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The population of a caribou herd varies sinusoidally with a period of one year. The population is at a minimum at the beginning...Asked by math help plz
The population of a caribou herd varies sinusoidally with a period of one year. The population is at a minimum at the beginning of the year when the population is about 4000 animals. The population is at a maximum of 7500 animals after 6 months. How many caribou are there expected to be at the end of 7 months? Round to the nearest whole number.
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A standing wave fixed at one end in 4th overtone has length 20m. If the temperature is 25 degrees celsius. What is the frequency?
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mathhelper
Using a sine curve,
range = 7500-4000 = 3500
sofar p = 1750sin kt , where t is the number of months
period =12 = 2π/k
k = π/6
then p = 1750sin (πt/6)
we need the max to be 7500, not 1750, so we raise the curve
by 7500-1750 = 5750
giving us:
p = 1750sin(πt/6) + 5750
but we want p to be 4000 when t = 0, so we need to shift horizontally
1750sin (π/6(0-d)) + 5750 = 4000
1750sin (π/6(-d)) = -1750
sin (π/6(-d)) = -1
I know sin(-π/2) = -1
π/6(-d) = -π/2
times 6/π
d = 3
p = 1750sin (π/6(t - 3)) + 5750
so when t = 7
p = 1750sin(π/6(7-3)) + 5750 = 7266
range = 7500-4000 = 3500
sofar p = 1750sin kt , where t is the number of months
period =12 = 2π/k
k = π/6
then p = 1750sin (πt/6)
we need the max to be 7500, not 1750, so we raise the curve
by 7500-1750 = 5750
giving us:
p = 1750sin(πt/6) + 5750
but we want p to be 4000 when t = 0, so we need to shift horizontally
1750sin (π/6(0-d)) + 5750 = 4000
1750sin (π/6(-d)) = -1750
sin (π/6(-d)) = -1
I know sin(-π/2) = -1
π/6(-d) = -π/2
times 6/π
d = 3
p = 1750sin (π/6(t - 3)) + 5750
so when t = 7
p = 1750sin(π/6(7-3)) + 5750 = 7266
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