Continuing with the last reading, your next reading will again be from our course notes, Sections 3.4–3.6. (If it’s been a while since you’ve looked at *vector* calculus, now may also be a good time to brush up on things like div, grad, and curl.) We’ve again included some simple calculations below, to test your own understanding of these ideas—as usual, please post in the comments if you have any trouble! This reading will be due before class next Tuesday. *Note that it is important to do these warm-up exercises, because your next homework will be all about exterior calculus!*

- Consider the scalar function \(\psi: \mathbb{R}^2 \to \mathbb{R}; (x,y) \mapsto \tfrac{1}{2}x^2 – 3y\), and compute the quantities \(d\psi\), \(\star d\psi\), \(d\star d\psi\), and finally \(\star d \star d \psi\). Express the results in terms of the basis 1-forms \(dx,dy\).
- Consider the scalar function \(\phi: \mathbb{R}^2 \to \mathbb{R}; (x,y) \mapsto e^{-(x^2+y^2)}\). Calculate the exterior derivative \(d\phi\), expressing the result in terms of the basis 1-forms \(dx,dy\).
- Using the same function \(\phi\) as above, evaluate the directional derivative \(\phi(X)\), where \(X := -(\tfrac{1}{4xe^{-(x^2+y^2)}}\tfrac{\partial}{\partial x} + \tfrac{1}{4ye^{-(x^2+y^2)}}\tfrac{\partial}{\partial y})\). Note that unlike the previous reading, here you are working with a “1-form
*field*“, i.e., you are effectively taking a different inner product at each point. Alternatively, you can imagine that you’re computing a different directional derivative at each point, where the direction is given by \(X\). - Integrate the 1-form \(\omega := \sin\theta\ d\theta\) over the unit circle.

To give a couple concrete examples of what these calculations look like: if \(f(x,y) := y/x – y^2\) and \(Z := \tfrac{\partial}{\partial x} – \tfrac{\partial}{\partial y}\), then

\[

df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = -\frac{y}{x^2} dx + (\frac{1}{x}-2y) dy

\]

and

\[

df(Z) = -\frac{y}{x^2} \underbrace{dx(Z)}_{=1} + (\frac{1}{x}-2y) \underbrace{dy(Z)}_{=-1} = -\frac{y}{x^2} – \frac{1}{x}-2y = \frac{2x^2 y – x – y}{x^2}.

\]

**Submission:** As usual, please send an email to kmcrane@cs.cmu.edu and nsharp@cs.cmu.edu no later than **10:00 AM on Tuesday, March 1st** including the string **DDGSpring2016 **in your subject line. Your email for readings should always include:

- a short (2-3 sentence) summary of what you read, and
- at least one question about something you found confusing / interesting / incomplete / not addressed.

In the example where $f(x,y)=\frac{1}{x}-y^2$, I think that the partial derivative $\frac{\partial f}{\partial x}=\frac{-1}{x^2}$, not $-\frac{y}{x^2}$, and similarly I have that $\frac{\partial f}{\partial y}=-2y$. Is there something that I am missing?

Yes, just a minor typo: \(f\) should be \(y/x-y^2\). Now fixed.