3/11/11

Alright, so for this blog I decided that I'm going to go over the concept of SLOPES and how to find them, when they are needed, etc.

1. So first, and example would be if you are asked to find the "slope of the normal line" and they give you a simple equation like this:
4x^3 + 12x^2 + 5x
and they'll say "at the point x=2"..
*Now in calculus, whenever you hear the word "slope", you're typically going to have to take the derivative of something. And in this case you will have to derive the equation they gave you and then plug in 2 for x to get the slope.
*So once you derive the equation you should get 12x^2 + 24x + 5
*Now plug in 2: 12(2)^2 + 24(2) + 5
*And once you simplify that, you should get 101 as your slope.
***Buttttttt!..since they asked for the slope of the NORMAL line, that means you have to take the negative reciprocal of that number you just got
*So the slope of the NORMAL line is -1/101

2. The next thing you could be asked to find is "the slope of the tangent line" for a certain equation...and you'll be given an x-value to plug in again here.
*You find this the same way as you did above^ except you keep the answer the way it was...As in, 101 would be the slope of the tangent line, while -1/101 is the slope of the normal line

3. Another time you'll have to find the slope is if you're asked to find "the average rate of change" or "average velocity"..Those are key words just to find the slope using this formula: f(b)-f(a)/b-a...And typically you'll just be given an interval [#,#] to work with.
Ex. Find the average rate of change of a particle for f(x)=3x^2+5x during the time interval [1,4]
*All you have to do for this is simply plug into the formula:
f(4)-f(1)/4-1
(68-8)/3
*slope = 20

4. Another time you'll need to find the slope is when in a problem you're asked to apply the Mean Value Theorem for an equation and a time interval..Remember for the Mean Value Theorem, you have to find the slope and then set it equal to the derivative of the equation and solve to find a value of c

5. You may also be asked to "find the slope of the horizontal tangent" for a certain equation...but, do not be fooled by the "slope" in the question! It does NOT mean you have to find the slope! "Slope of the horizontal tangent" simply means to take the derivative of the equation they gave you, set it equal to zero, and solve for x.
Ex. Find the slope of the horizontal tangent for f(x) = (6x^2+3x)/(5x+1)
*Oh, a fraction...the same rules above still apply here! Just use the quotient rule to differentiate. And you should get this:
(30x^2+12x+3)/(5x+1)^2
Now to find the slope of the horizontal tangent, all you have to do is set the top of the fraction equal to zero and solve for x
30x^2+12x+3 = 0
..And it doesn't factor..but you get the point on what to do haha