This week in Calculus we continued reviewing for our exams up until Wednesday. That's when we took the multiple choice part, and thursday we took the free response. After doing all of the study guides multiple times i thought i was finally getting a hang of everything, but i probably thought wrong.
Friday, we started learning again. I was pretty nervous, but implicit derivatives seem to be pretty easy....so far. I'm still waiting on the catch though. The first thing you need to know about implicit derivatives is that it involves x's and y's. The steps are shown below:
1. Take the derivative like normal of both sides.
2. Every time you take the derivative of y note it with dy/dx or y'.
3. Solve for dy/dx or y'.
If you want the slope you must plug in a x and a y-value. Remember if you're only given an x and not a y, plug in the x value to the original function to find the y.
An example of an implicit derivative is y^3 + y^2 - 5y - x^2 = -4
The first thing you should notice is there is x's and y's, meaning you must take the implicit derivative.
3y^2 dy/dx + 2y dy/dx - 5 dy/dx - 2x = 0
dy/dx(3y^2 + 2y - 5) = 2x
dy/dx = 2x/3y^2 + 2y - 5.
Pretty simple, right :)
Some of the things i think i've finally gotten the hang of are tangent lines and limits.
Some things i am still having trouble with is of course, optimization, and looking at the graphs and being able to tell what the derivative looks like or where the points of inflection are, etc.
GOODNIGHT :)
Subscribe to:
Post Comments (Atom)
for optimization..
ReplyDeletelets say that the promblem says:
solve for the total length and width of two squares that are not connected which have a perimeter of 24!
so we know that we have a perimeter of 24 right...well to make your secondary and primary equation you make two squares...we know that the area of a square is lenth times width! and that there is two lengths and two widths for each square so your seconday equation will be 2(2L+2w)...that's your perimater because you're adding them...and since we know that the sum of the two perimeters is 24 you set 2(2L+2w) equal to 0!
so now...dealing with optimization:
take your secondary equation and solve for...lets go with length
2[2(L)+2(w)] = 0
24=4(L)+4(w)
24-4(w)=4(L)
(L)=6-w
so now plug in L to the primary equation which is your area...so that's
A=(L)(w)
A=(6-w)(w)
A=6w-(w)^2
TAKE THE DERIVATIVE OF THAT:
A'=6-2w
-2w=-6
w=3
SO NOW TAKE WHAT YOU KNOW....W=3....and PLUG IT INTO THE SECONDARY FORMULA:
24=2(2(L)+2(w)
24=2(2(L)+2(3)
24=4L+12
12=4L
L=3
so knowing your W is 3 and your L is 3
[it's a square so they had to be the same]
you know that the width is three and the length is three!!
Ok, for optimization, when you're given a word problem that asks for something like the deminsions of a rectangle and to maximize the area, first you find the equtations you need. Most of the time there will be two equations in the problem: the primary and the secondary. The way I decifer between the two is I always look for the one that is set equal to a number. This one will always be your secondary. The other equation, the thing the problem is asking you to maximize will be your primary equation. After you find your primary and secondary equations, the steps are as follows:
ReplyDelete1. Solve secondary for one variable
2. Plug into primary
3. Take the derivative
4. Set equal to zero (to find first answer)
5. Plug back into secondary and sove (to get your second answer)
Also, when you are looking at a graph of an origional function, always remember that the maximums and minimums on that graph will always be the zeros on your f'(x) graph and the zeros on your origional function will be maximums and minimums on your f'(x) graph.
ReplyDeleteFor graphs, looks at the wiki i posted (problem 10) or http://s180.photobucket.com/albums/x15/ryanb985/?action=view¤t=scan0001.jpg
ReplyDeleteremember to keep ya primary and secondary equations striaght if you get em mixed up you gonna get it wrong
ReplyDeleteok so when you look at the graph of a function. you need to first try and figure out what the equation will look like. for example if you have a parabola, it would be something along the lines of f(x)=x^2 + 3. or anything with an x^2 in it. then, you must figure out what the derivative of that function looks like. whatever it is.. it will be equal to something like 3x, or any number times x. the graph of that would be some type of diagnol line. so pretty much, it's like guess and check, and the way you figure out if it's positive/negative or increasing/decreasing, is also by looking at the original function. if the parabola is positive, then your diagonal line will be increasing, and if the parabola is negative, the diagnol line will be decreasing.
ReplyDelete