# Find Differential Length

1. Jan 31, 2010

### mknut389

I got an answer for this, but it just doesn't seem correct. Maybe someone could tell me what I am doing wrong, or that I am just over thinking the problem and I am right.

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

Write an expression for the differential length vector dl at the point (1,2,8) on the straight line y=2x, z=4y, and having the projection dx on the x-axis

2. Relevant equations

dl=dxax+dyay+dzaz

3. The attempt at a solution

y=2x, z=4y, therefore z=8x
therefore

dxax+2dxay+8dxaz.

since there are no variables remaining after you take the derivative, dl is then
1ax+2ay+8az

Any help would be amazing, and please excuse me if I am just being a complete idiot.

Thanks

2. Jan 31, 2010

### Dick

That seems ok to me. So dl=dx(ax+2*ay+8*az). ax, ay and az are the unit vectors in each direction, right?

3. Jan 31, 2010

### mknut389

Yes, ax, ay, az are the direction unit vectors. Thanks

4. Dec 1, 2010

### yummEE321

..... I haven't really understood the problem, but i do understand the solution.. can anyone explain more about this problem especially the "projection dx on the x-axis" part..