AP Calculus AB : 4.6 Approximating Values of a Function Using Local Linearity and Linearization- Exam Style questions with Answer- FRQ

Question:

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 \(\frac{dW}{dt}=\frac{1}{25}(W-300)\) for the next 20 years. W is measured in tons, and t is measured in years from the start of 2010.
(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 \(\left ( time  t = \frac{1}{4} \right )\).
(b) Find \(\frac{d^{2}W}{dt^{2}}\) in terms of W. Use \(\frac{d^{2}W}{dt^{2}}\) 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  \(t = \frac{1}{4}\).
(c) Find the particular solution W  = W(t) to the differential equation \(\frac{dW}{dt}=\frac{1}{25}(W-300)\) with initial condition W(0) = 1400.

▶️Answer/Explanation

Ans:

(a)

at                 t = 0,                 W = 1400

so

\(\frac{dW}{dt}|_{t=0} = \frac{1}{25}(1100)=44 tons/ year = W'(0)\)

W(x + a) ≈ W(λ) + aW'(x)

\(W\left ( 0+\frac{1}{4} \right )\approx W(0)+\frac{1}{4}W'(0)\)

\(W\left ( \frac{1}{4} \right )\approx 1400 + 11\)

\(W\left ( \frac{1}{4} \right )\approx 1411 tons\)

There will be about 1411 tons of landfill after 3 months.

(b)

\(\frac{dW}{dt}=\frac{1}{25}W-12\)        

\(\frac{d^{2}W}{dt^{2}}=\frac{1}{25}\frac{dW}{dt}\)

\(\frac{d^{2}W}{dt^{2}}=\frac{1}{625}(W-300)\) 

so  \(\frac{d^{2}W}{dt^{2}}\) is always  positive

b/c  W > 300.

The answer in part a is an under estimate because since \(\frac{d^{2}W}{dt^{2}}\) is always positive for t > 0, the graph of W is  concave up, so  the linearization of W is an underestimate. 

(c)

\(\frac{dW}{dt}=\frac{1}{25}(W-300)\)

\(\int \frac{dw}{W-300}=\int \frac{1}{25}dt\)

\(e^{\xi n}\left | W-300 \right |=\frac{1}{25}t+c\)

can remove abs value 

b/c W > 300. (iner fn)

[W – 300] = (e 1/25 t

Initial condition w(0) = 1400

1400-300 = C e0

1100 = C

W – 300 = 1100 e 1/25 t

W = W(t) = 1100 e 1/25 t + 300

Question:

Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30 , the amount of grass clippings remaining in the bin is modeled by A(t) = 6.687 (0.931)t , where A (t) is measured in pounds and t is measured in days.
(a) Find the average rate of change of A (t) over the interval 0 ≤ t ≤ 30 . Indicate units of measure.
(b) Find the value of A'(15) . Using correct units, interpret the meaning of the value in the context of the problem.
(c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0 ≤ t ≤ 30.
(d) For t > 30,  L(t), the linear approximation to A at t = 30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.

▶️Answer/Explanation

Ans:

(a)

\(\frac{A(30)-A(0)}{30-0}=\frac{-5.904}{30}\approx -.197\frac{pounds}{day}\)

(b)

A'(t) = – .478(.931)t

A'(15) = -.164    \(\frac{pounds}{day}\)

The amount of grass clippings in the bin is decreasing (decomposing) at a rate of .164 pounds per day at time = 15 days

(c)

Average amount \(=\frac{1}{30}\int_{0}^{30}A(t)dt = 2.75263511\)

A(t)  = 2.75263511 = 6.687 (.931)t

this occurs at t = 12.414 days

(d)

L(t) is the tangent line to A(t) at t = 30

A(30) = .783 ⇒ (30, .783) = (t, A(t))

A'(30) = -.056               let  -.056  = M

(y-y1) = m (x, x1) so for this problem, 

(A(t) – .783) = -.056(t-30)

When there are .5 pounds of grass, A(t) = .5, 

(.5-.783) = -.056(t-30)

t = 35.054 days

Question

 The slope of a function f (x) at any point (x, y) is \(\frac{x-3}{x^{2}-3x-4}\) The point \(\left ( 5,\frac{4}{5}\ln 6 \right )\)  is on the graph of f (x).
(A) Write an equation of the tangent line to the graph of f (x) at x = 5.
(B) Use the tangent line in part (a) to approximate f (4.5) to the nearest thousandth.
(C) Find the antiderivative of \(\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{x-3}{x^{2}-3x-4}\) with the condition \(f(5)=\frac{4}{5}\ln 6\).
(D) Use the result of part (c) to find f (4.5) to the nearest thousandth.

▶️Answer/Explanation

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