AP Calculus BC Unit 7 Differential Equations Questions

Practice AP Calculus BC Unit 7 questions on verifying, estimating, solving, and interpreting first-order differential equations.

Covered syllabus points

7.1 Modeling Situations with Differential Equations 7.2 Verifying Solutions for Differential Equations 7.3 Sketching Slope Fields 7.4 Reasoning Using Slope Fields 7.5 Approximating Solutions Using Euler’s Method 7.6 Finding General Solutions Using Separation of Variables 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables 7.8 Exponential Models with Differential Equations 7.9 Logistic Models with Differential Equations

[Maximum number: 1]

As an ice block melts, the rate at which its mass, M, decreases is directly proportional to the square root of the mass. Which equation describes this relationship?

$\sqrt{M(t)}=k t$

$\frac{d M}{d t}=k \sqrt{t}$

$\frac{d M}{d t}=k \sqrt{M}$

$\frac{d M}{d t}=\frac{k}{\sqrt{M}}$

[Maximum number: 5]

Consider the differential equation $\frac{d y}{d x}=2 x-y$ .

(a)

On the axes provided, sketch a slope field for the given differential equation at the six points indicated.

Figure for Question (a) — AP Calculus BC

[ 2 ]

(b)

Find the values of the constants m and b for which y=m x+b is a solution to the differential equation.

[ 3 ]

$y=m x+b \Rightarrow \frac{d y}{d x}=\frac{d}{d x}(m x+b)=m$

\begin{aligned} & 2 x-y=m \\ & 2 x-(m x+b)=m \\ & (2-m) x-(m+b)=0 \\ & 2-m=0 \Rightarrow m=2 \\ & b=-m \Rightarrow b=-2 \end{aligned}

Therefore, m=2 and b=-2.

2:\arrayl1: slopes where x=0
1: slopes where x=1array.

2:\arrayl1: d^2 yd x^2
1: concave up with reason array.

2:\arrayl1: considers.d y/d x|_(x, y)=(2,3)
1: conclusion with justification array.

3:\arrayl1: d/d x(m x+b)=m
1: 2 x-y=m
1: answer array.

[Maximum number: 9]

The depth of seawater at a location can be modeled by the function H that satisfies the differential equation $\frac{d H}{d t}=\frac{1}{2}(H-1) \cos \left(\frac{t}{2}\right)$ , where H(t) is measured in feet and t is measured in hours after noon ( t=0 ). It is known that H(0)=4.

(a)

A portion of the slope field for the differential equation is provided. Sketch the solution curve, y=H(t), through the point (0, 4).

[ 1 ]

(b)

[ 3 ]

For 0<t<5, it can be shown that H(t)>1. Find the value of t, for 0<t<5, at which H has a

critical point. Determine whether the critical point corresponds to a relative minimum, a relative

maximum, or neither a relative minimum nor a relative maximum of the depth of seawater at the

location. Justify your answer.

Because H(t)>1, then $\frac{dH}{dt}=0$ implies $\cos \left(\frac{t}{2}\right)=0$ .
This implies that $t=\pi$ is a critical point.}

Considers sign of
$\frac{d H}{d t}$

1 point

Identifies $t=\pi$

1 point

For $0<t<\pi, \frac{d H}{d t}>0$ and for $\pi<t<5, \frac{d H}{d t}<0$ . Therefore,
$t=\pi$ is the location of a relative maximum value of H.

Answer with justification

1 point

Scoring notes:

- The first point is earned for considering $\frac{d H}{d t}=0, \frac{d H}{d t}>0, \frac{d H}{d t}<0, \cos \left(\frac{t}{2}\right)=0, \cos \left(\frac{t}{2}\right)>0$ ,

or $\cos \left(\frac{t}{2}\right)<0$ .

- The second point is earned for identifying $t=\pi$ , with or without supporting work. A response

may consider H=1 or t=1 as potential critical points without penalty.

- The third point cannot be earned without the first point. The third point is earned only for a correct

justification and a correct answer of "relative maximum."

- The justification can be shown by determining the sign of $\frac{d H}{d t}$ (or $\cos \left(\frac{t}{2}\right)$ ) at a single value in

$0<t<\pi$ and at a single value in $\pi<t<5$ . It is not necessary to state that $\frac{d H}{d t}$ does not change

sign on these intervals.

- The third point can also be earned by using the Second Derivative Test. For example:

Therefore, $t=\pi$ is the location of a relative maximum value of H.

(c)

Use separation of variables to find y=H(t), the particular solution to the differential equation $\frac{d H}{d t}=\frac{1}{2}(H-1) \cos \left(\frac{t}{2}\right)$ with initial condition H(0)=4.

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

Figure for Question (c) — AP Calculus BC

[ 5 ]

Separate variables:
$\frac{dH}{H-1}=\frac{1}{2}\cos\left(\frac{t}{2}\right)dt.$
Integrating gives
$\ln|H-1|=\sin\left(\frac{t}{2}\right)+C.$
Using H(0)=4, $\ln 3=C$ . Since H>1, |H-1|=H-1, so
$\ln(H-1)=\sin\left(\frac{t}{2}\right)+\ln 3.$
Therefore
$H(t)=1+3e^{\sin(t/2)}.$

[Maximum number: 2]

Let y=f(x) be the particular solution to the differential equation $\frac{d y}{d x}=(3-x) y^{2}$ with initial condition f(1)=-1.

Use Euler's method, starting at x=1 with two steps of equal size, to approximate f(1.4).

Show the work that leads to your answer.

Use Euler's method, starting at x=1 with two steps of equal size, to approximate f(1.4). Show the

work that leads to your answer.

\begin{aligned} & f(1.2) \approx f(1)+\left.(1.2-1) \cdot \frac{d y}{d x}\right|_{(1,-1)} \\ & =-1+0.2(2)=-0.6 \end{aligned}

\begin{aligned} & f(1.4) \approx f(1.2)+\left.(1.4-1.2) \cdot \frac{d y}{d x}\right|_{(1.2,-0.6)} \\ & \approx-0.6+0.2(3-1.2)(-0.6)^{2}=-0.4704 \end{aligned}

An approximation of f(1.4) is -0.47.

Scoring Notes for Part D

- To earn P8, a response must demonstrate the first step of Euler's method, with the correct initial

condition, correct step size, and correct (or imported) expression for the derivative.

Note: Any subsequent error in simplification or rounding will not affect the scoring for P8.

- The two steps of Euler's method may be explicit expressions or may be presented in a table. For

example:

$x$	$y$	$\frac{d y}{d x} \cdot \Delta x\left(\right.$ or $\left.\frac{d y}{d x} \cdot 0.2\right)$
1	-1	0.4
1.2	-0.6	0.1296
1.4	-0.4704

Note: In the presence of a correct answer, a table does not need to be labeled to earn both P8 and P9.

In the presence of no answer or an incorrect answer, such a table must be correctly labeled to earn

P8.

- A response of $-0.6+0.2(3-1.2)(-0.6)^{2}$ earns P9, regardless of any subsequent errors in

simplification or rounding.

- A response that imports an incorrect value for $f^{\prime}(1)$ from part A or part B is eligible to earn P9 with

All question bank results loaded

Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (c)

Question 4