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All,

In doing multivariable calculus, I have encountered line integrals. I have no problem calculating these, but I do have a question about the intuition behind two ideas (one of which, dot product, is early on in MV calculus).

First, the role of gradients in a conservative vector field. I do understand how to determine if a vector field is conservative: it is the gradient of a potential function. And I can do the math from there. But what I don't understand is the intuition behind this. What does the gradient have to do with the fact that I don't have to calculate a curve in the conservative vector field? I'm having a hard time understanding the WHY even though I get the HOW.

Second, on dot products. These are easy to do, and I understand that you need to dot product the force and the dr before taking the integral. But, again, why? The dot product, I guess, shows the relationship between two vectors - if it is zero, the vectors are orthogonal; and the dot product increases the closer the vectors get to parallel. But what is the intuition behind why the dot product can be used to determine work, i.e., w = F * d. Again, I can solve this, but I don't understand the intuition.

Thanks.

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  • $\begingroup$ You have two different questions in the same post, I suggest you seperate and post individually. I could possible give a satisfactory answer of the second question $\endgroup$ – Buraian Dec 21 '20 at 10:15
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Gravity is a good example of a conservative force.

If you expend energy to go up a hill, you get that energy back when you come down the hill. Regardless of the path you take from A to B, the only thing that matters is the net change in altitude between the two points. If you travel back to your starting point there is no net change in altitude and no net-energy spent.

Conservative force fields tend to be centripetal. e.g $F(x,y,z) = -x\mathbf {\hat i}+-y\mathbf {\hat j}+ -z\mathbf {\hat k}$ is a conservative force with all of the force vectors pointing toward the origin. It could be that all of the force vectors are parallel, and still be a conservative force field. e.g. $F(x,y,z) = 1\mathbf {\hat i}+2\mathbf {\hat j}+ 3\mathbf {\hat k}$

What about the dot product? Again using gravity as our model for our force field. When you are traveling at constant altitude, you are spending no energy to defeat gravity. You direction of travel is perpendicular to the force and $\mathbf {a\cdot b} = 0$ when $\mathbf a$ is perpendicular to $\mathbf b$. As you climb, the dot product will give you the vertical component of your travel, and thus the energy required to climb.

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  • $\begingroup$ thanks so much for your helpful reply. At the risk of being obtuse, I'm only half-understanding still. I understand your hill example, but I'm still not sure why "a conservative VF is the gradient of another function" reconciles with "gradients are a vector in the direction of steepest ascent," the textbook definition. $\endgroup$ – Abcderia Dec 9 '20 at 17:59
  • $\begingroup$ and thanks for the dot product illustration. I do understand the dot product is zero in your example. But why does the dot product give me the vertical component? I thought the vertical component of a climb would be Fcos(theta), with theta being the angle of ascent? $\endgroup$ – Abcderia Dec 9 '20 at 18:08
  • $\begingroup$ When we have a conservative force field, we can think of some potential function. How much potential energy we have we are at what position (or altitude in the case of gravity). The gradient of this potential function shows at any point, the direction to travel to gain potential energy, and the magnitude of gain for travel in that direction. The path integral is the change in the potential function between the starting and the ending points of the path. $\endgroup$ – Doug M Dec 9 '20 at 18:12
  • $\begingroup$ The dot product is $\|\mathbf{a}\|\|\mathbf{b}\|\cos \theta$ With the climb example $\|\mathbf{a}\|$ would be your speed of travel, $\|\mathbf{a}\|\cos \theta$ would be the rate of climb. And $\|\mathbf b\|$ would be the force of gravity. Same formula as you have in your comment. $\endgroup$ – Doug M Dec 9 '20 at 18:16
  • $\begingroup$ Aha. That example of the gain in potential energy suddenly makes it all sensible to me. So to tax your time more, I get from basic physics how increasing my height increases potential energy, and it increases along the gradient, the direction of steepest ascent, or "straight up." If i were to draw a gravitational field, as we are taught to recognize, would the gravitational field of the earth be graphed with vectors pointing straight down parallel to the y-axis? if so, would a curve of, say, a rocket going straight up be graphed in a line going straight up parallel to y-axis? Is DP then, -1? $\endgroup$ – Abcderia Dec 9 '20 at 18:34

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