Derivatives & integrals of special HL functions (inv / reciprocal trig, log and power): notes and practice questions

The continuous random variable $X$ has probability density function

$f(x) = \begin{cases} \frac{k}{1+4x^2}, & 0 \le x \le \frac{1}{2} \\ 0, & \text{otherwise.} \end{cases}$

(a) Find the value of $k$.

(b) Find $E(X)$.

Find $\int x \sec^2 x \, dx$.

The region $R$ is enclosed by the curve $y = x \sec^2 x$, the $x$-axis, and the lines $x=0$ and $x=\frac{\pi}{3}$. Show that the area of $R$ is $\frac{\pi\sqrt{3}}{3} - \ln 2$.

Find $\frac{dy}{dx}$ for the following functions:

(a) $y=\arcsin(e^{-x})$

(b) $y = e^{-2x} \arccos(x)$

This question asks you to investigate the motion of a buoy bobbing up and down in the water.

A buoy bobs up and down in the water.

A fixed origin $\text{O}$ is the equilibrium position of the buoy (the water level).

The buoy's displacement, $y$ metres, from $\text{O}$ at time $t$ seconds is given by

$y = 6\sin\left(2t + \frac{\pi}{6}\right), \text{ for } 0 \le t \le \pi.$

Determine

the amplitude of the buoy's motion;

the buoy's initial displacement from $\text{O}$;

the value of $t$ when the buoy first passes through $\text{O}$.

Now consider the general case of a buoy bobbing up and down.

The buoy's acceleration is always directed towards a fixed origin $\text{O}$ at its equilibrium position.

The buoy's acceleration, $a$, at a displacement, $y$, from $\text{O}$ satisfies the differential equation

$a = -\omega^2 y, \text{ where } \omega > 0.$

The buoy's displacement, $y$, from $\text{O}$ at time $t$ is given by

$y = H\sin(\omega t + c), \text{ where } t \ge 0,\ H, \omega > 0 \text{ and } -\pi \le c \le \pi.$

By finding expressions for $\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}t^2}$, verify that $y = H\sin(\omega t + c)$ satisfies the differential equation $a = -\omega^2 y$.

Use the chain rule to show that $a = v\frac{\mathrm{d}v}{\mathrm{d}y}$, where $v$ is velocity.

By solving the differential equation, $v\frac{\mathrm{d}v}{\mathrm{d}y} = -\omega^2 y$, show that $v^2 = \omega^2(H^2 - y^2)$.

Hence, or otherwise, find the buoy's maximum speed.

The continuous random variable $Y$ denotes the buoy's displacement, $y$, from $\text{O}$ at time $t$.

The probability density function $f$ of $Y$ is defined by

$f(y) = \begin{cases} \frac{1}{\pi\sqrt{H^2 - y^2}}, & -H < y < H \\ 0, & \text{otherwise.} \end{cases}$

Show that $\mathrm{P}\left(0 \le Y \le \frac{H\sqrt{3}}{2}\right) = \frac{1}{3}$.

For $-H < y < H$, the function $f(y)$ can be expressed in the form $\frac{m}{|v(y)|}$, where $m > 0$ and $v(y)$ is the buoy's velocity at a displacement, $y$, from $\text{O}$.

Find the value of $m$.

Determine $\mathrm{E}(Y)$, justifying your answer.

Interpret the result found in part (f)(i) in the context of the buoy's motion.