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

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