2 vector problems(am i doing them right)

[ilikephysics]

1. Write out the vector V=2i+3j at the point (x,y) = (1,2) in terms of the unit vectors in plane polar coordinates. Do this again for the same vector at the origin. Are your results different? Why?

Answer:
From my notes, I see that x,y = (r, theta) and x=r cos theta, y=r sin theta.
So, for (1,2) the vector is 2 cos 2 i + 3 sin 2 j

At the origin, (0,0), the vector is zero. Am I on the right track?

2.Find the gradient of the function phi(x,y)=2x^2y at the point (x,y)=(1,2), in plane polar coordinates.

Answer:

The gradient is 4xyi+2x^2j. At (1,2) it's 8i+2j. How would I put it in plane polar cooridnates? Would the answer be (8 cos 2) i + (2 sin 2) j?

Thanks.

[vadik]

r means the distance from origin to the end point of the vector. Theta is the angle between x-axis ant the straight line from (0,0) to the end point. So, when you have a notation in polar coordinates (r, theta) you can convert it to cartesian calculating x=rcos(theta) and y=rsin(theta). But when you turn from cartesian to polar, it will go like this: r=sqrt(x^2+y^2) and theta = arctan(x/y)
The first problem goes like this: vector 2i+3j at the point (1,2) (if i understood right)has its end point at (3,5), so r=sqrt(34) and theta=arctan(3/5).
The same vector at the origin has its end point at (2,3)

[ilikephysics]

How did you get the endpoint (3,5)? How would I right the sqrt(34) and arctan(3/5) as a vector? Did I do #2 right?

[vadik]

if I understand right "a vector at a point" means that we take the point and let the vector start from it. So if a vector 2i + 3j starts from point (1,2), thet its endpoint will be at (1+2,2+3)=(3,5).I made a picture, so you will understand what i mean. When you write a vector in polar coordinates you write just (r, theta), where in your case r=sqrt(34)~5.83 and theta=arctan(3/5)=0.54, so in polar coordinates : (sqrt(34), arctan(3/5))~(5.83, 0.54).
#2 you should do the same way. It's right until you convert cartesian to polar.