 # Suppose you are climbing a hill whose shape is given by the equation z=100 – 0.05x^2 – 0.1y^2, where x,y and z are measured in meters, and you are standing at a point with coordinates (60, 50, 1100). The positive x-axis points east and the positive y-axis points north. If you walk due south, will you start to ascend or descend? At what rate?

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.

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:

$f_y(x, y) = – 0.02y$

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

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.