Directional derivative without reparametrization

[ZPlayer]

Hi, Everyone,

I have the following problem. Given function is g (s, t) = s ^ 2 * exp (t). Find directional derivative of the function at the point (2,0) in direction of vector v = i + j. The book says to find del g with respect to s and t and dot it with normalized v. But I disagree because g is a function of s and t (not x and y) whereas v is defined as i + j. Can somebody help?

Thank you

 

[HallsofIvy]

"x and y" and "s and t" are just SYMBOLS. You can call the coordinates anything you like as long as you are consistent. Obviously this problem is asking you do the problem in "s,t space". That is, just treat "s" and "t" as another name for "x" and "y".

 

[ZPlayer]

Ok. The next problem in the book is:

"Find directional derivative of the function at the given point in the direction of the vector v. The function is g(r, theta) = exp (-r) * sin (theta), the point is (0, Pi/3) and v = 3i - 2j."

In solutions manual, they solve it just like you suggested, i.e. dot del g with normalized v. However, I think the choice of variables in this case makes it apparent that r implies radius and theta is angle, so the function is formulated in polar coordinates. However, the direction vector is in Cartesian coordinates. Do you think that they answer is correct? If yes, please explain why.

Thank you.

 

[HallsofIvy]

WHAT answer is correct? $\nabla g$ is a vector. In Cartesian coordinates it is given by
$$\nabla g= \frac{\partial g}{\partial x}\vec{i}+ \frac{\partial g}{\partial y}\vec{j}$$
In polar coordinates it has a very different formula in terms of the partial derivatives of r and $\theta$! You can calculate them, but since they are somewhat more complicated, the best way to do this problem is to convert $g(r, \theta) = e^{-r}sin(\theta)$ into Cartesian coordinates.