Kakšna je trenutna hitrost objekta, ki se giblje v skladu s f (t) = (t ^ 2, tcos (t (5pi) / 4)) pri t = (pi) / 3?

Odgovor:

#v (pi / 3) = 1 / 3sqrt (4pi ^ 2 + 9cos ^ 2 (pi / 12) + pisin ^ 2 (pi / 12) + 6picos (pi / 12) sin (pi / 12)) #

Pojasnilo:

Enačba #f (t) = (t ^ 2; tcos (t (5pi) / 4)) # daje koordinate objekta glede na čas:

#x (t) = t ^ 2 #

#y (t) = tcos (t (5pi) / 4) #

Najti #v (t) # morate najti #v_x (t) # in #v_y (t) #

#v_x (t) = (dx (t)) / dt = (dt ^ 2) / dt = 2t #

#v_y (t) = (d (tcos (t (5pi) / 4)) / dt = cos (t (5pi) / 4) -tsin (t (5pi) / 4) #

Zdaj morate zamenjati # t # z # pi / 3 #

#v_x (pi / 3) = (2pi) / 3 #

#v_y (pi / 3) = cos (pi / 3- (5pi) / 4) -pi / 3 cdot sin (pi / 3- (5pi) / 4) #

# = cos ((4pi-15pi) / 12) -pi / 3 cdot sin ((4pi-15pi) / 12) #

# = cos ((- - 11pi) / 12) -pi / 3 cdot sin ((- 11pi) / 12) #

# = cos (pi / 12) + pi / 3 cdot sin (pi / 12) #

To vem # v ^ 2 = v_x ^ 2 + v_y ^ 2 # najdeš:

#v (pi / 3) = sqrt (((2pi) / 3) ^ 2 + (cos (pi / 12) + pi / 3 cdot sin (pi / 12)) ^ 2) #

# = sqrt ((4pi ^ 2) / 9 + cos ^ 2 (pi / 12) + pi ^ 2/9 cdot sin ^ 2 (pi / 12) + (2pi) / 3 cdot cos (pi / 12) sin (pi / 12)) #

# = 1 / 3sqrt (4pi ^ 2 + 9cos ^ 2 (pi / 12) + pisin ^ 2 (pi / 12) + 6picos (pi / 12) sin (pi / 12)) #