AP CALC AB/BC NUMBA 5 BY: MAR AND BIG D

AT the beginning of 2010, a landfill contained 1400 tons of solid waste. The increasing function W models the total amount of solid waste stored at the landfill. Planners estimate that W will satisfy the differential equation dW/dt=1/25(W-300) for the next 20 years. W is measured in tons, and t is measured in years from the start of 2010.

PART A Use the line tangent to the graph of W at t = 0 to approximate the amount of solid waste that the landfill contains at the end of the first 3 months of 2010 (time t = ¼) The line tangent to W is where the point (0, W(0)) = (0,1400). dW/dt(0) = 1/25( ) = 44 Tangent line is W – 1400 = 44(t – 0) W = 44t When t = ¼, W = 44(1/4) = 1411 tons

PART B Find d 2 W/dt 2 in terms of W. Use d 2 W/dt 2 to determine whether your answer in part (a.) is an underestimate or an overestimate of the amount of solid waste that the landfill contains at time t = 1/4 d 2 W/dt 2 = 1/25(1/25(W-300)) = 1/625(W-300) We know d 2 W/dt 2 is positive on the interval 0< t <20 because W is increasing on that same interval and W(0) = Therefore W is concave up on the interval 0< t <20 and the answer to part (a) is an underestimate

PART C Find the particular solution W = W(t) to the differential equation dW/dt = 1/25(W-300) with initial condition W(0) = 1400 Separate the variables to get dW/W-300 = dt/25. When you integrate that you get ln|W-300| = t/25 + C You are given the point (0,1400) so when you plug those values in to the equation you get ln| | = 0 + C C = ln(1100) The new equation is ln|W-300| = t/25 + ln|1100| Then “e” both sides to get W-300 = e (t/25 + ln(1100)). We can discard the absolute value signs since we know W – 300 >0 from the initial condition Finally you simplify to get W = e (t/25 + ln(1100)) + 300

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