Main Question [5 marks]

Gradient of a Vector Function

The gradient of a scalar function $f(x, y, z)$ is defined as the vector obtained by applying the vector differential operator $\nabla$ to the function. It is denoted by $\nabla f$ or $\text{grad } f$.

$$\nabla f = \frac{\partial f}{\partial x}\vec{i} + \frac{\partial f}{\partial y}\vec{j} + \frac{\partial f}{\partial z}\vec{k}$$

The gradient points in the direction of maximum rate of increase of the function.

Finding the Directional Derivative

Given: $f(x, y, z) = x^3 - xy^2 + z$, point $P(1, 0, 0)$, direction $\vec{v} = 2\vec{i} - \vec{j} + \vec{k}$

Step A: Find the gradient $\nabla f$

$$\frac{\partial f}{\partial x} = 3x^2 - y^2, \quad \frac{\partial f}{\partial y} = -2xy, \quad \frac{\partial f}{\partial z} = 1$$

$$\nabla f = (3x^2 - y^2)\vec{i} + (-2xy)\vec{j} + (1)\vec{k}$$

Step B: Evaluate $\nabla f$ at $P(1, 0, 0)$

$$\nabla f \big|_{(1,0,0)} = (3(1)^2 - 0)\vec{i} + (-2(1)(0))\vec{j} + (1)\vec{k} = 3\vec{i} + 0\vec{j} + \vec{k}$$

Step C: Find the unit vector $\hat{v}$

$$|\vec{v}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$

$$\hat{v} = \frac{2\vec{i} - \vec{j} + \vec{k}}{\sqrt{6}}$$

Step D: Find the directional derivative

$$D_{\hat{v}}f = \nabla f \cdot \hat{v} = (3\vec{i} + 0\vec{j} + \vec{k}) \cdot \frac{(2\vec{i} - \vec{j} + \vec{k})}{\sqrt{6}}$$

$$D_{\hat{v}}f = \frac{(3)(2) + (0)(-1) + (1)(1)}{\sqrt{6}} = \frac{6 + 0 + 1}{\sqrt{6}}$$

$$\boxed{D_{\hat{v}}f = \frac{7}{\sqrt{6}}}$$

---

Sub-question 1 [5 marks]

Volume of a Solid of Revolution

The volume of a solid of revolution is the volume generated when a plane region is revolved about a given axis. Using the disc/washer method, it is given by:

$$V = \pi \int_a^b [R(x)]^2 \, dx \quad \text{(disc method)}$$

Finding the Volume

Given: Region bounded by $y = \sqrt{x}$, $y = 1$, and $x = 4$, revolved about the line $y = 1$.

Step A: Identify the region and radius

When revolved about $y = 1$, the radius of revolution at any point is:

$$R(x) = |y - 1| = |\sqrt{x} - 1|$$

Step B: Find limits of integration

• $y = \sqrt{x}$ meets $y = 1$ when $\sqrt{x} = 1 \Rightarrow x = 1$
• Upper limit: $x = 4$
• So limits are from $x = 1$ to $x = 4$

Step C: Set up and evaluate the integral (Disc Method)

$$V = \pi \int_1^4 (\sqrt{x} - 1)^2 \, dx$$

$$V = \pi \int_1^4 (x - 2\sqrt{x} + 1) \, dx$$

$$V = \pi \left[\frac{x^2}{2} - 2 \cdot \frac{2x^{3/2}}{3} + x\right]_1^4$$

$$V = \pi \left[\frac{x^2}{2} - \frac{4x^{3/2}}{3} + x\right]_1^4$$

Step D: Substitute limits

At $x = 4$:

$$\frac{16}{2} - \frac{4(8)}{3} + 4 = 8 - \frac{32}{3} + 4 = 12 - \frac{32}{3} = \frac{36 - 32}{3} = \frac{4}{3}$$

At $x = 1$:

$$\frac{1}{2} - \frac{4(1)}{3} + 1 = \frac{1}{2} - \frac{4}{3} + 1 = \frac{3 + 6 - 8}{6} = \frac{1}{6}$$

$$V = \pi\left(\frac{4}{3} - \frac{1}{6}\right) = \pi\left(\frac{8 - 1}{6}\right)$$

$$\boxed{V = \frac{7\pi}{6} \text{ cubic units}}$$