The question aims to find the **direction** if the **person** starts to **walk** to the **south,** whether the person will **ascend** or **descend,** and at what **rate.**

This question is based on the concept of **directional derivatives.** The **directional derivative** is the **dot product** of the **gradient** of the **function** with its **unit vector.**

## Expert Answer

The given **function** for the **shape** of the **hill** is given as:

\[ f(x, y) = 100 – 0.05x^2 – 0.01y^2 \]

The **coordinate point** where you are currently **standing** is given as:

\[ P = (60, 50, 1100) \]

We can find whether the person **walking** due **south** is **ascending** or **descending** by finding the **directional derivative** of f at **point P** along the direction of **vector v**. The **directional derivative** of **f** is given as:

\[ D_u f(x, y) = \triangledown f(x, y) . u \]

Here, **u** is a **unit vector** in the **direction** of **vector v**. As we are moving due **south,** the direction of the** vector v** is given as:

\[ v = 0 \hat {i} – \hat {j} \]

The **unit vector** **u** will become:

\[ u = \dfrac{ \overrightarrow {v} }{ |v| } \]

\[ u = \dfrac {1} {1} [0, -1] \]

The **gradient** of the function** f** is given as:

\[ \triangledown f(x, y) = [ f_x(x, y), f_y(x, y) ] \]

The **x-gradient** of the function** f** is given as:

\[ f_x(x, y) = – 0.1x \]

The **y-gradient** of the function** f** is given as:

\[ f_y(x, y) = – 0.02y \]

Hence, the **gradient** becomes:

\[ \triangledown (x, y) = [ – 0.1x, – 0.02y ] \]

Substituting the values of **x** and **y** from **point** **P** in the above equation, we get:

\[ \triangledown (60, 50) = [ – 0.1 (60), – 0.02 (50) ] \]

\[ \triangledown (60, 50) = [ – 6, – 1 ] \]

Now substituting the values in the equation with **directional derivative,** we get:

\[ D_u f(60, 50) = [ -6, -1 ] . d \frac {1} {1} [ 0, -1 ] \]

\[ D_u f(60, 50) = 0 + 1 = 1 \]

Since $D_u f \gt 0$, the person moving due **south** will **ascend** at the **rate** of **1 m/s.**

## Numerical Result

The **directional derivative** of the function **f** at point **P** is greater than **zero** or **positive,** which means that the person is **ascending** while walking due **south** at the rate of** 1 m/s.**

## Example

Suppose you are **climbing** a **mountain** and its shape is given by the equation $z = 10 – 0.5x^2 – 0.1y^2$. You are standing on the point **(40, 30, 500)**. The positive **y-axis** points **north** whiles positive **x-axis** points **east.** If you walk towards **south,** will you **ascend** or **descend?**

The **directional derivative** is given as:

\[ D_u f(x, y) = \triangledown f(x, y) . u \]

The **gradient** of the function is given as:

\[ \triangledown (x, y) = [ -1x, -0.2y ] \]

Substituting the values of **x** and **y** from point **P** in the above equation, we get:

\[ \triangledown (40, 30) = [ – 0.1 (40), – 0.02 (30) ] \]

\[ \triangledown (40, 30) = [ – 4, – 6 ] \]

Now, substituting the values in the equation with **directional derivative,** we get:

\[ D_u f(60, 50) = [ -4, -6 ] . d \frac {1} {1} [ 0, -1 ] \]

\[ D_u f(60, 50) = 0 + 6 = 6 \]

If the person is walking towards the **south,** the person will be walking **uphill** or **ascending.**