QCAA Mathematical Methods Further applications of differentiation

$y = x^3 - 3x^2$
$\frac{dy}{dx} = 3x^2 - 6x$
$\frac{d^2x}{dy^2} = 6x - 6$

$\frac{d^2y}{dx^2} = 6x - 6$
When $x = -1$
$\frac{d^2y}{dx^2} = 6 \times -1 - 6$
$= -12$

$\frac{d^2y}{dx^2} = 0$
$6x - 6 = 0$
$6x = 6$
$x = 1$
Substitute into the graph equation:
$y = (1)^3 - 3(1)^2$
$y = -2$
Coordinates are $(1, -2)$.

$f'(x) = 6e^{2x+1}$

$g(x) = \frac{\ln(x)}{x}$
Let $u = \ln(x)$ and $v = x$
$\therefore \frac{du}{dx} = \frac{1}{x}$ and $\frac{dv}{dx} = 1$
$g'(x) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
$= \frac{x \times \frac{1}{x} - \ln(x) \times 1}{x^2}$
$= \frac{1-\ln(x)}{x^2}$

$g'(e) = \frac{1-\ln(e)}{(e)^2} = 0$

$h(x) = x \sin(x)$

Let $u = x$ and $v = \sin(x)$
$\therefore \frac{du}{dx} = 1$ and $\frac{dv}{dx} = \cos(x)$
$h'(x) = u\frac{dv}{dx} + v\frac{du}{dx}$
$= x \times \cos(x) + \sin(x) \times 1$
$= x \cos(x) + \sin(x) \times 1$

$= x \cos(x) + \sin(x)$
Let $u = x$ and $v = \cos(x)$
$\therefore \frac{du}{dx} = 1$ and $\frac{dv}{dx} = -\sin(x)$
$h''(x) = u\frac{dv}{dx} + v\frac{du}{dx} + \cos(x)$
$= x \times -\sin(x) + \cos(x) \times 1 + \cos(x)$

$= 2 \cos(x) - x \sin(x)$

The rate of absorption is given by:
$\frac{dA}{dt} = 20t - 12t^2$

$\frac{d^2A}{dt^2} = 20 - 24t$

$\therefore 20 - 24t = 0$
$t = \frac{20}{24} = \frac{5}{6}$ hours

Verify this corresponds to a maximum rate.
Using the second derivative test, we investigate the sign of the derivative of $\frac{d^2A}{dt^2}$, i.e. $\frac{d^3A}{dt^3}$
$\frac{d^3A}{dt^3} = -24$
This is negative, therefore the rate of absorption is a maximum.
The time the chemical is increasing most rapidly since delivery is $\frac{5}{6}$ hours.
$= \frac{5}{6} \times 60$
$= 50$ minutes
The required time is 10:50 am.