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(a) has to be 0 as u and the gradient vector are perpendicular to each other.

(b) must be false as the gradient vector is perpendicular to the surface, and v is not perpendicular to the surface.

(c) I have no idea how to approach this question.

Are my reasonings for (a) and (b) correct? How do I approach part (c)?

Any help will be greatly appreciated, thanks in advance.

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    $\begingroup$ To me $\bf{v}$ Looks close enough to perpendicular to say it is false for the reason that it points opposite to the actual gradient. $\endgroup$ – Triatticus Jan 3 '18 at 4:51
  • $\begingroup$ @Triatticus Although v may be perpendicular, I said false because it is not perpendicular to the surface. If you visualize the surface that the contour map is depicting, v would not be perpendicular to it, meaning that it cannot be the gradient vector. Is this right? What do you mean by "points opposite to the actual gradient"? $\endgroup$ – StopReadingThisUsername Jan 3 '18 at 5:57
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    $\begingroup$ The gradient isn't perpendicular to the surface, it's perpendicular to the level curves of the surface, otherwise all directional derivatives would be zero $\endgroup$ – Triatticus Jan 3 '18 at 6:02
  • $\begingroup$ @Triatticus Ahh you're right. My bad. Thanks for pointing that out. By the way, do you have any idea how to do part (c)? $\endgroup$ – StopReadingThisUsername Jan 4 '18 at 1:49
  • $\begingroup$ Well the line integral is a generalization of the 1D Riemann integral and essentially gives you the area of a curve under the graph and above the $x-y$ plane. The function looks continuous that you can guess approximate values between the level sets. The function is approximately monotone along the path, it's unlikely that you'll obtain $0$ as an answer $\endgroup$ – Triatticus Jan 4 '18 at 3:08

Here are a couple of ways to go about making some informed decisions in estimating a line integral given a picture:

We note the path is a simple path, it is a semi-circle of radius $r=1$ and total arc length of $\pi$, this can help in the estimation step. We also note that $f$ is approximately monotone decreasing along the path in such a way as to make estimation of values between the contour lines somewhat simple or at least probable.

First lets use a trick from single variable calculus to constrain our integral between its max and min values. We will take two integrals, both treating $f$ as a constant set by the beginning and ending of the path (because $f$ is approximately monotone we have a highest and lowest point). I will take these points as $f(3,3)\approx 4.1$ the beginning of the curve above the $f=4$ contour, and $f(3,1)\approx 1.3$ the point on the curve above the $f = 1$ contour. We then compute the two approximate integrals to help us bound the answers a bit.

Integral 1: $$\int_C f(x,y)ds = \int_C f(3,3)ds \approx \int_C (4.1)ds = 4.1 \cdot \pi \approx 12.88 $$

Integral 2:

$$\int_C f(x,y)ds = \int_C f(3,1)ds \approx \int_C (1.3)ds = 1.3 \cdot \pi \approx 4.08 $$

At least we have cut off anything less in magnitude than $4$, so the answers $-3,0,3$ are all too small.

To decide if the answer is closer to $9$ or $6$, I appeal to the drawing, the path spends more time traveling through the segment (is longer) $3<f<4$ than the segment $2<f<3$, this pushes the value closer to $9$. And lastly the integration path travels backwards towards decreasing values of $f$ and this could make the value of the integral negative, I would suspect that the value of the integral is most likely $\pm9$ with a preference towards $-9$

  • $\begingroup$ Thank you for your answer! Why is there a preference towards $-9$? I thought the result had to be positive as all f values are positive and the arc length must be positive...? $\endgroup$ – StopReadingThisUsername Jan 4 '18 at 7:27
  • $\begingroup$ No problem, I only hope it was close enough, estimation can be difficult in those situations $\endgroup$ – Triatticus Jan 4 '18 at 12:11
  • $\begingroup$ The only reason for a negative answer is think about what happens when you switch a and b in a normal Riemann integral, it switches the sign of the result even if the area under the curve is positive. I hope you get some professional input from whomever gave this question as I too am curious if I was close in my approximations. I just made the best judgement I figured would work based on what I know about line integrals $\endgroup$ – Triatticus Jan 4 '18 at 20:00

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