# Math Help - Directional derivative

1. ## Directional derivative

Is this question asking for us to find the saddle point,maximum,minimum?I'm confused,how would I go about doing this?

Suppose you are climbing a hill of height z whose shape is given by the equation z(x,y)=1000-0.01x^2-0.04y^2 and you are standing at the point with the coordinates(240,80,168).

In which direction is the line of greatest steepness? Give your answer as unit vector.

Use a directional derivative to find the rate of climb in this direction.

2. Originally Posted by Raidan
Is this question asking for us to find the saddle point,maximum,minimum?I'm confused,how would I go about doing this?

Suppose you are climbing a hill of height z whose shape is given by the equation z(x,y)=1000-0.01x^2-0.04y^2 and you are standing at the point with the coordinates(240,80,168).

In which direction is the line of greatest steepness? Give your answer as unit vector.

Use a directional derivative to find the rate of climb in this direction.
the steepness is greatest in the direction $\nabla (z)(240,80,168)$ (write that as a unit vector).

the rate of climb in that direction is given by $\| \nabla (z)(240,80,168) \|$

3. ## Directional derivative...

I still dont know how to get this..can someone please help me in doing by steps? Thanks

4. Originally Posted by Raidan
Suppose you are climbing a hill of height z whose shape is given by the equation z(x,y)=1000-0.01x^2-0.04y^2 and you are standing at the point with the coordinates(240,80,168).

In which direction is the line of greatest steepness? Give your answer as unit vector.

Use a directional derivative to find the rate of climb in this direction.
The directional derivative is:

$D_v f = \frac{\vec{v} . \nabla f}{|\vec{v}|}$

f is your function, and v is the direction in which you are calculating the derivative.

The direction in which the line is steepest is when $v = \nabla f$

Hence the directional derivative in that direction is:

$D_v f = \frac{\nabla f . \nabla f}{|\nabla f|} = \frac{|\nabla f|^2}{|\nabla f|} = |\nabla f|$

So

1) Find $\nabla f$ in terms of x, y and z.

2) Plug in your point (240,80,168)

3) Find the magnitude of the vector you get.

5. Originally Posted by Mush
The directional derivative is:

$D_v f = \frac{\vec{v} . \nabla f}{|\vec{v}|}$

f is your function, and v is the direction in which you are calculating the derivative.

The direction in which the line is steepest is when $v = \nabla f$

Hence the directional derivative in that direction is:

$D_v f = \frac{\nabla f . \nabla f}{|\nabla f|} = \frac{|\nabla f|^2}{|\nabla f|} = |\nabla f|$

So

1) Find $\nabla f$ in terms of x, y and z.

2) Plug in your point (240,80,168)

3) Find the magnitude of the vector you get.

Thanx,I solved it and got sqrt(65) as the magnitude,is this value correct? and directional derivative as -4.8i-6.4j-k

6. Originally Posted by Raidan
Thanx,I solved it and got sqrt(65) as the magnitude,is this value correct? and directional derivative as -4.8i-6.4j-k
Is the k component not 0?

There are no z terms in the equation, so it's a constant function with respect to z. So the k term of the gradient is simply the derivative of a constant = 0.

I get $(-4.8, -6.4, 0)$, and hence the Directional derivative is 8.

7. Originally Posted by Mush
Is the k component not 0?

There are no z terms in the equation, so it's a constant function with respect to z. So the k term of the gradient is simply the derivative of a constant = 0.

I get $(-4.8, -6.4, 0)$, and hence the Directional derivative is 8.

ahh..I see I stand corrected,thanx for that..