A solid cylinder of height h cm and volume V cm^3 is cut from a solid sphere of radius 7 cm. The rim of each base of the cylinder touches the surface of the sphere.

1. Apply Pythagoras' theorem to the right triangle formed in the cross-section and rearrange for $r^2$.

In the cross-section, the right triangle has sides $r$, $\frac{h}{2}$ and hypotenuse $7$.

By Pythagoras' theorem,
$r^2+\left(\frac{h}{2}\right)^2=7^2$
$r^2+\frac{h^2}{4}=49$
$r^2=49-\frac{h^2}{4}$

1. Substitute $h=4$ into the given formula for $V$ and evaluate.

For $h=4$,
$p=\pi(4)\left(49-\frac{4^2}{4}\right)$
$=4\pi(49-4)$
$=180\pi$
$=565.486\ldots$
Correct to the nearest whole number,
$p=565.$

1. Use the completed table of values for the graph.

From the table and part (b)(i), the points to plot are

2. Plot the points accurately on the given axes.

On the grid with horizontal axis $h$ and vertical axis $V$, plot:

Use the given axis scales: $h$ from $0$ to $12$ and $V$ marked in hundreds.

3. Draw the required curve over the interval.

Join the plotted points with a smooth curve for

The curve should rise from $(0,0)$ to a maximum between $h=7$ and $h=10$, then fall towards $(12,490)$.

1. Read the maximum value of $V$ from the highest point of the graph.

From the graph of $V=\pi h\left(49-\dfrac{h^2}{4}\right)$, the highest point of the curve occurs at about $h\approx 8.1$.

The corresponding value of $V$ is approximately
$V\approx 830.$

1. Relate the given equation to the graph equation for $V$.

From the graph,
$V=\pi h\left(49-\frac{h^2}{4}\right).$
Since
$196-h^2=4\left(49-\frac{h^2}{4}\right),$
then
$\pi h(196-h^2)=4V.$

2. Convert the given equation to a horizontal line on the graph and read the intersections.

Given
$2400=\pi h(196-h^2),$
we have
$2400=4V.$
So
$V=600.$
Draw the horizontal line $V=600$ on the graph and read the two intersection points with the curve.

From the graph,
$h\approx 4.3 \quad \text{or} \quad h\approx 11.3.$

1. Draw the suitable tangent line on the existing graph.

On the graph, mark the point $(0,500)$ on the $V$-axis. Draw a straight line through $(0,500)$ so that it is a tangent to the curve $V=\pi h\left(49-\frac{h^2}{4}\right)$.

From the graph, the tangent touches the curve at about
$h\approx 6.8,\quad V\approx 800.$

2. Find the gradient of the tangent line and identify it as $k$.

Since the line has equation
$V=kh+500,$
its gradient is $k$.

Using the two points approximately $(0,500)$ and $(6.8,800)$,
$k=\frac{800-500}{6.8-0}$
$k\approx 44.1.$

Hence
$k\approx 44.$