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?(Wondering)

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.

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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?(Wondering)

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.

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the steepness is greatest in the direction $\displaystyle \nabla (z)(240,80,168)$ (write that as a unit vector).

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

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

Quote:

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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.

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The directional derivative is:

$\displaystyle 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 $\displaystyle v = \nabla f $

Hence the directional derivative in that direction is:

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

So

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

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

3) Find the magnitude of the vector you get.

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Originally Posted by Mush

The directional derivative is:

$\displaystyle 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 $\displaystyle v = \nabla f $

Hence the directional derivative in that direction is:

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

So

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

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

3) Find the magnitude of the vector you get.

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

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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(Wondering)

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

Quote:

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

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ahh..I see I stand corrected,thanx for that..(Happy)