Here are the answers to Part B of our AP Calculus AB free response questions.

**Question #3**

**Consider the differential equation $\dfrac{dy}{dx} = x^3 − 2y^2$.**

**(a)** Sketch the slope field for the differential equation on the graph provided below.

**(b)** Find $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$. Determine the concavity of the possible solution curves to the differential equation in the second quadrant based on your answer.

**(c)** Given the differential equation $\dfrac{dy}{dx} = 2x^3y^2$, solve for $y$ in terms of $x$.

**Answer to Question #3:**

**(a)**

**(b)**

$\dfrac{d^2 y}{dx^2}$ $= 3x^2 – 4y(x^3 – 2y^2)$

$= 3x^2 – 4x^3y + 8y^3$

In the second quadrant, $x < 0$, and $y > 0$. Therefore $3x^2 – 4x^3y + 8y^3 > 0$ in quadrant II. Solution curves are all concave up.

**(c)**

$\dfrac{1}{y^2}dy = 2x^3 dx$

$-y^{-1} = \dfrac{1}{2} x^4 + C$

$y = \dfrac{1}{-\frac{1}{2}x^4 + C}$

**Question #4**

**Tim and John get in an argument over the limit below.**

$\lim\limits_{x\to 0} \dfrac{2x}{\cos(x)}$

**Tim says the limit exists while John says the limit does not exist.**

**(a)** Identify derivative of $x^2$ using the limit definition.

**(b)** Identify the derivative of $\sin(x)$ using the limit definition.

**(c)** Identify $\lim\limits_{x\to 0} \dfrac{2x}{\sin(x)}$

**(d)** Who is correct? Give a reason with your answer.

**Answer to Question #4:**

**(a)**

$\dfrac{d(x^2)}{dx}$ $= \lim\limits_{∆h\to 0} \dfrac{(x + ∆h)^2 − x^2}{∆h}$

$= \lim\limits_{∆h\to 0} \dfrac{2x∆h + ∆h^2}{∆h}$

$= \lim\limits_{∆h\to 0} 2x + ∆h = 2x$

**(b)**

$\dfrac{d(\sin(x))}{dx}$ $= \lim\limits_{∆h\to 0} \dfrac{\sin(x + ∆h) − \sin(x)}{∆h}$

$= \lim\limits_{∆h\to 0} \dfrac{\sin(x)\cos(∆h) + \cos(x)\sin(∆h) − \sin(x)}{∆h}$

$= \lim\limits_{∆h\to 0} (−\sin(x))\left(\dfrac{1 − cos(∆h)}{∆h}\right)$ $+ \lim\limits_{∆h\to 0} \cos(x) \left(\dfrac{\sin(∆h)}{∆h}\right)$

$= \lim\limits_{∆h\to 0} \dfrac{1 − \cos(∆h)}{∆h}$ $= 0,\lim\limits_{∆h\to 0} \dfrac{\sin(∆h)}{∆h}$ $= 1$

$\dfrac{d(\sin(x))}{dx} = \cos(x)$

**(c)**

$\lim\limits_{x\to 0} \dfrac{x^2}{\sin(x)} \to \dfrac{0}{0}$

$\lim\limits_{x\to 0} \dfrac{2x}{\cos(x)}$ $= \dfrac{0}{1} = 0$

**(d)**

Tim is correct, as the limit is equal to $0$.

**Question #5**

**A particle moves along the $x$-axis over time, $t$, in seconds by the equation below. $t ≥ 0$**

$s(t) = t^3 − 6t^2 + 9t − 18$

**(a)** Find the velocity and acceleration functions for the particle.

**(b)** When is the particle at rest?

**(c)** When is the particle moving in the forward direction?

**(d)** When is the particle increasing in speed?

**(e)** When is the particle moving the fastest in the backwards direction?

**(f)** Find the total distance travelled by the particle after $5$ seconds.

**Answer to Question #5:**

**(a)**

$v(t) = 3t^2 − 12t + 9$

$a(t) = 6t – 12 = 6(t − 2)$

**(b)**

$v(t) = 0 = 3(t − 3)(t − 1)$

$t = 3$, $t = 1$

**(c)**

$v(t)$ is positive when particle moves forward

$0 ≤ t< 1$, $3 < t < \infty$

**(d)**

Both $v(t)$ and $a(t)$ must have same sign.

$1 < t < 2$ Both negative

$3 < t < \infty$ Both positive

**(e)**

$a(t) = v'(t)$

$a(t) = 0$,$t = 2$ Critical Point

$a(t)$ changes from negative to positive about $t = 2$, therefore $t = 2$ is a local minimum on $v(t)$.

**(f)**

Distance $= \displaystyle\int_0^5 |v(t)|dt$

$\displaystyle\int_0^5 v(t)dt − 2\displaystyle\int_1^3v(t)dt = 28$ units

**Question #6**

**Use the table below to answer all parts of the question. Assume that $f(x)$ is continuous and differentiable on all $x$. $f'(x) = 0$ only at one value for all $x$.**

**(a)** Identify the critical point(s) and point(s) of inflection for $f(x)$. Give a reason for each answer.

**(b)** Identify any local extrema for $f(x)$, stating reasons for each extremum identified.

**(c)** Using a tangent line approximation from $f(1)$, identify an appropriate approximation for b.

**(d)** How many roots does $f(x)$ have on $−4 ≤ x ≤ 13$? State a reason for your answer.

**Answer to Question #6:**

**(a)**

Critical Point: $(7,9)$

$f'(7) = 0$

Point of Inflection: $(1,0)$

$f”(1) = 0$

**(b)**

Local maximum at $x = 7$ as $f'(x)$ changes from positive to negative about $x = 7$.

**(c)**

$f'(1) = 1$, $f(1) = 0$

$y − 0 = (1)(x − 1)$

$y ≈ (3 − 1) = 2$

$b ≈ 2$

**(d)**

$f(x)$ has at least two roots on $−4 ≤ x ≤ 13$ due to intermediate value theorem, which guarantees the existence of one root between $x = 7$ and $x = 13$ as $f(x)$ is continuous on the domain and $f(x)$ changes from positive to negative. $x = 1$ is a root as well, as $f(1) = 0$.