Sketch the curve with equation $\frac{x^2}{400} + \frac{(y-10)^2}{900} = 1.$ [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 $(0 < h < 30)$ 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 cm^3 per minute. Find the rate at which the water level is increasing when the depth is 15 cm. State the value of $h$ at which the rate of increase of the water level is the least.

1. Identify the shape, centre and semi-axes of the curve.

Rewrite the given equation as $\frac{x^2}{20^2} + \frac{(y-10)^2}{30^2} = 1$.
This equation represents an ellipse with its centre at $(0, 10)$.
The horizontal semi-minor axis is $20$ and the vertical semi-major axis is $30$.

2. Determine the coordinates of the turning points and $y$-intercepts.

For the turning points on the vertical axis (which also serve as the $y$-intercepts since $x=0$):
$\frac{(y-10)^2}{30^2} = 1 \implies y - 10 = \pm 30 \implies y = 40 \text{ or } y = -20$.
These points are $(0, 40)$ and $(0, -20)$.

For the horizontal extreme points (at $y=10$):
$\frac{x^2}{20^2} = 1 \implies x = \pm 20$.
These points are $(20, 10)$ and $(-20, 10)$.

3. Sketch the curve with the key features clearly labelled.

Draw a smooth, closed elliptical curve that is taller than it is wide.
Plot the centre visually at $(0, 10)$, and ensure the curve passes through the maximum and minimum points $(0, 40)$ and $(0, -20)$. Label the $y$-intercepts $40$ and $-20$.
Mark the leftmost and rightmost points at $(-20, 10)$ and $(20, 10)$. (As instructed, there is no need to label the $x$-intercepts).

1. Express $x^2$ in terms of $y$ to prepare for the volume integral.

$\frac{x^2}{400} + \frac{(y-10)^2}{900} = 1$\n$\frac{x^2}{400} = 1 - \frac{(y-10)^2}{900}$\n$x^2 = 400 - \frac{4}{9}(y-10)^2$\n$x^2 = 400 - \frac{4}{9}(y^2 - 20y + 100)$\n$x^2 = -\frac{4}{9}y^2 + \frac{80}{9}y + \frac{3200}{9}$

2. Set up the definite integral for the volume of revolution about the y-axis.

$V = \pi \int_{0}^{h} x^2 \, \mathrm{d}y$\n$V = \pi \int_{0}^{h} \left( -\frac{4}{9}y^2 + \frac{80}{9}y + \frac{3200}{9} \right) \, \mathrm{d}y$

3. Integrate the expression with respect to $y$ and evaluate from $y = 0$ to $y = h$.

$V = \pi \left[ -\frac{4}{27}y^3 + \frac{40}{9}y^2 + \frac{3200}{9}y \right]_{0}^{h}$\n$V = \pi \left( -\frac{4}{27}h^3 + \frac{40}{9}h^2 + \frac{3200}{9}h - 0 \right)$\n$V = \frac{4\pi}{27} (-h^3 + 30h^2 + 2400h)$

1. Express the rate of change of volume with respect to depth, $\frac{\mathrm{d}V}{\mathrm{d}h}$, in terms of $h$.

Since the volume is obtained by rotating the curve about the $y$-axis, $V = \int_0^h \pi x^2 \, \mathrm{d}y$. By the Fundamental Theorem of Calculus, the derivative with respect to $h$ is the cross-sectional area:
$\frac{\mathrm{d}V}{\mathrm{d}h} = \pi x^2$
From the equation of the curve, substituting $y = h$:
$\frac{x^2}{400} + \frac{(h-10)^2}{900} = 1$
$x^2 = 400\left(1 - \frac{(h-10)^2}{900}\right) = \frac{4}{9}\left(900 - (h-10)^2\right)$
$\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{4\pi}{9}\left(900 - (h-10)^2\right)$

2. Evaluate $\frac{\mathrm{d}V}{\mathrm{d}h}$ at $h = 15$.

Substitute $h = 15$ to find the rate of change of volume with respect to height at this instant:
$\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{4\pi}{9}\left(900 - (15-10)^2\right)$
$\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{4\pi}{9}(900 - 25) = \frac{4\pi}{9}(875) = \frac{3500\pi}{9}$

3. Use the chain rule to find the rate of change of the water level, $\frac{\mathrm{d}h}{\mathrm{d}t}$.

We are given the rate of water flowing in is $\frac{\mathrm{d}V}{\mathrm{d}t} = 1000$.
Using the chain rule for connected rates of change:
$\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}$
$1000 = \frac{3500\pi}{9} \times \frac{\mathrm{d}h}{\mathrm{d}t}$
$\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{9000}{3500\pi} = \frac{18}{7\pi} \approx 0.819 \text{ cm/min}$

1. Identify the condition for the least rate of change

By the chain rule for connected rates of change, $\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}} = \frac{1000}{\pi x^2}$.
For the rate of increase $\frac{\mathrm{d}h}{\mathrm{d}t}$ to be the least, the cross-sectional area $\pi x^2$ (which is the denominator) must be at its maximum.

2. Determine the value of $h$ that maximizes the cross-sectional area

From the equation of the curve, we can express $x^2$ in terms of $h$ (since $y = h$ at the water surface):
$x^2 = 400\left(1 - \frac{(h-10)^2}{900}\right)$
This value is greatest when the subtracted squared term $\frac{(h-10)^2}{900}$ is $0$.
$h - 10 = 0 \implies h = 10$