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Hitungan: Untuk f (t) = (sint-cost, t ^ 2) berapakah jarak antara f (2) dan f (5)? - 2020

Anonim

Menjawab:

# sqrt (2-2cos7) #

Penjelasan:

# f (t) = (sint-cost, t ^ 2) #

# f (2) = (sin2-cos2,4) #
# f (5) = (sin5-cos5,25) #

Menurut Pythagoras, perbedaan antara titik-titik ini adalah;
jarak = # sqrt ((sin5-sin2) ^ 2 + (- cos5 - cos2) ^ 2) #
= # sqrt ((sin5-sin2) ^ 2 + (- cos5 + cos2) ^ 2) #
= # sqrt (sin ^ 2 5-2sin5sin2 + sin ^ 2 2 + cos ^ 2 5-2cos5cos2 + cos ^ 2 2) #
= # sqrt (sin ^ 2 5 + cos ^ 2 5-2sin5sin2-2cos5cos2 + cos ^ 2 2 + sin ^ 2 2) #
= # sqrt (1-2sin5sin2-2cos5cos2 +1) #
= # sqrt (2-2sin5sin2-2cos5cos2) #
= # sqrt (2-2 (sin5sin2 + cos5cos2)) #
= # sqrt (2-2cos (5 + 2)) #
= # sqrt (2-2cos (7)) #