Consider a vector field defined as the gradient of some well-behaved scalar function:
$$\mathbf{v}(x,y,z) = \nabla T(x,y,z).$$
What is the value of $\oint_C \mathbf{v} \cdot d\mathbf{l}$?
1. Zero
2. Non-zero, but finite
3. Can't tell without a function for $T$
Note:
* CORRECT ANSWER: A
* Closed loop integral of a gradient is zero.
* Fall 2016: [92] 4 4 0 0
## Announcements
* Homework 1 solutions posted immediately after class
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- Agrawal, Prakash
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For me, the first homework was ...
1. entirely a review.
2. mostly a review, but it had a few new things in it.
3. somewhat of a review, but it had quite a few new things in it.
4. completely new for me.
I spent ... hours on the first homework.
1. 1-2
2. 3-4
3. 5-6
4. 7-8
5. More than 9
## Numerical Integration
<img src="./images/numerical_midpoint.gif" align="center" style="width: 600px";/>
Consider this trapezoid
<img src="./images/trapezoid_shape.png" align="left" style="width: 300px";/>
What is the area of this trapezoid?
1. $f(c)h$
2. $f(d)h$
3. $f(c)h + \frac{1}{2}f(d)h$
4. $\frac{1}{2}f(c)h + \frac{1}{2}f(d)h$
5. Something else
Note:
* Correct Answer: D
The trapezoidal rule for a function $f(x)$ gives the area of the $k$th slice of width $h$ to be,
$$A_{k} = \frac{1}{2}h\left(f(a+(k-1)h) + f(a+kh)\right)$$
What is the approximate integral, $I(a,b) = \int_a^b f(x) dx$, $I(a,b) \approx$
1. $\sum_{k=1}^N \frac{1}{2}h\left(f(a+(k-1)h) + f(a+kh)\right)$
2. $h\left(\frac{1}{2}f(a) + \frac{1}{2}f(b) + \frac{1}{2}\sum_{k=1}^{N-1}f(a+kh)\right)$
3. $h\left(\frac{1}{2}f(a) + \frac{1}{2}f(b) + \sum_{k=1}^{N-1}f(a+kh)\right)$
4. None of these is correct.
4. More than one is correct.
The trapezoidal rule takes into account the value and slope of the function. The next "best" approximation will also take into account:
1. Concavity of the function
2. Curvature of the function
3. Unequally spaced intervals
4. More than one of these
5. Something else entirely
<img src="./images/hung_spheres.png" align="left" style="width: 300px";/>
Two small spheres (mass, $m$) are attached to insulating strings (length, $L$) and hung from the ceiling as shown.
How does the angle (with respect ot the vertical) that the string attached to the $-q$ charge ($\theta_1$) compare to that of the $-2q$ charge ($\theta_2$)?
1. $\theta_1 > \theta_2$
2. $\theta_1 = \theta_2$
3. $\theta_1 > \theta_2$
4. ????
Note:
CORRECT ANSWER: B
Draw the FBD, Newton 3 is important here.