Vector Function

[Saladsamurai]

Okay. My reason for posting this is that I need help actually formulating the 'math part' of it. I can get the right answer by 'inspection.' And from the way the book is written, I believe that is how the authors expect you to find it. But for self gratifying reasons, I wish to generalize the answer a little bit:

1. The problem statement, all variables and given/known data

Write the formula for a vector function in 2 dimensions whose direction makes an
angle of 45 degrees with the x-axis and whose magnitude at any point (x,y) is (x+y)2.


3. The attempt at a solution

Now just looking at it, we can say that to insure that the vector at any point (x,y) is at 45 degrees to the x-axis, we can let the funciton equal something to the effect of:

\mathbf{F}(x,y) = k(\bold{i} + \bold{j})

Now k(x,y) is the 'scalar portion' of F and must have the effect that its product with the magnitude of the
vector portion of F, namely v=(i + j), must equal (x+y)2.

Now since the magnitude of v=(i + j) is \sqrt{2}, k(x,y) must have \sqrt{2} in its denominator.

This will make 'unit vector' in the v direction or

\bold{u}_v = \frac{\bold{v}}{\sqrt{2}}=\frac{\bold{i} + \bold{j}}{\sqrt{2}}

Therefore, multiplying v by k(x,y) = (x+y)2/\sqrt{2} gives the desired result.

\Rightarrow \bold{F}(x,y) = \frac{(x+y)^2}{\sqrt{2}}(\bold{i} + \bold{j})


I am just curious how other would approach this problem, or if this is the most efficient method from a mathematical standpoint.

~Casey

[Feldoh]

Dunno if it helps but....

I assumed a vector initially completely in the i direction: \mathbf{F} = (x+y)^2 \mathbf{i}.

Then just did a coordinate rotation:

\left(\begin{array}{cc}cos(45) & -sin(45)\\sin(45) & cos(45)\end{array}\right) \left(\begin{array}{cc}(x+y)^2\\0\end{array}\right ) = \frac{(x+y)^2}{\sqrt{2}}\left(\begin{array}{cc}\ma thbf{i}\\\mathbf{j}\end{array}\right)

[Saladsamurai]

Thanks Feldoh! That's a very interesting approach. I like it!

:smile: