Calculus Question?

Question

W(t)=95sqrt(t) (sin (t/6))^2 =rate at which water is being pumped into the tank in gallons per hour

R(t)=275 (sin (t/3))^2 =rate at which water is being removed from the tank in gallons per hour

the tank holds 1200 gallons at t=0

1.How many gallons of water are in the tank at time t=18?

2.At what time t, from 0 to 18, is the amount of water in the tank at an absolute minimum?

3.For t>18, no water is pumped into the tank, but water continues to be removed at a rate R(t) until the tank becomes empty. Let k be the time at which the tank becomes empty. What is the equation (unsolved) involving an integral expression that can be used to find the value of k?



I don't expect anyone to answer all three answers, but I could use help with any one of the parts, so an answer to one of the questions (with work) would be great. Thanks!

Answer 1

The amount of water in the tank at any given time is

A(t) = 1200 + W(t) - R(t)

1. A(18) = 1200 + W(18) - R(18)

I put that into my calculator and it says 1170.503 gallons

2. We set the derivative of A(t) = 0

A(t) = 0

I did this using the calculator's nderiv function, and got

t = 4.032 or 9.623, but the value of the deriv. changed from neg. to pos. at the first answer, so that one is a minimum, and the second answer is a maximum. So I'm getting t = 4.032 hours.

3. For part 3, I would use my answer from part A. After the 18 hours we have 1170.503 gallons. Then subtracting only:

1170.503 - *integral from 18 to k* (R(t)dt = 0. Sorry, I don't know how to type in an integral symbol.

Answer 2

Madclowndisease: age 16
math experience: HS Pre-calculus

Answers:
P.S. this is a bullshit problem because when dealing with any trig functions your going to get unsteady outputs.

1. this is a simple calculator problem.
1200-(W(x)-R(x))
x=18
Ans. 1200-13.44=1186.56

2.Use the vertex formula for binomial equations and you have your answer:

Ans: I used a graphing calculator quick so:
approximately: x=4

3.for your equation thats simple -1200=(W(x)-R(x))
but to your dismay when dealing with sin the highest absolute value number you can get for sine is 1(-1,+1)
so that means the lowest you can go is within the first 18 hrs
but due to the inverse variation of the sine function being squared you will OVERFILL at about t=160.48(1199.09) and the again at t=178.28(1228)

All of this may be wrong because i just used the equation YOU gave me. I would not rely on trig functions when dealing with water pumps.