AP Calculus BC 7.4 Differential Equation Solutions Overview

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