. So $v(1) = 9$ m/s and $a(1) = -12$ m/s$^2$.
Check off the points you earned:
(b) When is the particle at rest?
Show the answer and scoring guide
$3t^2 - 18t + 24 = 3(t - 2)(t - 4) = 0$, so $t = 2$ and $t = 4$.
Check off the points you earned:
(c) Is the particle speeding up or slowing down at $t = 5$? Justify.
Show the answer and scoring guide
$v(5) = 3(3)(1) = 9 > 0$ and $a(5) = 12 > 0$. The signs are the same, so it is speeding up.
Check off the points you earned: