Home / AP Calculus AB : 7.4 Reasoning Using Slope Fields- Exam Style questions with Answer- FRQ

AP Calculus AB : 7.4 Reasoning Using Slope Fields- 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:

Consider the differential equation \(\frac{dy}{dx}=e^{y}(3x^{2}-6x).\) Let y = f (x) be the particular solution to the differential equation that passes through (1 , 0) .
(a) Write an equation for the line tangent to the graph of f at the point (1, 0) . Use the tangent line to approximate f (1, 2).
(b) Find y = f(x), the particular solution to the differential equation that passes through (1, 0). 

▶️Answer/Explanation

Ans:

(a)

\(\frac{dy}{dx}= e^{0}(3-6)\)

\(\frac{dy}{dx}= -3\)

y – 0 = -3 (x-1)

y = -3x + 3

y (1.2) = -3(1.2) + 3

y = -.6

(b)

\(\int \frac{1}{e^{y}}dy = \int 3x^{2}-6x dx\)

e-y                       -e-y = x3 – 3x2 + c

                          e-y = -x3 – 3x2 + c

                         -y = -en ( -x3 – 3x2 + c)

                        0 = – en(-(1)3 + 3 + c)

                        0 =  – en (-1+3+c)

                       0 = – en (2 + c)

                   2 + c = 1 

                    c = -1

y = – en ( -x3 + 3x2 + c)

Question:

Consider the differential equation \(\frac{dy}{dx}= (3-y)cos x.\) Let y = f(x) be the particular solution to the differential equation with the initial condition f(0) = 1 . The function f is defined for all real numbers.
(a) A portion of the slope field of the differential equation is given below. Sketch the solution curve through the point (0, 1).

(b) Write an equation for the line tangent to the solution curve in part (a) at the point (0, 1) . Use the equation to approximate f(0.2) .
(c) Find y = f (x), the particular solution to the differential equation with the initial condition f(0) = 1 .

▶️Answer/Explanation

Ans:

(a)

(b) 

\(\frac{dy}{dx}= (3-y)cos x = m\)

\(\frac{dy}{dx}= (3-1)cos (0) = m\)

\(m = 2\)

\(y – 1 = 2(x-0)\)

\(y -1 = 2x\)

\(y = 2x + 1\)

\(y – 1 = 2(0.2 – 0)\)

\(y – 1 = 0.4\)

\(y = 1.4\)

\(f(0.2) ≈ 1.4\)

(c)

\(\int \frac{dy}{3-y}=\int cos x dx\)                    \(3 – y = u\)

\(-\int \frac{du}{u}=sin x + c\)                           \(du = -dx\)

\(- In |u| = sin x  + c\)

\(- In |3-y| = sin x + c\)                                   \(- In |3-y| = sin x – In 2\)

\(- In |3-1| = sin (0) + c \)                                       \(In |3-y| = In 2 – sin x\)

\(- In |2| = 0 + c\)                                                          |3 – y| = eIn2 – sin x

\(c = – In 2\)                                                                  \(\left | 3-y \right |= \frac{e^{In2}}{e^{sinx}}\)

                                                                                    |3-y| = 2e-sinx

                                                                                    3-y =  2e-sinx

                                                                                      -y = 2e-sinx 3

                                                                                      y = -2e-sinx + 3

Question

Consider the differential equation \(\frac{dy}{dx}=\frac{x}{y} where y\neq 0\)

(a) The slope field for the given differential equation is shown below. Sketch the solution curve that passes through the point ( 3,- 1 ),  and sketch the solution curve that passes through the point ( 1, 2) .(Note: The points ( 3, -1) and (  1, 2) are indicated in the figure.)

(b) Write an equation for the line tangent to the solution curve that passes through the point (  1, 2) .
(c) Find the particular solution y = f ( x) to the differential equation with the initial condition f ( 3 ), = -1 and state its domain

▶️Answer/Explanation

.

(a)\(g(1)=\int_{1}^{1}f(t)dt=0 \)and \(g'(1)=f(1)=0 \)(b) Since g'(x) = f(x), g is increasing on the interval 1≤ x ≤ because \(f(x) > 0 for 1 < x < \pi.\) (c) For \(0 < x < 2\pi, g'(x) = f(x) = 0 \)when x = 1, \pi. g’ f changes from negative to positive only at x = 1. The absolute minimum must occur at x = 1 or at the right endpoint. Since g(1) = 0 and \(g(2\pi)=\int_{1}^{2\pi}f(t)dt=\int_{1}^{\pi}f(t)dt+\int_{\pi}^{2\pi}f(t)<0\) by comparison of the two areas, the absolute minimum occurs at \(x = 2π. \)(d) Yes, the graph of g is tangent to the x-axis at x = 1 since g(1) = 0 and g'(1) = 0.

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