Question

Can somone explain this to me:
Is there a direction "u" in which the rate of change of f(x,y)=x^2-3xy+4y^2 at P(1,2) equals 14?

Answer 1

The answer in the back of the book is sqrt(185), but i dont know what the problem is asking or why the answer is sqrt(185)

Answer 2

@dumbcow
@myininaya
can you help?

Answer 3

@satellite73

Answer 4

Is that the exact text of the problem? And are you sure that's the right answer (if u is a direction in 2 dimensions, it should be a vector, not a scalar)

Answer 5

the wording is the same as the book, and the answer says"No, the maximum rate of change is sqrt(185)<14"

Answer 6

@UnkleRhaukus

Answer 7

\[f(x,y)=x^2-3xy+4y^2\]\[\frac{\partial f(x,y)}{\partial x}=2x-3y\]\[\frac{\partial f(x,y)}{\partial y}=-3x+8y\] \[\frac{\partial f(x,y)}{\partial u}=\frac{\partial }{ \partial u } \left(x^2-3xy+4y^2\right)\]

Answer 8

but could you explain what the problem is asking, and how it is that the answer they have satifies the problem

Answer 9

find u such that
\[\frac{∂}{∂u}(x2−3xy+4y2)=14\]

Answer 10

at the point (x,y)= (1,2)

Answer 11

Well shoot, I WAS writing a nice neat reply, but my internet died so I'll give you the abridged version.

Answer 12

i just need to undersatnd, what is the problem asking as how, the answer they have satifies the problem. I dont understand what is going on in this problem

Answer 13

A function in multiple dimensions can have its derivative taken in any direction, not just x,y,z, etc. The simple formula for a directional derivative in two dimensions is:
Given a UNIT vector u = [a,b], the directional derivative of f(x,y) is given by:\[\delta f / \delta u =a*(\delta f/ \delta x) + b*(\delta f/ \delta y)\]This definition is easily generalized to any number of dimensions.

There are two useful properties of the directional derivative that are pertinent to your problem. The first is that the gradient of f(x,y) at any point p gives a vector whose direction is that of the greatest rate of change. The second is that the norm of f(x,y) at any point p gives the greatest rate of change of the function at that point.

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So for example, with your problem:
f(x,y) = x^2 - 3xy + 4y^2

The gradient is: [2x - 3y, 8y - 3x]

Then the gradient at P = (1,2) is [-4, 13]. Hence, the greatest rate of change of the function at point P occurs along the direction of this vector.

The norm of the function then is \[\sqrt{(-4)^2 + (13^2)} = \sqrt{16 + 169} = \sqrt{185}\]
Since 185^(1/2), the maximum rate of change of the function at (1,2), is less than 14, there can be no vector u for which the rate of change is 14.



Hope this helps, let me know if you have any other questions!

Answer 14

thank you so much, that cleared up a lot, especially when you talked about the norm giving the greatest rate of change

Answer 15

so the gradient f is in the direction of the greates rate of change,
and
the norm give us that value (which is the greates rate of change)

Answer 16

Precisely.

Answer 17

@NightShade
By the way, I was playing around with this a little more last night (I was bored) and I figured it might be helpful to post a general solution in case you get a problem like this where such a u DOES exist.

In general, suppose you want to find a directional vector u = [a,b] at point p = (x,y) in which the rate of change is c. We can set up a system of equations:

From the simple form of the directional derivative:\[c = a*\delta f/ \delta x + b*\delta f/ \delta y\] From the requirement that u be a unit vector:\[\sqrt{a^2 + b^2} = 1 \rightarrow a^2 + b^2= 1\] Since x, y, and c are constant, we have two equations with two unknowns and can solve for a and b.

For example, with the problem you were given:
P = (x,y) = (1,2)
c = 14
u = [a,b]

Then\[14 = a(2x - 3y) + b(8y - 3x) = 2ax - 3ay + 8by - 3bx = x(2a-3b) + y(8b-3a)\]\[= 2a - 3b + 16b - 6a = 13b - 4a\] Note that a is the x component of u and b is the y component of u, and their coefficients are in fact the values of the corresponding elements of the gradient of f evaluated at P.
Then we have a system of equations:\[13b - 4a = 14\]\[a^2 + b^2 = 1\]Then:\[b = [14-4a]/13\]\[a^2 + ([14-4a]/13)^2 = 1\]\[a^2 + (16/169)a^2 + (112/169)a + 196/169 = 1\]\[185a^2 + 112a + 196 = 169\]\[185a^2 + 112a + 27 = 0\]Thus, we can use the quadratic equation to solve for a and then solve for b using one of our original two equations to find u. (You'll find that if you use the quadratic equation here, you get no real solutions - hence there is no u for which the rate of change is 14).

Answer 18

Very cool, especially with that last bit, about there being no u for which the rate of change is 14 sice we get no real solutions from the quad. eq.