# Find a unit vector

#### al1850

Find a unit vector in the direction in which f decreases most rapidly at Q

An interesting tidbit: the direction of quickest ascent is the direction of the gradient (similar to the idea that the gradient of a function is normal to any level set of that function).

So, find the gradient. Or if you are not familiar with the gradient, multiply the function by the del operator (d/dx, d/dy, etc.)

Now that we have the direction of quickest ascent, plug in the point Q.

Finally, to normalize the gradient, take the root of the dot product of the gradient with itself and divide the gradient by this number.

Is that enough?

#### al1850

could you please show me the working, still don't get it.

No problem.

So we first find the gradient of f. To do this, we multiply (d/dx,d/dy) by f.

Well, the derivative of f wrt x is 2y/(x-y)^2. (with respect to)
The derivative of f wrt y is -2y/(x-y)^2.
Thus the direction of quickest ascent is ( 2y/(x-y)^2 , -2y/(x-y)^2 )

Let us plug in Q(3,-1).

We get ( -1/8, 1/8 ). This is the direction of quickest ascent. To find the direction of quickest descent, we take the negative.

Thus a vector in the direction in which f decreases most rapidly at Q is (1/8,-1/8). But this is not a unit vector. A quick sense would suggest that the two components are equal, and thus the unit vector should be the legs of an isosceles right triangle with hypnotnuse 1. Thus our vector is ( (2^.5)/2 , -(2^.5)/2 ) or one/root2 , -one/root2.

If that quick sense doesn't make sense, then we can also dot the vector by itself and take the root of the product. The dot product of 1/8,-1/8 with itself is 1/32. Rooted --> 1/(32)^.5 = 1/(4*2^.5)

To make -1/8,1/8 a unit vector, divide both components by 1/(32^.5). This yields ( (2^.5)/2 , -(2^.5)/2 ), the same answer as before.

Thus, the final answer is ( (2^.5)/2 , -(2^.5)/2 ).

Anything I can do to make it more clear?

#### al1850

Now I got it. Thank you man, that explains all.Very detailed!