The Official Math/Physics/Whatever Homework questions thread

[solsek]

Originally Posted by lastcardcharlie

|a + bi| = sqrt(a^2 + b^2) = r and b/a = tan(theta).

Thanks for the fast reply. So I haven't taken a math class since my junior year of high school and now I'm a junior in college so please forgive me with all these dumb questions. The problem is 3 + 4i, and I just need to convert it to re^itheta. I get r = 5, and tan(theta) = 4/3. How do I convert this value in terms of pi? You don't need to give solutions, just a guide in the correct thought process and procedure would be great! This is for my physical chemisty class btw. Our assignment says to not use calculators and to simply everything.

[Never Win]

Okay, hopefully someone can explain this to me. It is about solving a system of equations for three equations. Here is the example in the book: 2x+y-z=2
x+3y+2z=1
x+y+z=2
Solution: Begin by eliminating the term with x, this time from equations 2 and 3 as follows.
2x+y-z=2
-5y-5z=0
-y-3z= -2
My question is how did the book arrive at the last two equations?? There are a couple more steps to solve the problem, but I have to understand the first step before I move on. If anyone could help I would greatly appreciate it.

[lastcardcharlie]

Originally Posted by solsek

...and tan(theta) = 4/3. How do I convert this value in terms of pi?

Sounds like theta is required to be in radians is all, since there's 2pi radians in a circle. I don't know if there's some cute way of doing it, or if you simply do tan^-1(4/3) in radian mode on your calculator.

[solsek]

Thanks, looks like its just a preference issue. Would leaving the exponent as i(tan-1(4/3))pi suffice as theta? Is that a standard notation?

[Wyman]

Originally Posted by Never Win

(#1): 2x+y-z=2
(#2): x+3y+2z=1
(#3): x+y+z=2

Solution: Begin by eliminating the term with x, this time from equations 2 and 3 as follows.

(#1): 2x +y -z = 2
-2*(#2): -2x-6y-4z=-2
(#1) - 2*(#2): -5y-5z = 0
(#1) - 2*(#3): -y -3z= -2

You're allowed to create new equations by

1) multiplying an old equation by a constant.
Idea: If Left Side = Right Side, then certainly
c*(Left Side) = c*(Right Side)

2) Adding two equations together (on both sides)
Idea: If A = B and C = D, then
A + C = B + D

So combining those ideas together, we can multiply an equation by a constant and add the resulting equation to an existing one. In the case above, we used this technique to cancel the x's in eqns #1 and #2.

Hope this helps.

-BW

[Wyman]

Originally Posted by solsek

Thanks, looks like its just a preference issue. Would leaving the exponent as i(tan-1(4/3))pi suffice as theta? Is that a standard notation?

yes

edit: lol that i had to wait to post this, since the forum requires > 25 seconds between posts. Seriously, I got flagged for that in a math homework help thread. FML.

edit #2 -- siegmund is right. you added an extra pi and I missed it. theta = arctan(4/3).

[Siegmund]

theta=arctan(4/3) may well be acceptable. Not the extra pi. arctan(4/3) is a number -- about .927 radians, or 53.1 degrees, or 0.295*pi radians.

It doesn't look like their intention is for you to search for a rational equivalent.

[lastcardcharlie]

Originally Posted by solsek

Would leaving the exponent as i(tan-1(4/3))pi suffice as theta?

Just i(tan^-1(4/3)). Note that this form does not specify whether theta is to be considered in radians or degrees. They're just different units of measurement, like miles and kilometres. x degrees = x(pi/180) radians.

[Never Win]

Originally Posted by Wyman

You're allowed to create new equations by

1) multiplying an old equation by a constant.
Idea: If Left Side = Right Side, then certainly
c*(Left Side) = c*(Right Side)

2) Adding two equations together (on both sides)
Idea: If A = B and C = D, then
A + C = B + D

So combining those ideas together, we can multiply an equation by a constant and add the resulting equation to an existing one. In the case above, we used this technique to cancel the x's in eqns #1 and #2.

Hope this helps.

-BW

Yes, thanks for clearing this up for me.

[solsek]

Thanks everyone for the help. Helped out a lot since I had 15 of these problem to work out.

[Myrmidon7328]

Cool problem from my math test today. I had about 10 minutes to do it.

How many integers between 0 and 10,000 have digits that sum to 13?

[Wyman]

Originally Posted by Myrmidon7328

Cool problem from my math test today. I had about 10 minutes to do it.

How many integers between 0 and 10,000 have digits that sum to 13?

Spoiler:

561? Or am I teh suck at adding?

[ManUTDFan]

Why for small angles does sin(theta) = theta? Can someone explain the reasoning behind, or point to a website that can show some graph/pictures why? I remember something about pendulums and why we can use sin(theta) = theta for solving for variables in a pendulum problem but I don't have the slightest idea how to explain it, our physics professor just said assume sin(theta) = theta for small angles and never explain why in detail.

Also quick question...pretty simple one, just want to make sure if I am on the right track - lol first assignment with no lectures about it and its due tomorrow

Simplify: 2e^(3ln(2)) = (2e^3)(e^ln2) = (2e^3)(2) = 4e^3

Sound about right?

[Wyman]

How much, if any, calculus do you know?

edit: the short answer is just -- graph y=x and y=sin(x), and see that they are very close near x=0. The longer answer can involve tangent lines, linearization, and/or Taylor series.

[ManUTDFan]

Originally Posted by Wyman

How much, if any, calculus do you know?

If this is referring to me, I took calculus in high school and never bothered to take it in college (always putting it off until next semeter haha, bad idea, forgot everything). That was about 5 years ago and I am retaking it because some medical schools require college calculus ldo. Btw, what I am working on is like a "warming up math skills" homework assignment, the class isn't a calculus class, just uses math to derive certain equations that we will end up using.

edit: Thanks for the help, I will look up some of the long answer hints you gave me and see if I can jog my memory.