Differentiation – Application

1. Introduce a variable (often \(x\) or \(h\)) and express the quantity (e.g. volume) as a function of that variable.
2. Differentiate to find \(\dfrac{dQ}{dx}\) (where \(Q\) is the quantity).
3. Set \(\dfrac{dQ}{dx} = 0\) to find stationary points.
4. Use the second derivative or gradient change to decide if it is a maximum or minimum.

The diagram shows a solid cone which has a slant height of \(15\) cm and a vertical height of \(h\) cm.

(i) Show that the volume, \(V \,\text{cm}^3\), of the cone is given by \[ V = \frac{1}{3}\pi(225h - h^3). \] [The volume of a cone of radius \(r\) and vertical height \(h\) is \(\dfrac{1}{3}\pi r^2 h\).]

(ii) Given that \(h\) can vary, find the value of \(h\) for which \(V\) has a stationary value. Determine, showing all necessary working, the nature of this stationary value.

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Part (i): Expressing the Volume in Terms of \(h\)

Let the radius of the base be \(r\) cm. We know:

• slant height \(= 15\) cm
• vertical height \(= h\) cm
• radius \(= r\) cm

These form a right-angled triangle, so by Pythagoras: \[ r^2 + h^2 = 15^2 = 225 \quad\Rightarrow\quad r^2 = 225 - h^2. \] Volume of the cone: \[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(225 - h^2)h = \frac{1}{3}\pi(225h - h^3). \] Hence the required expression is shown.

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Part (ii): Stationary Value of the Volume

Step 1: Differentiate \(V\) with respect to \(h\).

\[ V = \frac{1}{3}\pi(225h - h^3) \] \[ \frac{dV}{dh} = \frac{1}{3}\pi(225 - 3h^2) = \pi(75 - h^2). \]

Step 2: Find stationary points by setting \(\dfrac{dV}{dh}=0\).

\[ \pi(75 - h^2) = 0 \quad\Rightarrow\quad 75 - h^2 = 0 \quad\Rightarrow\quad h^2 = 75 \quad\Rightarrow\quad h = \sqrt{75} = 5\sqrt{3}\ \text{cm} \] (We take the positive value because \(h\) is a length.)

Step 3: Determine the nature of the stationary value (max / min).

Find the second derivative: \[ \frac{dV}{dh} = \pi(75 - h^2) \quad\Rightarrow\quad \frac{d^2V}{dh^2} = -2\pi h. \] At \(h = 5\sqrt{3}\): \[ \frac{d^2V}{dh^2} = -2\pi (5\sqrt{3}) = -10\pi\sqrt{3} < 0. \] Since the second derivative is negative, the stationary point is a maximum.

• Use geometry (Pythagoras, similar triangles) to link variables first.
• Express the quantity (e.g. \(V, A, S\)) in terms of a single variable.
• Differentiate carefully and show \(\dfrac{dQ}{dx} = 0\) clearly.
• Use \(\dfrac{d^2Q}{dx^2}\) to justify “maximum” or “minimum”.
• State units in your final answer where appropriate.