NOW MATH. yuck
integration which is something i love
LRAM-left hand approximationx[f(a)+f(a+x)+...f(b)]
RRAM-right hand approximationx[f(a+x)+...f(b)]
MRAM-midpoint approximationx[f(mid)+f(mid)+...]
Trapezoid-approximates using a trapezoidx/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
An example of integration is:
x^2-3 [1,4] n=3
First you find your delta x which is 1
LRAM 1[f(1)+f(2)+f(3)]1[-2+1+6]1[5]=5
RRAM 1[f(2)+f(3)+f(4)]1[1+6+13]=20
MRAM 1,2,3,42+1/2=3/23+2/2=5/24+3/2=7/21[-3/4+13/4=37/4]=47/4
Trapezoid1/2[f(1)+2f(2)+2f(3)+f(4)]1/2[-2+2+12+13]=25/2
WEll thats basically it im really having alot of trouble with the non calculator portions of the ap things that i love butt if someone can spicifically help me with related rates that would be amazing and i might actually like you since i hate all
For related rates the first thing you should do is figure out what is given. We had a problem the other day on the non-calculator portion of the exam that dealt with related rates and a triangle. They gave you the hypotenuse, which is 5 ft. They have you one leg which is x equals 3. They also gave you the rate at which that leg is decreasing which is dx/dt which is 2. So we are solving for Da/dt, which is the rate the area is changing.
ReplyDeleteSo we know:
x=3
y=4
dx/dt=2
What we don't know:
dy/dt
Da/dt
to find dy/dt, you plug in your given into the pythagorean theorum. so you get 5^2=x^2 +y^2, which gives you 25=x^2+y^2. Do solve for dy/dt you have to ge y by itself, so you get y^2= 25-x^2. Now take the derivative, 2y dy/dt= -2dx/dt. Now plug in what we know: 2(4)dy/dt=2(3)(-2). After solving that gives you dy/dt= 3/2.
We are not done yet lol. To solve for Da/dt, use the area of a triangle formula which is A=1/2xy. So take the derivative, Da/dt=1/2 dx/dt(y)+ dy/dt(x). Now plug in what you know, Da/dt= 1/2(-2)(4)+(3/2)(3). Da/dt equals = -7/4.. hope this helps.
related rates
ReplyDeleteThink of it like one of the problems that were doin in physics. Read your problem and figure out everything that you are given. If it is something that is related to time or speed or velocity or any vector quantity. (if its increasing,moving) its a rate. After that you then instead take your dirivative of your function with respect to time and plug in all that your given. Solve for your missing.
the first thing you should do for finding related rates is find all the stuff that is already given to you.
ReplyDeleteafter that you need to figure out what you are LOOKING FOR.
They gave you the hypotenuse, which is 5 ft.
so if the problem gave you a conical tank with x ft of water and blah blah blah, then asked you the rate at which the area of the water in the tank was increasing.. you would be looking for dA/dt.
so after you figure out what's given and what you need to find.. you plug in everything that's given into the formula you need. (the formula usually isn't given, you just have to know it)
^^^so for the conical tank problem, you would need to know the formula for the area of a cone.
plug in all that's given, take derivative *because you are looking for dA/dt* and solve.
related rates:
ReplyDeletethe first thing you should do is figure out what is given.
If you are given the hypotenuse, which is 5 ft. one leg which is x equals 3. the rate at which that leg is decreasing which is dx/dt which is 2. solve for Da/dt, which is the rate the area is changing.
what we know so far:
x=3
y=4
dx/dt=2
What we don't know yet:
dy/dt
Da/dt
to find dy/dt, you plug in your given into the pythagorean theorum. so you get 5^2=x^2 +y^2, which gives you 25=x^2+y^2. solve for dy/dt, you get y^2= 25-x^2. Now take the derivative, 2y dy/dt= -2dx/dt. plug in what we know: 2(4)dy/dt=2(3)(-2). solving gives you dy/dt= 3/2.
now we have to solve for Da/dt, use the area of a triangle formula: A=1/2xy. take the derivative, Da/dt=1/2 dx/dt(y)+ dy/dt(x). plug in what you know, Da/dt= 1/2(-2)(4)+(3/2)(3).
Da/dt equals = -7/4.
GOOD LUCK!