For this problem, it is helpful to use a graphing utility like desmos.com/calcul

Question

For this problem, it is helpful to use a graphing utility like desmos.com/calculator Ian is climbing every day, using a camp at the base of a snowfield. His only supply of water is a trickle that comes out of the snowfield. The trickles dries at night, because the temperature drops and the snow stops melting.

lan entertains himself by using measurements of the water in his cooking pot to model the flow as 2T(t - 16) 4 w(t) = 15 1 + cos where t is the time in hours after midnight and w(t) is the rate at which water drips into his pot, in gallons per hour (a) At 4p Ian puts his empty pot under the trickle. About how long does it take to fill the 2-quart pot? Round your answer to one decimal place. What time is it when the pot has become full? (b) What is the total amount of water that flows out of the snowfield in a single 24-hour day? Write a simplified expression for the exact answer, along with the answer rounded to one decimal place.

Explanation / Answer

w(t) = 15(1 + cos(2*pi(t - 16)/24))

here t is time in hours after midnight, w(t) is rate at which water drips in gallons per hour

a. w(t) = dV/dt [ where V is volume in gallons]

15(1 + cos(2*pi(t - 16)/24)) = dV/dt

dV = 15(1 + cos(2*pi(t - 16)/24))dt

integrating from t = t1 to t = t2

V = 15((t2 - t1) + 24*sin(2*pi(t2 - 16)/24)/2*pi - 24*sin(2*pi(t1 - 16)/24)/2*pi)

so, given t1 = 4 pm = 16 hrs

t2 = ?

V = 2 quart = 0.5 gallons

0.5 = 15((t2 - 16) + 24*sin(2*pi(t2 - 16)/24)/2*pi - 24*sin(2*pi(16 - 16)/24)/2*pi)

0.5/15 + 16 = t2 + 3.81971*sin(2*pi(t2 - 16)/24)

t2 + 3.81971*sin(2*pi(t2 - 16)/24) - 16.0333 = 0

so solving

t2 = 16.01667 hrs

so time taken = t2 - t1 = 0.01667 hours = 1.002 minutes

the pot is full by 4:01 pm

b.

V = 15((t2 - t1) + 24*sin(2*pi(t2 - 16)/24)/2*pi - 24*sin(2*pi(t1 - 16)/24)/2*pi)

let t1 = 0

t2 = 24

then V = 15(24 + 24*sin(2*pi(24 - 16)/24)/2*pi - 24*sin(2*pi(0 - 16)/24)/2*pi)

V = 15(24 + 12*sin(2*pi/3)/pi + 12*sin(4*pi/3)/pi) = 15(24 + 12*sin(120)/pi + 12*sin(240)/pi) = 360 gallons

c. V = 15((t - t0) + 24*sin(2*pi(t - 16)/24)/2*pi - 24*sin(2*pi(t0 - 16)/24)/2*pi)

d. t1 = 5 hours

V = 0.5 = V = 15((t2 - 5) + 24*sin(2*pi(t2 - 16)/24)/2*pi - 24*sin(2*pi(5 - 16)/24)/2*pi)

0 = t2 + 3.8197*sin(0.261799(t2 - 16)) - 4.84142

solving t2 = 8.28065 hours

time taken = t2 - t1 = 3.28065 hours

time in the day = 8:16:50.34 in the day when the pot fills up

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