Decreasing Temperature at Point (3,1) Using Gradient Vector - Homework Problem

[Loppyfoot]

Homework Statement

The equation T(x,y) = x²y-y² + 180 is a temperature equation at point (x,y).

If One is standing at point (3,1), what direction should he move in order to decrease the temperature?

Secondly, if one moves a unit distance in this direction found above, how much should he/she expect the temperature to drop?

Homework Equations

The Gradient Vector <fx[/,f_y>

The Attempt at a Solution

So at the point (3,1) he is at a temperature of 188^o.

The gradient of this function is: <2xy,x^2-2y>

so if I plug (3,1) into the gradient I get: <6,7>, which represents the rate of change of the temperature, correct?

What should I do to find what direction he needs to move?

 

[micromass]

You should find a vector v such that the directional derivative D_vf(3,1) is negative.
Note that

D_vf(a)=v_1\frac{\partial f}{\partial x}(a)+v_2\frac{\partial f}{\partial y}(a)+v_3\frac{\partial f}{\partial z}(a)

So what should you do? You should first find the partial derivatives of f and evaluate them in a=(3,1). Then you should find v such that the above expression is negative...

 

[Loppyfoot]

Ok so the gradient vector is:

<2xy,x^2-2xy>, so once evaluated at point a, I get <6,7> for gradf(3,1).

Should I do anything with a unit vector of (3,1), to find the directional derivative?

 

[micromass]

It doesn't matter for the directional derivative if you're working with unit vectors. You can always transfrom a vector in a unit vector with thesame derivative.

So the gradient is (6,7). Now you'll need to find a vector (v,w) such that

6v+7w\leq 0

If you found this vector, then you can easily transform this into a unit vector by \frac{(v,w)}{\|(v,w)\|}.

 

[Loppyfoot]

I'm having trouble visualizing how to find the vector (v,w)...

 

[micromass]

There's no need to visualize this. You just need to find v and w such that 6v+7w&lt; 0. For example, if v=1, for what value does 6+7w&lt;0? There are many such w, but only 1 should do...

 

[Loppyfoot]

Oh Ok. So let's say v=1, then w= -1. So the vector would be <1,-1>.

So using that vector <1,-1> I'm then confused about how to solve the direction he should be headed towards? I shouldn't do the dot product because that would produce a scalar.. hm..

 

[micromass]

So (1,-1) is indeed a good vector. The only problem now is that this vector is centered at (0,0). We want it centered at (3,1). For this, just do (1,-1)+(3,1)=(4,0). So if you go from (3,1) to (4,0), then the temperate should decrease...

 

[LCKurtz]

Homework Statement

The equation T(x,y) = x²y-y² + 180 is a temperature equation at point (x,y).

If One is standing at point (3,1), what direction should he move in order to decrease the temperature?

Secondly, if one moves a unit distance in this direction found above, how much should he/she expect the temperature to drop?

Homework Equations

The Gradient Vector <fx[/,f_y>

The Attempt at a Solution

So at the point (3,1) he is at a temperature of 188^o.

The gradient of this function is: <2xy,x^2-2y>

so if I plug (3,1) into the gradient I get: <6,7>, which represents the rate of change of the temperature, correct?

<sub>No, that is not correct. The gradient vector points the direction of the maximum rate of change. The value for the maximum rate of change of T is the length of the gradient vector

If you want to move in a direction where the temperature decreases most rapidly you would move in the direction opposite to the gradient vector. And the rate at which temperature is changing in that direction is minus the length of the gradient vector at that point.</sub>

 

[Loppyfoot]

So since the gradient is <6,7>, the length of the gradient will be (36+49)^.5. So the length of the gradient is 9.22, but since it should be pointing in the direction of greatest decrease, the length of the gradient should be -9.22.

correct?

 

[LCKurtz]

So since the gradient is <6,7>, the length of the gradient will be (36+49)^.5. So the length of the gradient is 9.22, but since it should be pointing in the direction of greatest decrease, the length of the gradient should be -9.22.

correct?

No, not quite. The length of the gradient vector at the point is approximately 9.22. It is nonnegative like the length of a vector always is and it doesn't change.

The length of gradient at a point gives the maximum rate of change of T and the gradient itself points in the direction of max rate of change. -T points in the opposite direction and the rate of change in that direction is -9.22. But the lengths of T and -T are equal and positive.