Calculus Unit 2: The Derivative
Choosing to make the derivative graph straight means you're
--> PRETENDING THE RATE THE FUNCTION CHANGES IN CONSTANT
Multiplication and division rules
-Don't know multiplication and division shortcuts -Multiply out, then use shortcuts -Same with dividing, individual way earlier, etc.
Horizontal tangent line problem
-FOIL out squared term, then combine like terms, get derivative of that -*Horizontal tangent line indicates a derivative of zero* -Set it equal to zero, isolate x to determine that! Maximum/minimum value touches x-axis
First derivative graph example
-Graphing f'(x) tells us the value of slope at any input (ex . input of -5, value of slope = -7) -Think of this as your friend telling you the slope for any input and writing that point down (plotting) -Ex. f'(x) output of zero means a slope of zero ---> Instantaneous rate of change!
Graph of f(x)
-Instantaneous rate of change at any point on the curve can be given to us by f'(x)=4x-4 For instance, f'(2)=4(2)-4= an instantaneous rate of change of 4 (see point 2, that slope makes sense --> large and positive) f'(-2)=4(-2)-4=-12 ---> negative, large rate of change in terms of magnitude -See lines displayed on curve (like earlier) to show the instantaneous rate of change of each point.
About this topic
-Large section on the test for it, each question worth 4 points -Understand it really well, do your homework (matching, etc.) WITH SKETCHING and you'll be able to tell if you're ready -Hit or miss type thing -ASK YOUR QUESTIONS NOW BEFORE IT IS TOO LATE! -Review H/W, note sections that were harder & wrong answers
Position, Velocity, and Acceleration H/W
-Look over, tons of different problems -Finding derivatives a lot -Be sure to list your variables and the equations you'll need, go step by stpe -DOWNWARD VELOCITY IS NEGATIVE --> UNDERLINE WHEN READING PROBLEMS -whole idea of factoring out -16 as well -made a lot of mistakes, so note that & your writing on the last page, etc.
Next graph - new point
-Mark a second point x+∆x, f(x+∆x). -Create a second line that hits that point and the point x, f(x). ∆x=the amount we moved
Test/review
-See topics you will/have review/reviewed -2 classes, then review day -YOU MUST LEARN THIS WELL, RECENT CONCEPT TOO, REVIEW YOUR HOMEWORK!! AND CLASS EXAMPLES. Understand process and weird things that might come with that.
Example problem - first one
-Soon we will be able to look at this function and give the derivative in under ten seconds (we will learn shortcuts), but we can always use this formal definition if shortcuts don't apply, etc. -Begin by plugging f(x+∆x) in to f(x). You want to FOIL the squared term out and multiply out the linear one. You can combine all your new terms now, and plug it into the definition of: f'(x)= lim [f(x+∆x)-f(x)/∆x] ∆x->0 Remember that you're already given f(x)!! When you subtract f(x) (PUT IN PARENTHESIS), you will see that all of it cancels out with terms that don't have ∆x in them. You are now left with 3 terms that have ∆x in them, all over ∆x (haven't plugged in anything yet, reasonably so) Since all of the numerator terms and the denominator have ∆x in them, you can factor it out of the numerator and cancel it with the denominator. Now you have: lim (4x+2∆x-4) ∆x-->0 Do direct substitution, make ∆x = 0 ---> you get *f'(x)=4x-4* -PLEASE LOOK AT YOUR NOTES FOR ACTUAL NUMBERS AND MATH/PROCESS!!!! -Remember this is subjective
Fourth graph
-Start out with a positive slope that gets less positive and moves towards zero. Eventually you hit zero, then get negative. You get less and less negative, approach zero, hit it, then the graph is over. See graph for elaboration and f'(x) to see how the slope works ---> cusp within the derivative MORE IMPORTANT: -Current down at the negative point but the curvative changes --> curved up. This is called an inflection point. The slope stays negative before and after the inflection point, but there's a change in the way the graph is trending.
Third graph
-Start with a negative constant slope, so we draw a HORIZONTAL LINE However, there's a cusp at zero, so we jump up to positive slope after putting a hole there because it's undefined. We then have a constant positive slope up until the finish
formal definition derivative problems 2 and 3
-being given f(x) -plugging in f(x+∆x), simplifying accordingly after putting into definition of derivative -3rd problem: MUST FIND LCD. the LCD is (SEE BACK FOR WORK) (x)(x+∆x). Keep this in the denominator so you can combine the fractions but you can cancel out numerator variables that repeat in the denominator two times. simply left with -∆x, so multiply by reciprocal (get rid of denominator) and you'll get that cancelled as well. after simplifying accordingly, you're going to be left with an expression that can be solved via direct substitution, and you'll have f'(x) -SEE NOTES!
d/dx cos(x)
-sin(x)
Class drill -- THIS IS VERY VALUABLE INFORMATION, SO REVIEW IT!
1 page of drawing derivatives on test 1st graph: slope is never zero, so be aware of that when you're drawing! it goes from positive to positive, but tends towards zero in the middle *read out slope numbers at each point* 2) plot any minimums/maximums first --> graph is still positive but but less, almost goes back to zero -Inflection point on y-axis --> a point on a graph where a change in curvature occurs 3) see your opened and closed circles --> infinite amt. of tangent lines at a cusp
General steps for this process
1) Obtain slope via derivative 2) PLUG IN to slope equation with information you know and obtain final function y-intercept
First problem - process (finding k)
1) Set functions equal (they touch at that point, as they have the same output values) 2) Set their derivatives equal --> obtain both of them Ex. -6=-2x ---> K IS A CONSTANT!! --> solving for x, which is 3. Plug that into the tangent line to determine y, which is -17. PLUG BOTH INTO f(x) and get k=-8/ *Put it back into the function where it was missing* *f(x)=8-x²*
What two things do you need to write the equation of a line?
1) Slope of the line 2) A point
Two ways to write slope
1) y=mx+b (slope-intercept form) 2) y-y₁=m(x-x₁) [point-slope form] Derived from slope formula, moving denominator to other side
tangent line
A line that hits the curve at exactly one point
secant line
A line that hits the curve at two points
Acceleration
A(t)=V'(t)=S''(t) Derivative of velocity/derivative of the derivative of position
Just like any time before
Always look over your actual notes rather than just the Quizlet. They give context, explanation, and process, and are especially important in this scenario of doing tedious problems.
Derivative of a constant ---> why?
Always zero, since the slope of a horizontal line is zero
Derivative:
An output, a y-value -Calculate it going left to right
Derivative in this context
As always, it's the rate of change for example, miles/hour, ft/s, et cetera Distance = meters ---> apply a derivative ---> meters/sec (velocity) ---Getting rate of change Apply a derivative to velocity ---> acceleration (adding another denominator) meters/sec/sec Overview: Distance/position function --> take derivative ---> creating velocity ---> take derivative ---> acceleration
Instantaneous Rate of Change
At an exact point, what is the slope? -The slope at the tangent line for that point Ex. Driving to school, what is your speed at one exact moment?
HW review
Complicated formal definition process --> see LCD, final answer, process, etc.
Taking a quantity and making it into a rate:
Distance ---> Velocity ---> Acceleration
Polynomials, etc.
Divide them up and find the derivative of each term, remember that about the constants ex. 2 ---> 2x⁰ --> 0x⁻¹ -Might have to change things around to make it a form you can use
More complicated example
Do the same: set functions equal, then get their derivatives and set those equal to each other. 2x-k=5 --> Set this equal to k: k=2x-5 *Then substitute into f(x)* f(x)=x²-(2x-5)x f(x)=x²-2x+5x = -x²+5x --Set equal to tangent line, cancel out 5x and get that f(x) = -x²=-4 ---> multiply both sides by -1: f(x)=x²=4, so x=±2 NOT DONE YET---PLUG INTO TANGENT LINE FOR BOTH X-VALUES, GET YOUR TWO Y-VALUES FOR EACH, THEN YOU CAN PLUG INTO F(X), GET K, ISOLATE --> TWO ANSWERS! MAKE SURE TO LIST TWO DIFF. FINAL F(X) DETERMINATIONS USING K VALUES OF -1 AND -9
Hierarchy of cuntions
F(x) f(x) f'(x) f''(x) f'''(x) *F(x) comes before what you'd normally consider your regular function --> keep taking derivative (see paper) -Differentiation ---> things happen in this order -Going up would be integrating = anti-derivative
Falling object model
FEET: s(t)=-16t²+v₀t+s₀ or (meters) s(t)=-4.9t²+v₀t+s₀ *Tells us the height of a falling object above ground as time moves on*
Next example
Filling in table Remember that d/dx sin(x) = cos(x), so work based off of that Slope is 1 when x=0, so plot one, et cetera -Look at how you're making lines at each input to resemble the tangent line Take tangent line, place an f(x) point, see how that affects output (ALSO PLOT DERIVATIVE) Above all, we can plot derivatives just by looking at the graph -Examine this closely please
Power Rule for Differentiation
If f(x)=xⁿ, then f'(x)=nxⁿ⁻¹ -See all examples in our notes ---> bringing down n and multiplying by the whole expression, not just x -Not noting examples here b/c there are so many, also look at the class document
Example d -- why does it take less time than the previous model?
Initial velocity of 0 ft/s in this case since he isn't jumping off and an initial height above ground (s₀) of 32 ft like before set that equal to 16t² and get that t is equal to the square root of 2, which is about 1.4 s -It takes less time when you're walking straight off because you're starting with no velocity and assuming the max height is lower while also traveling down in a straight line
∆x -->
Make ∆x smaller-->closer to the tangent line (ideally zero, but can't do that)--> you're going to get infinitesimally close to 0
Example c
Maximum ---> the derivative of a maximum is zero, which means a velocity of zero in this case Set 0 equal to v(t) or s'(t), then get t Once you get t, you can plug it into the original position function and determine your output in feet
Common test error
Must include limit at the end when working w/ formal definition ---> cannot get to answer without applying the limit -Need to show limit in order to get to answer
Rational exponents
Numerator=power, denominator=root (remember!!) -See examples for these too, H/W when you put negative exponents in denominator of a fraction -See x²/³, putting it in a form you can work with!!
Cusp
Pointy spot of graph where it changes slope immediately -Infinite amount of tangent lines for a cusp ---> NO DERIVATIVE because there's more than one tangent line (derivative = slope of THE tangent line that passes through a point) ---> HOLE -Changing direction instantaneously, which isn't possible in the real world
First graph analysis
Rate of change is changing Calculus=changing rates of change Instantaneous rate of change at all the points is marked by a tangent line (more ---> less negative, 0, then less ---> more positive based on each of their slopes_
Example problem - a --> what does our negative time value tell us?
Review fully, finding t after plugging in your given info. -Factor it out, then use parenthesis to get that t= 2 s and t=-1 s t=-1s --> other x-intercept, just not in the domain we're considering --> see drawing: it tells us how long it would take to go up the other side of parabola to get to the maximum (thinking about the real-life theme here, which is something we're going to need to do a good bit)
10/13 Homework
Review that --- finding slope at a particular point, then horizontal tangent line (setting derivative equal to zero, getting x value, THEN PLUGGING BACK INTO ORIGINAL AND GETTING Y-VALUE -->POINTS!!) -Look over well
notes on 2 videos
Review well, not everything was noted because most of it was redundant, but still important to look over closely
position
S(t)
Finding equation of a tangent line at a given point -- example
SEE example -Get derivative (using x-value) -Plug in your value to determine m -Then use one of the two slope forms mentioned above (slope-intercept or point-slope) to figure out the value for b. Plug in your original x and y values to determine this. Slope here is 4, b=2, so y=4x+2 y-(-2)=4(x-(-1)) y+2=4x+4 y=4x+2, same thing!
Given functions and their tangent lines:
Same function value but also have the same derivative/slope at the points we're working with -Graph on calculator --> see the two diff. curves and them both hitting the tangent lines, as we got two answers for k here
H/W
See how to make position and time graph into velocity ---> calculate it, convert to mi/hour from mi/minute and note your rate of change!! For instance, since nothing changes from minute 6 to 8, there's a velocity of zero there. Consistent velocity at the begin and for the last two minutes (low) ---Look at your math!! *Once graph is always going to be a distance graph, and the other will be velocity*, so be able to go both ways and convert from velocity --> distance graph as well
Intervals increasing/decreasing
See responses, can do interval notation, PARENTHESIS VS. BRACKETS - OPEN VS. CLOSED INTERVALS, f'(x) above or below x-axis respectively
Finding slope of graph at a given point
Simply determine derivative, plug input into that to get the instantaneous rate of change ANYWHERE on the function
Parabola -- point
Slope of that exact point = will be the slope of the tangent line that passes through that point
Minimum point
Slope values going from negative ---> zero ---> positive -g(x) has a minimum point at x=1 and x=-2 (slope increases after it hits both of these points) x=-2 ---> a little confused, ask question in class about that
Maximum point
Slope values going from positive-->zero--> negative -Remember that you can tell your extremes via them being on the x-axis -Maximum point at x=-1 because the slope goes from positive--->zero---> negative when it reaches the point
Introduction to derivatives
Slope: m=y₂-y₁/x₂-x₁ = ∆y/∆x -Also known as rate of change -Derivative = the MOST important process in all of Calculus (this and next class we'll get background knowledge, see basic derivation, and learn the formal definition, which is a little tedious)
Why does this make sense?
The derivative tells us the slope at any point
inflection point
The point where a change in curvature of a graph occurs (from concave down --> up, et cetera) -see notes for further elaboration
Velocity
V(t)=S'(t) DERIVATIVE OF POSITION, as mentioned earlier
Example b
Velocity at impact is our final velocity -Use equations, derivative of position, plug in the time and calculate -Makes sense that it's negative because you're going in a downwards direction
Process
When you move your point down, you're making ∆x smaller and smaller --> pick a point that's closer to your first one -Trying to make the slope of the secant line match the slope of the tangent line (move point further down to match the other one)
Can you take the derivative of a derivative?
Yes, because it's a function NOTES: took derivative of a derivative 3 times, it worked just fine, using more and more ' every time!
However,
a slope needs two points, and for tangent lines, we only have one point.
see 10/21 class notes for...
an overview of velocity, position, and acceleration h/w Main questions on test: 1) How long it takes to reach ground (quad. form, factoring w/ s(t)) 2) Speed at ground etc...
LCD process
combining the two or more denominators into one, then cancelling new numerators out accordingly with your existent denominators (remember that denominator likes to be kept in factored form!)
d/dx sin(x)
cos(x)
When the function is falling,
derivative is negative (below x-axis)
When the function has a maximum or minimum then
derivative is zero (touches x-axis)
When dividing a function by x,
divide each individual component of the function by x, then use your power rule for differentiation -Negative exponents are positive in denominator, same thing!
Derivative of f(x) is commonly written as:
f'(x) f PRIME of x
First graph for drawing derivative graphs
f'(x) ---> approximate, trend is correct Less and less positive, getting closer to zero
derivative notation
f'(x) --> y' --> dy/dx --> d/dx f(x) [derivative of fn. with respect to x]
Slope of the tangent line AKA instantaneous rate of change
f'(x)= lim [f(x+∆x)-f(x)/∆x] ∆x->0
Formal definition of derivative
f'(x)= lim [f(x+∆x)-f(x)/∆x] ∆x->0 It is simply the instantaneous rate of change at a point
Second graph
f(x) gets less negative and approaches zero, then gets more and more positive slowly (hits zero) See graph to line up vertex w/ XInt, etc. -Should be able to go back and forth from derivative ---> f(x), not on test though
The derivative of f(x)...
is a *function* which tells us the *instantaneous rate of change* *anywhere on the original function* -Note underlined words/phrases
When our estimation gets closer and closer to the point of tangency,
it becomes exact since the two points will be infinitesimally close --> finding the slope of a SINGLE POINT here
Slope of the second line
m=y₂-y₁/x₂-x₁ = ∆y/∆x = f(x+∆x)-f(x)/x+∆x-x "Clean up" denominator ---> f(x+∆x)-f(x)/∆x ---> this is the slope of the secant line
Be comfortable with...
negative exponents, rational exponents, and simplifying expressions.
see examples
of these, also note graphical stuff and how that relates to sin/cos function and their derivatives
Speed is always
positive, velocity (speed with direction) is positive or negative depending on its direction
In calculus, we're dealing with instantaneous
rates of change of a curve at a point Slope = dy/dx Super small ∆y/super small ∆x -see multitude of ways to note this
To finish the unit
review sheet review on the hardest problems as well as those you made mistakes on. you made an error on formal definition which was tiny but changed your answer, an error with not paying attention on a tangent line function, and then two graphing errors & a derivative error *See corrections, what you did wrong, process...* *Sketching needs the most work, but do full review to ensure you know everything. Sunday test*
negative exponents process
same as putting exponent in denominator ---> do that then convert to power/root if applicable, make sure to also incorporate the constants involved -see examples
d/dx tan(x)
sec²(x)
What these variables mean and their units
s₀ = initial height above ground ---> could be negative! (ex. in a ditch) --> ft or m t=time in seconds position: feet or meters velocity: ft/sec or meters/sec acceleration: -32 ft/sec² or -9.8 m/sec²
Idea of...
taking a secant line and making it approach a tangent line -Make it approximate the tangent line by moving the ∆x in towards it
When the function is rising,
the derivative is positive (above x-axis)
Slope of the secant line is approximately
the slope of the tangent line However, we want our estimate to be better --> move estimation point a little closer to the point of tangency --> KEEP MOVING CLOSER AND CLOSER
Conclusions --> if you need elaboration
use common sense and/or the previous graphs
Why do we add the variable k?
we can draw infinite functions tangent to a line. -NARROW IT DOWN BY TELLING US EXACT FORMAT OF FUNCTION -see graph/fn. example there
When reviewing these past two lessons,
you're having some difficulties. Therefore, you must look over any medium-difficulty homework problems, you MUST get the examples from class, and be aware of the process but do not produce a concrete vision of how to do anything. Rather, you want to be aware of variables and general processes and ways to go about doing things, and use your notes to show how you've applied those. Look at your paper when reviewing, and place extra emphasis on these past two lessons and getting to know terms but more importantly problem-solving processes, like finding k problems and these pos/velocity/acceleration real-life examples and how your variables are applied in those.
Also look over
your homework (in addition to class notes, et cetera) ---> don't need to do this everyday, but it's good from time to time
