High order derivatives: notes and practice questions

The function $f$ is defined by $f(x) = e^x \sinh x$, where $x \in \mathbb{R}$.

Find the Maclaurin series for $f(x)$ up to and including the $x^3$ term.

Hence, find an approximate value for $\int_0^1 e^{x^2} \sinh(x^2)dx$.

The function $g$ is defined by $g(x) = e^x \cosh x$, where $x \in \mathbb{R}$.

Show that $g''(x) = 2g'(x)$.

Hence, find the values of $g'''(0)$ and $g^{(4)}(0)$.

Using the result from part (c), find the Maclaurin series for $g(x)$ up to and including the $x^4$ term.

Hence, or otherwise, determine the value of $\lim_{x \to 0} \frac{2e^x \cosh x - 2 - 2x - 2x^2}{x^3}$.

(a) Prove by mathematical induction that $\frac{d^n}{dx^n}\left(\frac{x}{1-x}\right) = n!(1-x)^{-(n+1)}$ for $n \in \mathbb{Z}^+$.

(b) Hence or otherwise, determine the Maclaurin series of $f(x) = \frac{x}{1-x}$ in ascending powers of $x$, up to and including the term in $x^4$.

(c) Hence or otherwise, determine the value of $\lim_{x\to0} \frac{\left(\frac{x}{1-x} - x\right)^2}{x^4}$.

A packaging company is designing a new cylindrical can. The can must have a fixed volume of $54\pi$ cm$^3$. The company wants to minimize the amount of material used, which corresponds to minimizing the total surface area of the can.

Let the radius of the can be $r$ cm and its height be $h$ cm.

Show that the total surface area, $S$ cm$^2$, of the can is given by $S = 2\pi r^2 + \frac{108\pi}{r}$.

The total surface area of the can has a local minimum value when $r = a$.

(i) Find an expression for $\frac{dS}{dr}$.

(ii) Hence, find the exact value of $a$.

(i) Find an expression for $\frac{d^2S}{dr^2}$.

(ii) Use the second derivative of $S$ to justify that $S$ is a minimum when $r = a$.

(iii) Find the minimum surface area of the can.

A curve is given by the equation $e^y = \cos(x)$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

(a) Use implicit differentiation to show that $\frac{dy}{dx} = -\tan(x)$.

(b) Show that $\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + 1 = 0$.

(c) Find an expression for $\frac{d^3y}{dx^3}$ in terms of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

(d) Hence, find the Maclaurin series for $y = \ln(\cos(x) )$ up to and including the term in $x^4$.

A chemical reaction is monitored over time. The concentration of a reactant, $C$, in $\text{mol dm}^{-3}$, at time $t$ minutes, is modelled by the function $C(t) = k + Ae^{bt}$.

Initially, at $t=0$, the concentration of the reactant was $15.0\,\text{mol dm}^{-3}$. After $10$ minutes, the concentration had dropped to $7.0\,\text{mol dm}^{-3}$. It is known that the concentration approaches a minimum value of $2.0\,\text{mol dm}^{-3}$ as time increases.

(a) Find the value of

(i) $A$

(ii) $b$

(b) Use the model to estimate the concentration of the reactant after it has been reacting for $25$ minutes.

(c) Write down the equation of the horizontal asymptote of the graph of $C$.

(d) State the meaning of the asymptote found in part (c) in the context of this problem.

(e) Find an expression for the rate of change of the concentration of the reactant after $t$ minutes.

(f) Find an expression for $\frac{d^2C}{dt^2}$.

(g) Hence explain how the concentration of the reactant will vary if the reaction is left for a long time.

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.