Transcript AP CALC AB/BC NUMBA 5

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(1400 - 300) = 44
Tangent line is W – 1400 = 44(t – 0)
W = 44t + 1400
When t = ¼, W = 44(1/4) + 1400 = 1411 tons
PART B
• Find d2W/dt2 in terms of W. Use d2W/dt2 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
d2W/dt2 = 1/25(1/25(W-300)) = 1/625(W-300)
We know d2W/dt2 is positive on the interval 0< t <20
because W is increasing on that same interval and W(0) =
1400. 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|1400-300| = 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
THE END