Namemath111861 Bonus: Complete All Work On A Separate Sheet

Namemath111861 Bonuscomplete All Work On A Separate Sheet Of Paper

Namemath111861 Bonuscomplete All Work On A Separate Sheet Of Paper

Complete all work on a separate sheet of paper; write in pencil. Make sure to include all work for credit. Each problem is worth one point. Due date: January 27, 2017.

Problem 1: A mass hanging on a spring is set in motion and its velocity is given by v(t) = 2π cos(πt), for t ≥ 0. Assume the positive direction is up and that s(0) is known. Determine the position function s(t) and find the first three times the mass reaches its low points.

Problem 2: A 1000-liter tank has 200 liters of water initially. It begins to be filled at a rate modeled by Q′(t) = 3√t liters per minute, where t is in minutes. Find a function modeling the total amount of water in the tank and determine how long it will take to fill the tank completely.

Paper For Above instruction

The problems presented involve application of calculus concepts such as differential equations, integration, and motion analysis. To approach the first problem, we need to find the position function s(t) based on the velocity function v(t) = 2π cos(πt). Recognizing that velocity is the derivative of position with respect to time, v(t) = s'(t), we can find s(t) by integrating v(t).

Integrating v(t) gives us:

s(t) = ∫ v(t) dt = ∫ 2π cos(πt) dt.

Using substitution, let u = πt, so du = π dt, hence dt = du/π. Substituting, we get:

s(t) = 2π ∫ cos(u) (du/π) = 2π (1/π) ∫ cos(u) du = 2 ∫ cos(u) du = 2 sin(u) + C.

Reverting back to t, s(t) = 2 sin(πt) + C. To find the constant C, we use the initial condition s(0). Assuming s(0) = s₀ (a known value), then at t = 0:

s(0) = 2 sin(0) + C = 0 + C = s₀, so C = s₀.

Thus, the position function is:

s(t) = 2 sin(πt) + s₀.

Next, to find the times when the mass reaches its low points, we analyze s(t). The low points occur when s(t) attains a minimum, which happens when sin(πt) = -1. Since sin(πt) reaches -1 at (2n+1)π/2, but here it is scaled to πt, the sine function equals -1 at:

πt = 3π/2 + 2πn, where n is an integer.

Solving for t:

t = (3/2) + 2n, where n = 0, 1, 2,...

Therefore, the first three times the mass reaches its low points are at:

  • t = 1.5 minutes (n=0)
  • t = 3.5 minutes (n=1)
  • t = 5.5 minutes (n=2)

For Problem 2, we are given that the water flow rate into the tank is Q′(t) = 3√t liters per minute, and the initial volume is 200 liters. To find the total amount of water in the tank at time t, we integrate Q′(t) with respect to time and add the initial volume:

V(t) = ∫ Q′(t) dt + V₀ = ∫ 3√t dt + 200.

Since √t = t^{1/2}, the integral becomes:

V(t) = 3 ∫ t^{1/2} dt + 200.

Integrating t^{1/2} gives:

∫ t^{1/2} dt = (2/3) t^{3/2} + C.

Thus, the volume function is:

V(t) = 3 * (2/3) t^{3/2} + 200 = 2 t^{3/2} + 200.

To find how long it takes to fill the tank (which has a capacity of 1000 liters), set V(t) = 1000:

2 t^{3/2} + 200 = 1000.

Subtract 200 from both sides:

2 t^{3/2} = 800.

Divide both sides by 2:

t^{3/2} = 400.

Now, solve for t:

t^{3/2} = 400.

Raise both sides to the power of 2/3:

t = (400)^{2/3}.

Calculating, 400^{2/3} = (400^{1/3})^2. The cube root of 400 is approximately 7.368, so:

t ≈ (7.368)^2 ≈ 54.27 minutes.

Therefore, it will take approximately 54.27 minutes to fill the tank completely.

This application demonstrates the integration of rate functions to determine total accumulation over time, a fundamental concept in calculus with broad applications in engineering, environmental science, and resource management.

References

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