AP Calculus BC Unit 7: Differential Equations

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

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.

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

Solution curve 1 point

Scoring notes:

- The solution curve must pass through the point (0,4), extend to at least t=4.5, and have no

obvious conflicts with the given slope lines.

- Only portions of the solution curve within the given slope field are considered.

Total for part (a)

1 point

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.

Total for part (b) 3 points

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
$$H(t)=1+3e^{\sin(t/2)}.$$