Recent content by rstor

What had confused me was that to my understanding you can write the conjugate of a+b as a-b OR -a+b however the complex conjugate of a+bi can only be written as a-bi and not -a+bi

I would like to confirm my understanding between conjugates and complex conjugates. When rationalizing the denominator (when dealing with square roots) it is shown that we can multiply the fraction by its conjugate. I have seen the conjugate being defined as one of the signs of the binomial...

Thank you! I thought I had done something wrong. One thing that is bothering me (from my opening post) is when coming up with the following equation: ##| \vec A |(cos \phi + 2cos111°) = 0## I could write that ##|\vec A|=0## (by zero product property) and if I substitute this into the other...

Thank you for confirming. So for figure (b), they have ##\phi=45.8°##. As labeled in the diagram, should it not be ##\phi=44.2°##.?

For configuration (a), if we follow the given diagram, ##\phi=45.8°## and the angle between the initial side and the terminal arm of ##\vec A## is ##44.21°##. Is my understanding for this part correct?

I am still having problems with this question. There is no diagram provided in the original question, however I have attached the diagram from the student solution manual. Continuing from my first post, I understand that one possible solution is ##\vec A## having a direction of ##45.8°## east...

Your answers had me look back to the solution manual. It defines the angle ##/phi## between the terminal arm and the y-axis for vector A (and states the answer as 45.8 degrees east of north) whereas I was looking at ##/phi## between the initial side and the terminal arm of vector A. Thank you all!

My solution for finding the direction of the smaller pull is slightly off from the text solution. I am unsure why. Assuming the larger vector is ##\vec B##, the smaller ##\vec A##, and the resultant ##\vec C##. My solution for the direction of this smaller pull (for the smaller pull in quadrant...

I find that I have no interest in learning an epsilon delta method for calculus. It is my understanding that unless I am going into real analysis, omitting this is okay. Occasionally I ask those in other fields such as graduates of computer science, engineering, or those studying physics at...

An earlier attempt at this question, I had reached to something similar to part of your solution as follows: ##b=\frac {\epsilon} {5-\sqrt{25+ε}}## ##b=\frac {\epsilon} {5-\sqrt{25(1+\frac{ε}{25})}}## ##b=\frac {\epsilon} {5-5\sqrt{1+\frac{ε}{25}}}## At this point I was unsure of what could be...

Thank you for taking the time to elaborate. I can see how if we assume that ##\epsilon=0## that the denominator equals zero as you mentioned in your earlier post: ##5-\sqrt{25+0}=0## I suppose one could alternatively apply the standard value function to the denominator and also conclude that it...

I am confused. The author indicates that ##\varepsilon ≠0.## In the previous section (1.5) one might use algebraic manipulation to conclude if an expression is infinitesimal, finite but not infinitesimal, or infinite using the rules outlined on page 31. Also I understood from my previous post...

In Keisler Elementary Calculus page 39, example 4 it shows how to compute the standard parts of the following expression: Example 4: If ##\epsilon## is infinitesimal but non zero, find the standard part of ##b=\frac {\epsilon} {5-\sqrt{25+ε}}## Before calculating the standard parts the...

I believe I now see my mistake: I was blindly following the rule of adding finite hyperreals without fully understanding why their sum could possibly be infinitesimal. From what I understand from Orodruin's reply: When looking at the original question, the possibility of the numerator being...

I haven't reached to the next section (1.6) as yet which deals with standard parts. The reason I came to how there could be two possible answers is based on page 31 where it outlines a list of rules. It says that b and c are hyperreal numbers that are finite but not infinitesimal. It then...