Find Answer for Gradient Question Starting at (3,2)

[bryanosaurus]

I am given z = 32 - x^{2} - 4y^{2}
Starting at the point (3,2) in i + j direction,
find if you are going up or down the hill and how fast.

The way I thought to proceed was that the gradient would tell me if I was going down or up hill and that \left|\nabla z \right| would give me "how fast". My answer of \sqrt{292} isn't correct however, so I'm obviously doing something wrong. Can anyone point me in the right direction of how to proceed?

 

[Hootenanny]

I am given z = 32 - x^{2} - 4y^{2}
Starting at the point (3,2) in i + j direction,
find if you are going up or down the hill and how fast.

The way I thought to proceed was that the gradient would tell me if I was going down or up hill and that \left|\nabla z \right| would give me "how fast". My answer of \sqrt{292} isn't correct however, so I'm obviously doing something wrong. Can anyone point me in the right direction of how to proceed?

The magnitude of the gradient effectively gives you the magnitude of the greatest increase/decrease of the function, rather than the rate of change of the function in a given direction, which is what you were asked. Instead of simply calculating the gradient of the function, you need to evaluate the directional derivative of the function at the given point, in the given direction.

 

[Ben Niehoff]

What gradient did you compute?

A few notes:

1. "The (i + j) direction" is misleading, because (i + j) is not a unit vector!

2. The directional derivative in the direction of a unit vector \mathbf{\hat u} is given by:

\mathbf{\hat u} \cdot \nabla f

 

[bryanosaurus]

What gradient did you compute?

A few notes:

1. "The (i + j) direction" is misleading, because (i + j) is not a unit vector!

2. The directional derivative in the direction of a unit vector \mathbf{\hat u} is given by:

\mathbf{\hat u} \cdot \nabla f

The gradient I computed was:
-2xi - 8yj

If I am supposed to calculate \mathbf{\hat u} \cdot \nabla f, what unit vector am I supposed to use? As you said, i + j isn't a unit vector...

 

[HallsofIvy]

Do you not know how to construct a unit vector in the direction of, say, vector u? Just divide by its length.

 

[bryanosaurus]

Do you not know how to construct a unit vector in the direction of, say, vector u? Just divide by its length.

I just realized that was what I was over looking. Thanks, if I use

i + j / |i + j|

i get the correct answer.