Unit 4: Contextual Applications of Differentiation Flashcards
Position Symbol
s(t)
Velocity Symbol
v(t) = s’(t)
Acceleration Symbol
a(t) = v’(t) = s(t)
Particle changing direction
velocity changes sign –> find v(t) = 0
Velocity quantity
vector quantity: magnitude and direction
Velocity and Acceleration signs
Same sign = increasing speed
Different signs = decreasing speed
+ = right
- = left
Distance vs. Displacement
Distance = the sum of the distances traveled on each subinterval without a change of direction [find where the v(t) changes sign, absolute values of the differences and add them up]
Displacement = the difference between its terminal and initial positions.
Describing the particle’s motion
state the interval, the direction, incr./decr. speed, and the reason.
ex. for 0<t<2, moving right, speed decreases because v(t)>0 and a(t)<0.
Related Rates Tip
never substitute for things that change until after you take derivatives!
related rates triangles
x^2 + y^2 = z^2