In a tidal river the time between high and low tide is 8 hours. At high tide the depth of the water is 20 feet and at low tide the depth of the water is 6 feet.

Question

In a tidal river the time between high and low tide is 8 hours. At high tide the depth of the water is 20 feet and at low tide the depth of the water is 6 feet.

Write the equation of the sinusoidal function where it is low tide at t = 0.

Graph the function for a period of 16 hours. How long will it take to the nearest tenth of an hour for the depth of the water to first reach 12 feet?

Answer 1

(0,6)

(8,20)

(16,6)

y = C*sin( k*pi*x) + 6 since (0,6) is given

C(sin(k*pi*8) + 6 = 20

C(sin(k*pi*8) = 14 <--- label this equation ALPHA

Also,

C(sin(k*pi*16) + 6 = 6

C(sin(k*pi*16) = 0

since C cannot be zero, k*pi*16 must be 180 = pi

k*16 = 1

k = 1/.16

C(sin( (1/16)*pi*8) = 14

C ( sin (pi/2) ) = 14

C = 14

D(t) = 14 sin( (pi/16) t ) + 6 where D is the depth and t is time

12 = 14 sin((pi/16)t) + 6

6/14 = sin((pi/16)t)

arcsin(3/7)*16/pi = t

t = 2.25572...

So roughly 2 hours and 15 minutes, the depth will be 12 feet

Answer 2

In a tidal river the time between high and low tide is 8 hours. At high tide the depth of the water is 20 feet and at low tide the depth of the water is 6 feet.

Write the equation of the sinusoidal function where it is low tide at t = 0.

I'm going to do this problem in the most general way.

The tide oscillates from low tide to high tide and back in 16 hrs so, the period, T=16 hrs.

The depth oscillates from D=6 ft at low time and t=0 to D=20 ft at t=8 hrs.

The amplitude of the sine wave is (20-6)/2=7 ft

D=B+7sin(wt+C)

T=period=16 hrs so f=1/T=(1/16) hr^-1

and w=2πf=2(π)(1/16 hr^-1)=π/8 rad/hr

D=B+7sin(πt/8+C)

Since we know values at two points in time, we can get two equations in two unknowns B and C:

If D=6 at t=0,

[1] 6=B+7sinC

If D=20 at t=8

20=B+7sin[π(8)/8+C]

20=B+7sin(C+π)

20=B+7sinCcosπ+cosCsinπ

[2] 20=B-7sinC

From [1], B=6-7sinC

[1] into [2]

20=(6-7sinC)-7sinC

20=6-14sinC

14=-14sinC

sinC=14/-14=-1

C=-π/2

B=6-7sin(-π/2)=6+7=13

D=13+7sin(πt/8-π/2)

Check: D(0)=13-7=6 ft

D(8)=13+7sin[π(8)/8-π/2]=13+7sin(π/2)=20

Solution: D=13+7sin(πt/8-π/2) feet

Answer 3

14/8=1.75'/hr

to raise 6' it'll take 6/1.75=3.4hrs