Sketch the curve with equation $\displaystyle\frac{x^2}{400}+\frac{(y-10)^2}{900}=1.$ [There is no need to indicate the x-intercepts on the curve.] The region lying in the first quadrant bounded by the curve, the x-axis, the y-axis and the line $y=30$ is rotated completely about the y-axis to form a fish bowl into which water flows. Given that units on the axes are centimetres and $h$ cm $(0<k<30)$ denotes the depth of water in the fish bowl, find, in terms of $h$, the volume of water in the bowl. Water flows into the fish bowl at the constant rate of $1000\ \text{cm}^3$ per minute. Find the rate at which the water level is increasing when the depth is $15\ \text{cm}$. State the value of $h$ at which the rate of increase of the water level is the least.

1. Identify the shape and center of the curve from its equation.

The equation $\displaystyle\frac{x^2}{20^2} + \frac{(y-10)^2}{30^2} = 1$ represents an ellipse centered at $(0, 10)$ with a horizontal semi-axis of $20$ and a vertical semi-axis of $30$.

2. Find the y-intercepts.

Substitute $x = 0$ into the equation:
$\frac{0^2}{400} + \frac{(y-10)^2}{900} = 1$
$(y-10)^2 = 900$
$y - 10 = \pm 30$
$y = 40 \text{ or } y = -20$
The y-intercepts are $(0, 40)$ and $(0, -20)$.

3. Determine the horizontal extremities to aid in sketching.

Substitute $y = 10$ into the equation to find the maximum width:
$\frac{x^2}{400} + \frac{0}{900} = 1$
$x^2 = 400$
$x = \pm 20$
The horizontal extremities are $(20, 10)$ and $(-20, 10)$.

4. Sketch the curve.

Draw a smooth vertical ellipse passing through $(0, 40)$, $(20, 10)$, $(0, -20)$, and $(-20, 10)$. As instructed, there is no need to calculate or indicate the specific x-intercepts.

1. Express $x^2$ in terms of $y$ to prepare for integration.

$\frac{x^2}{400} + \frac{(y-10)^2}{900} = 1$\n$\frac{x^2}{400} = 1 - \frac{(y-10)^2}{900}$\n$x^2 = 400 - \frac{4}{9}(y-10)^2$

2. Expand the expression for $x^2$ and set up the integral for the volume of revolution about the $y$-axis.

$x^2 = 400 - \frac{4}{9}(y^2 - 20y + 100)$\n$x^2 = 400 - \frac{4}{9}y^2 + \frac{80}{9}y - \frac{400}{9}$\n$x^2 = \frac{3200}{9} + \frac{80}{9}y - \frac{4}{9}y^2$\n$V = \pi \int_0^h x^2 \, dy$\n$V = \pi \int_0^h \left( \frac{3200}{9} + \frac{80}{9}y - \frac{4}{9}y^2 \right) dy$

3. Evaluate the definite integral to find the volume in terms of $h$.

$V = \pi \left[ \frac{3200}{9}y + \frac{40}{9}y^2 - \frac{4}{27}y^3 \right]_0^h$\n$V = \pi \left( \frac{3200}{9}h + \frac{40}{9}h^2 - \frac{4}{27}h^3 \right)$

1. Express the derivative of the volume with respect to the water depth, $\frac{\mathrm{d}V}{\mathrm{d}h}$.

From part (ii), the volume of water is $V = \pi \int_0^h x^2 \,\mathrm{d}y$, where the curve equation gives $x^2 = 400 - \frac{4}{9}(y-10)^2$.
By differentiating $V$ with respect to $h$, or recognising it as the circular cross-sectional area at depth $h$, we get:
$\frac{\mathrm{d}V}{\mathrm{d}h} = \pi \left( 400 - \frac{4}{9}(h-10)^2 \right)$

2. Set up the connected rates of change equation using the chain rule.

By the chain rule, $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}$.
Given the constant rate of water flow $\frac{\mathrm{d}V}{\mathrm{d}t} = 1000$, we substitute this into our relationship:
$1000 = \pi \left( 400 - \frac{4}{9}(h-10)^2 \right) \frac{\mathrm{d}h}{\mathrm{d}t}$

3. Substitute $h = 15$ to find the corresponding rate of increase of the water level.

Substitute $h = 15$ into the equation:
$1000 = \pi \left( 400 - \frac{4}{9}(15-10)^2 \right) \frac{\mathrm{d}h}{\mathrm{d}t}$
$1000 = \pi \left( 400 - \frac{4}{9}(25) \right) \frac{\mathrm{d}h}{\mathrm{d}t}$
$1000 = \pi \left( \frac{3600 - 100}{9} \right) \frac{\mathrm{d}h}{\mathrm{d}t}$
$1000 = \frac{3500\pi}{9} \frac{\mathrm{d}h}{\mathrm{d}t}$
$\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{9000}{3500\pi} = \frac{18}{7\pi} \approx 0.819 \text{ cm/min}$

4. Determine the condition for the least rate of increase of the water level.

We need the value of $h$ that minimises $\frac{\mathrm{d}h}{\mathrm{d}t}$.
Rearranging the rate equation gives $\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1000}{\frac{\mathrm{d}V}{\mathrm{d}h}}$.
The rate $\frac{\mathrm{d}h}{\mathrm{d}t}$ is least when the denominator (the cross-sectional area $\frac{\mathrm{d}V}{\mathrm{d}h}$) is at its maximum.

5. Find the value of $h$ that maximises the cross-sectional area.

The expression $\frac{\mathrm{d}V}{\mathrm{d}h} = \pi \left( 400 - \frac{4}{9}(h-10)^2 \right)$ is a downward-opening parabola.
Its maximum value occurs when the subtracted squared term is zero:
$(h - 10)^2 = 0 \implies h = 10$
(Alternatively, by inspecting the bowl's elliptical shape, its maximum width occurs at the centre of the ellipse $y = 10$.)
Thus, the rate of increase is the least when $h = 10$.