Differentiation — practice questions

The function is a quotient, so we use the quotient rule with $u = \ln x$ and $v = x^2$. $u' = \frac{du}{dx} = \frac{1}{x}$ and $v' = \frac{dv}{dx} = 2x$.

$\frac{dy}{dx} = \frac{v u' - u v'}{v^2} = \frac{x^2(\frac{1}{x}) - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x \ln x}{x^4} = \frac{1 - 2 \ln x}{x^3}$. [M1 for quotient rule]

For stationary points, set $\frac{dy}{dx} = 0$. $\frac{1 - 2 \ln x}{x^3} = 0 \implies 1 - 2 \ln x = 0$. [M1 for setting numerator to 0] $2 \ln x = 1 \implies \ln x = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e}$. [A1 for x-coordinate]

Now find the y-coordinate: $y = \frac{\ln(e^{1/2})}{(e^{1/2})^2} = \frac{1/2}{e} = \frac{1}{2e}$. The stationary point is $(\sqrt{e}, \frac{1}{2e})$. [A1 for y-coordinate]

To determine the nature, we find the second derivative, $\frac{d^2y}{dx^2}$. We differentiate $\frac{dy}{dx} = \frac{1 - 2 \ln x}{x^3}$. Let $u = 1 - 2 \ln x$ and $v = x^3$. Then $u' = -\frac{2}{x}$ and $v' = 3x^2$.

$\frac{d^2y}{dx^2} = \frac{x^3(-\frac{2}{x}) - (1 - 2 \ln x)(3x^2)}{(x^3)^2} = \frac{-2x^2 - 3x^2 + 6x^2 \ln x}{x^6} = \frac{-5x^2 + 6x^2 \ln x}{x^6} = \frac{6 \ln x - 5}{x^4}$. [M1 for second derivative]

Substitute $x = e^{1/2}$ into the second derivative: $\frac{d^2y}{dx^2} = \frac{6 \ln(e^{1/2}) - 5}{(e^{1/2})^4} = \frac{6(\frac{1}{2}) - 5}{e^2} = \frac{3 - 5}{e^2} = -\frac{2}{e^2}$.

Since $\frac{d^2y}{dx^2} < 0$, the stationary point is a local maximum. [A1 for correct conclusion from correct working]

Let $h$ be the depth of the water and $r$ be the radius of the water surface at time $t$. We are given $\frac{dV}{dt} = 0.1$. We want to find $\frac{dh}{dt}$ when $h=3$.

First, we need a relationship between $V$ and $h$. The formula for $V$ involves both $r$ and $h$. We can use similar triangles to relate $r$ and $h$. The ratio of radius to height for the cone is constant: $\frac{r}{h} = \frac{2}{6} = \frac{1}{3}$. So, $r = \frac{h}{3}$. [M1 for relating r and h]

Substitute this into the volume formula: $V = \frac{1}{3}\pi (\frac{h}{3})^2 h = \frac{1}{3}\pi (\frac{h^2}{9}) h = \frac{1}{27}\pi h^3$. [M1 for V in terms of h only]

Now, we can find $\frac{dV}{dh}$ by differentiating this expression with respect to $h$: $\frac{dV}{dh} = \frac{1}{27}\pi (3h^2) = \frac{1}{9}\pi h^2$. [M1 for differentiating V]

We use the chain rule for connected rates: $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$. We want to find $\frac{dh}{dt}$, so we rearrange: $\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}$.

When $h=3$, $\frac{dV}{dh} = \frac{1}{9}\pi (3^2) = \frac{1}{9}\pi (9) = \pi$. [A1 for value of dV/dh]

Now substitute the known values: $\frac{dh}{dt} = 0.1 \div \pi = \frac{0.1}{\pi}$.

The rate at which the water level is rising is $\frac{0.1}{\pi} \text{ m min}^{-1}$ (or approximately $0.0318 \text{ m min}^{-1}$). [A1 for final answer]