| Start | heres a fun practice problem you originally have 500 grams of an element |
| 0:30 | 8 hours later you have 400 grams if you split this up you get 250 grams, so whats half life how are we gonna find that, well the easiest you set up basic exponetial equation y=a times b to the x, now put the thought in here the original amount will be a if youre not sure you can always figure that out you can say, when x is 0, i have 500 |
| 1:01 | so 500 is a times b to the 0 b to the 0 is 1 anything to the zero is one besides 0 a is 500 so now we know a, you can take this equation and prove it y=500b to the x now you say okay and after 8 hours i have 400 |
| 1:32 | so 400 equals 500 times b to the 8 so you divide by 400 you get 4/5 so b to the 8th root of 4 5 or write it in a better way, 4/5 to the 1/8 |
| 2:05 | understand how im getting there?
you start with your two pieces of information at time 0 you have 500 at time 8 you have 400 so first you can use the 500 and 0 to find a is 500 and then you plug in a and do it again i know when x is 8 y is 400 now i can get for b that means my equationis really |
| 2:32 | y is 500
times
4/5
to the 1/8
to the x
you are just saying that 500
times 4/5 to the x/8
so far so good?
last step |
| 3:11 | now that i know that y is 500 times 4/5 to the x/8 i want to find the half life so i want to find when y=250 cause 250 is half of 500 so 250 is 500 |
| 3:30 | times 4/5 x/8 divide 250/500 you get 1/2 and now you just have to solve for x the easiest way to solve for x is with logs you can use natural logs, you can use common logs you can use any log youd like log of 1/2 |
| 4:01 | is log of 4/5 and x/8 3x/8 then you just solve for x you get 8 times a log of 1/2 |
| 4:30 | over the log of 4/5
equals x, howd we do on the half life problem
good, bad
you okay?
practice one more? lets practice one more ill leave that up there ill make up another one |
| 5:57 | you originally have a 1000 grams of an element |
| 6:03 | 6 hours later you have 200 grams, the half life is less then 6 hours after 6 hours you only have a 5th so what is its half life so we have y is a times b tot he x and you should know that a is 1000, but just in case start off with 1000 |
| 6:33 | x is 0 and anything to 0 is 1 a is 1000 we rewrite this equation as 1000 yes further more we would have 200 after 6 hours so 200 1000 b to the 6 divide it by 1000 you get 1/5 |
| 7:00 | b to the 6
b is 1/5
to the 1/6
the 6th root
so far so good?
easy, alright now we can take our equation you get y=1000 times 1/5 to the x/5 you take the x and multiply by 1/6 you get that and then you substitute in a you solve for |
| 7:31 | you have 500 grams the half life so 500 is 1000 times 1/5 to the x/6 divided by a 1000 you get 1/2 1/5 to the x/6 take the log of both sides you get log of 1/2 |
| 8:01 | log of 1/5 to the x/6 bring the x of 6 in front and solve for x and you get 6 log of 1/2 over log of 1/5 |
| 8:30 | equals x howd we do?
you get it? yay alright second set of things |
| 9:47 | tan of a is 9/13 and cosine of b is -5/11 where a is between pi and 3pi/2 and b is between pi/2 and pi find the sin of a plus b |
| 10:02 | cos2a and tan of a plus b this is mostly formulas and figure out where the angles go so i tell you that a is between pi and 3pi/2 that means that a somewhere here in the third quadrant |
| 10:31 | this is pi this is 3pi/2 the tangent of a is 9/13 sohcahtoa thats 9 and 13 you do pythagorean theorem and find that this is 250 square root of 250 howd we do on that |
| 11:08 | b is between pi/2 and pi so this is pi/2 and this is pi cosine is -5/11 little pythagorean theorem and then this is 96 and the square root of 96 |
| 11:32 | so far so good so now we just have to remmeber the formulas we will not give them on the test we might give you like one of them to help jog your memory what can i say i suggest you keep it in your head until youre done with math otherwise youre just gonna have to memorize it again |
| 12:04 | sin of a plus b sin of a plus b sin of a plus b |
| 12:31 | sin of a cos of b
plus cos a
sin of b
if im motivate ill put the formula on instagram
if im motivated
its a two hour train trip either way well
lets see what i can get
the sin of a
is -9/ the square root of 250
how do we know its negative?
because were in the 3rd quadrant |
| 13:02 | the cosine of b why it told you what cosb is its -5/11 cosA is -13/ square root of 250 the sin of b is square root of 96/11 and you can leave it right now |
| 13:33 | i dont think you have to really simplify from there we dont really care if we multiply out now i have to fins cos2a well the cosin of 2a is cosin squared a |
| 14:01 | minus sin square a you also can do cosa plus a you can memorize a couple of these formulas to figure out the rest the cos if a is negative 13 over the square root of 250 squared and the sin of a is -9/ square root of 250 |
| 14:33 | and you dont need to simplify that so 9 minus 81 over square root of 250 which is 88/square root of 250 i certainly dont expect you to rationalize and simplify from there its not important to me |
| 15:00 | oh yea 250, thank you now we have to find the tqangent of a plus b to find the tangent of a plus b, the simple way is you already have the sin of a plus b find the cosin of a plus b and divide |
| 15:30 | the cosin of a plus b is cosin a cosin b minus sinasinb so what are they? cosa is is -13/square root of 250 |
| 16:06 | cosb is minus 5/11 and sin of a is -9/square root of 250 and the sin of b is the square root of 96 over 11 and you take the top of sin and divde it by the cos and get the tangent |
| 16:33 | so its messy now we have the cosin is we wanted to find the tangent you just take the sin which is on this board take the sin and divide it by the cosin the sin divided by the cos is tangent |
| 17:00 | you can write the unsimplified expression over the other unsimplified expression that would be fine with me okay yes many professors dont care if you get to this step its usually good enough |
| 17:30 | you have to remind the tas of that one lets do another problem |
| 19:25 | you have two ships ship a leaves port at noon and sails 8 mph |
| 19:34 | ship b leaves port at 6:30 and sails about 10 mph if the angle between the ships is 60 degrees how far apart are they at 2pm and next semester youll learn how fast but not yet going this way, going that way |
| 20:00 | 60 degrees thats so how can we find c well we can use the law of cosine c sqyared is a squared plus b squared minus 2ab cos c |
| 20:30 | and what else do we know?
we know that a is how far the ship as gone at 2pm also 8mph in 2 hours so thats 16 b is going 10 mph for an hour and a half thats 15 so far so good? you know where those numbers come from? |
| 21:02 | 8mph for 2 hours is 16 miles so this is 16 10 mph per hour and 1/2 is 15 miles 10 mph for 1 hour is 10 miles plus an hour and 1/2 you get another 5 miles you get 15 miles minus 2 |
| 21:31 | times 2 times 16 imes 15 cos60 degrees that is 256 plus 225 -480 times cos60 which is a 1/2 |
| 22:01 | 2 times 16 times 15 is 480 cos60 is 1/2 481 minus 240 c squared c squared is 241 and c is square root of 241 that would be 15.7 maybe 15.8 |
| 22:30 | howd we do on this one?
so once again where do the 16 and 15 come from you go 8 mph for 2 hours so 8 miles per hour for 1 hour is 8 miles 2 hours at another 8 is 16 miles 10 mph for an hour and a half so 10 to the first hour five for the next hour is 15 |
| 23:01 | do we know the law of cosin or should we do another of those one more? alright |
| 25:08 | airplane a and airplane b
are initially at the same point
a flies at 500 mph and b flies at 300mph
the annually pi/6
how far are they after 3 hours?
alright well c squared is well lets see |
| 25:31 | airplane a
is going 200 mph
for 200 mph
for 3 hours
is 600 miles
everyone see where the 600 comes from?
200 mph for 3 hours b b goes 300 mph for 3 hours so b is going 900 miles |
| 26:01 | minus 2 times 600 times 900 cos150 degrees do scientific notation |
| 26:32 | and you get 360,000+810,000
minus 1,080,000
times negative radical 3/2
arent those fun numbers?
why is it negative radical 3/2 cos150 is negative degrees youre in the second quadrant its the same as cos30 but where on the other side so you add those together |
| 27:15 | and since square root of something awful like that i dont really care what numbers come out |
| 27:45 | whats another kind of thing you need to use
we can do one of those
everyone set with this?
|
| 28:56 | piecewise function x squared minus 4 is greater then or equal to zero |
| 29:02 | and 5 plus 3x is less then 0 find all values of x where f of x equals 0 this is very straight forward it will only take two or three minutes if we wanna know where f of x equals 0 f of x is x squared minus 4 x is positive 5 plus 3x s is negative so you shouold know where both of these parts equal to zero |
| 29:34 | x squared minus 4 equals 0 well when x squared equals 4 s is plus or minus 2 you would have to throw out the x is -2 becaues it has to be x is greater then 0 so you have to throw one of those out so you only get x=2 |
| 30:00 | now for the second equation you find out what is 0 so 5 =3x equals 0 and x is negative 5/3 negative 5/3 is less than 0, so thats also true |
| 30:56 | alright find the domain |
| 31:00 | the numerator can be anything it wants we dont care whats in the numerator denominator what about the cubed root you can get a negative number sure the only problem in the denominator is 0 the domain would be all reals except x is 5 wen x is 5 the denominator is 0 what if instead |
| 31:30 | i gave you 6-2x over the squrae root of x-5 so once again the numerator can be anything it wants but the denominator, you cant have a negative number |
| 32:01 | the denominator has to be greater then 0
and that the denominator is 0
x is greater then y
interval notation, you want to d that again?
thats 5 and infinity, thats 5 and negative infinity and 5 and 5, infinity |
| 32:39 | what i i turn this upside down and do h of x square root of x-5 over 6 minus 2x interval notation is nice but not necessary only necassary when theres graph stuff |
| 33:50 | now now we do care about the numerator because you cant do the square root of a negative number so the numerator |
| 34:02 | x has to be greater than or equal to 5, whats the difference between here and here here x cant be 5 because because you have 5 in the denominator you put in 5 and denominator you put 0 thats bad of the numerator is 0, thats fine of the numerator is 0 and 6-2x cant be 0 so x cannot equal 3 |
| 34:30 | you plug in 3 you get 0 in the denominator
square root of x-5 in the numerator so you
you can have 0 in the numerator
x could be equla to 5
or the middle one, you cant have 0 in the denominator so x cannot be 5
got the idea
yes no?
|
| 35:48 | how bout something like that sin cos negative 1 (3/7 first of all you notice its a positive number and we are in the first quadrant that means we have some number |
| 36:02 | the cos of this angle is 3/7 and now were just trying to find the sin of that angle you just want to find the sin of x use the pythagorean theorem and thats square root of 40 sinx is square root of 40/7 that wasnt so bad |
| 36:34 | one last one of these |
| 37:05 | cos now we do negative, where do we do inverse cos negative the second quadrant we have sin x cos of x is negative 2/9 so we use pythagorean theorem and that comes out square root 7 over 7 |
| 37:30 | 9 square minus 2 squared this is gonna be the tangent of x which is negative square root of 77 over 2 |