final review MA 141 Calc
12-16 questions : average of 3-4 for each testthan 1 trig substitution1 table of integrals
trig stuff I should probably know : cos(x)sin(x) tan(x) = sinx/cos/cot(x)= cosx/sinxsec(x)= 1/cosxcsc(x)= 1/sin(x)cos = adjacent/hypotenusesin = opposite/hypotenusetan = opp/adjcot = adj/oppsec = hyp/adjcsc = hyp/oppdegress = radian0 = 030 = pie/645 = pie/460 = pie/390 = pie/2180 = pie270 = 3pie/2360 = 2piecos^-1 = arccossin^-1 = arcsintan^-1 = arctanf(x)=b^xnatural exponential function =e^xy=logbX = x=b^yexp: log2,16=2^x=16logbB=1logb1=0logbB^x=xb^logbX=XlogbXY=logbX+logbYlogb(x/y)=logbX-logbYLogb(X^r)=rlogbXsin = y/1cos = x/1tan = y/xcsc = 1/ysec = 1/xcot = x/ycsc = 1/sinsin = 1/cscsec = 1/coscos = 1/seccot = 1/tantan = 1/cotsin^2+cos^2=1tan^2+1=sec^21+cot^2=csc^2sin (a ± b ) = sin a cos b ± cos a sin bcos (a ± b ) = cos a cos b -+ sin a sin btan(a±b)= tana±tanb / 1-+ tana tan bsin a sin b = 1/2 ((cos (a - b ) - cos (a + b ))cosa cosb =1/2 (cos(a-b)+cos(a+b))sin a cos b = 1/2 (sin (a + b ) + sin (a - b )cosa sinb =1/2 (sin(a+b)-sin(a-b))sina+sinb=2sin(a+b/2)cos(a-b/2)sin a - sin b = 2 cos (a+b/2) sin(a-b/2)cosa +cosb = 2 cos(a+b/2) cos(a-b/2)cosa=cosb = -2sin(a+b/2) sin (a-b/2)
graph the split domain function and use it to answer, limit and continuous : continuous at x=a iflim (x->a) f(x) = f(a) a function is said to be continuous on the interval (a,b) if it is continuous at each point in the interval.If f(x) is continuous at x=a then,lim (x->a) f(x)=f(a)lim (x-a-) f(x) = f(a)lim (x-a+) f(x) = f(a)if f(x) is continuous at x=b and lim g(x) = b then,lim f(g(x)) = f(lim g(x))exp: 5 if x</ -2, 3x-2 if -2<x<0, 6-x^2 if x>/01. make three different charts, and solve out x and y values for each one2. Make graph and graph points. Open dots when it is not greater/less than and EQUAL. 3. Solve for lim + and -, + coming from right backtracking, - coming from left moving forward4. continuous at point, check if f(x) existscheck if there is a limit for both sidesif they are the same and exist, than yes, if not on
find f^-1(x)What the the domain and range of f(x) or f^-1(x) : exp: 2x-5 /1+3x1. To find inverse, switch the y and xx=2y-5/1+3y2.solve back to get y alonex(1+3y) = 2y-5X+3xy=2y-53xy-2y=-x-53. factor out the ysy(3x-2) = -x-5y= -x-5/(3x-2)4. Find domain and range for bothf(x) = domain equals range for f-(x)f-(x) = domian equals range for f(x)f(x) domain equals -1/3, range f-(x) -1/3f-x domain equals 2/3, range f(x) 2/3
(f of g)(x) and (g of f)(x) : exp: f(x) = tan^2(3x) , g(x) = 1+3x-5x^21. f of g means you take f(g) and than plug in equation for every xf(-5x^2+3x+1) -> tan^2(x)(-5x^2+3x+1)2. g of f means you take g(f) and then plug in equation for every x
parametric equation : dx/dt & dy/dtfind dy/dx = dy/dt / dx/dtchain rule:dy/du * du/dx = dy/dx
chain rule : If we define F(x) = (f of g) (x) then the derivative of F9x) isF'(x) = f'(g(x)) g'(x)exp: (2x-5)^3 = 2(2x-5)^2 * 2
derivative of trig : sin(x) = cos (x)cos(x) = -sin(x)tan(x)) = sec^2(x)cot(x)) = -csc^2(x)sec(x)) = sec(x) tan(x)csc(x)) = -csc(x) cot(x)sin-1 = 1/sqtr(1-x^2)cos-1 = -1/sqtr(1-x^2)tan-1 = 1/1+x^2cot-1 = -1/1+x^2sec-1 = 1/ abs(x) sqtr(x^2-1)csc-1 = -1/ abs(x) sqtr(x^2-1)
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implicit differentiation : differentiate x normally and everytime there is a y, you differentiate it as dy/dxThan you combine all the dy/dx on one side and all the x on the other side, and than solve out so its just dy/dx=equation...
linear approximation (use of first derivative) : f(x) = we can find its tangent at x=a The equation of the tangent line, which we'll call L(x) for this discussion isL(x) = f(a) + f'(a) (x-a)exp 1: determine the linear approximation for f(x) 3sqtr(x) at x=8Use the linear approximation to approximate the value of 3sqtr(8.05) and 3sqtr(25)Since this is just the tangent line there really isn't a whole lot to finding the linear approximation.f'(x) = 1/3x^-2/3 = 1/3(3)sqtr(x^2)f(8) = f'(8) 1/12The linear approxmation is thenL(x) = 2+1/12(x-8) = 1/12x + 4/3L(8.05) = 2.00416667L(25) = 3.4166667(3)sqtr(8.05) = 2.00415802(3)sqtr(25) = 2.92401774exp: sqtr(1+x) at a=0 and use it to approximate sqtr(1.2)1. find the derivative of equation2. y = f(a) + f'(a)* (x-a) ( if a does not equal zero, you can just plug in estimate number and f(a) to get that answer. 3. the difference between the number and estimate equation = x4. plug number into equation to solve for the answer
the equation of the tangent line to the curve at the point, use definition of derivative : ...
related rates : cube = a^3recantular prism = aBccylinder = pi r^2 * hpyramid = (1/3) b*hcone = (1/3) pi r^2 hsphere = (4/3) pi r^3f'(x) represents the rate of change of f(x)exp: determine all points where the following function is not changing.g(x) = 5-6x-10cos(2x)g'(x) = -6 + 20sin(2x)now, the function will not be changing if the rate of change is zero so you have to determine where the derivative is zeroexp 2: determine where the following function is increasing and decreasingA(t) = 27t^5 - 45t^4 - 130t^3 + 150A'(t) = 135t^4 - 180t^3 - 390t^2 = 15t^2 (9t^2 -12t -26)Find where the function isn't changing. aka find the zeroThe function is not changing at 3 different values. (0, =1.159, 2.492) To determine where the function is increasing or decreasing we need to determine where the derivative is positive or negative. If the derivative is positive then the function must be increasing and if the derivative is negative then the function must be decreasing. exp3: two cars start out 500 miles apart. Car A is to the west of Car B and starts driving to the east towards car B at 3 mph at the same time car B starts driving at 50mph. After 3 hours at what rate is the distance betwen the two cars changing. First thing to do is sketch a figure.Triangle. Initial distance, distance driven by car A and Car B. x=distance separating A to intial B. y = distance driven by car B. z=distance separate both carx= 500 - (35(3) = 395y= 50(3) = 150Use the pythagorean theorem to find z z^2 = 395^2 + 150^2 = 178525Take square root of that = 422.5222Now, to answer the equation we will need to determine z' give that x'=-35 and y'=50. We use the pythagorean theorem again, so differentiate the equation using implicit differentiation z^2=x^2+y^2 = 2zz' = 2xx' + 2yy'z'(422.5222)=395(-35) + (150)(50) = z' -6325/422.5222 = =-14.9696. The decreasing rate is 14.9696 mph1. draw a diagram2. assign variables to each quantity in the problem that is a function of time.3. list all information that is given in the problem and the rate of change4.write an equation that associates the variables with one another. If there are variables for which we are not given the rates of change, we must find some relation from the nature of the question that allows us to write these variables in terms of variables for which the rates of change are given.5. using chain rule, differentiate each side of the equation6. substitute all information givenexp: paper cup has the shape of a cone with height 10 cm and radium 3cm at the top. If water is poured into the cup at a rate of 2 cm^3/sec, how fast is the water level rising when the water is 5cm deep1. draw picture2. find rate of change given dv/dt =2cm3. find the rate of change needed to find - dh/dt4. write out other informationh = 10 r = 35. Figure out what normal formula you need - volume of cone.6. Use proportions or similar triangle or shapes to eliminate one of the variables so only one remains. Solve out and take derivative of. 7. Now plug in dv/dt and solve out for answer.
max min problem (optimization problem) : n optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to solve one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval. The first step in all of these problems should be to very carefully read the problem. Once you've done that the next step is to identify the quantity to be optimized and the constraint. In identifying the constraint remember that the constraint is the quantity that must true regardless of the solution. In almost every one of the problems we'll be looking at here one quantity will be clearly indicated as having a fixed value and so must be the constraint. Once you've got that identified the quantity to be optimized should be fairly simple to get. exp 1: We need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won't need any fencing. Determine the dimensions of the field that will enclose the largest area. In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. Sketching the situation will often help us to arrive at these equations so let's do that.In this problem we want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are,Maximize A = xyConstraint 500 = x+2yOkay, we know how to find the largest or smallest value of a function provided it's only got a single variable. The area function (as well as the constraint) has two variables in it and so what we know about finding absolute extrema won't work. However, if we solve the constraint for one of the two variables we can substitute this into the area and we will then have a function of a single variable. So, let's solve the constraint for x. Note that we could have just as easily solved for y but that would have led to fractions and so, in this case, solving for x will probably be best.x=500-2ySubstitute this into the area functionA(y) = (500-2y)y = 500y -2y^2So, recall that the maximum value of a continuous function (which we've got here) on a closed interval (which we also have here) will occur at critical points and/or end points. As we've already pointed out the end points in this case will give zero area and so don't make any sense. That means our only option will be the critical points. So let's get the derivative and find the critical points.A'(y) = 500 - 4y y=125, plugged in is 31250ft^2x=500-2(125) = 250Method 1: method in finding absolute extremaMethod 2: use a variant of the first derivative test. Method 3: use the second derivative. exp 2: We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. If the box must have a volume of 50ft3 determine the dimensions that will minimize the cost to build the box.We want to minimize the cost of the materials subject to the constraint that the volume must be 50ft3. Note as well that the cost for each side is just the area of that side times the appropriate cost. The two functions we'll be working with here this time are,Maximize: C =10(2lw) + 6(2wh+lh) = 60w^2 + 48whConstraint : 50=lwh=3w^2 hh=50/3w^2Plugging into the cost gives,C(w) = 60w^2 + 48w (50/3w^2) = 60w^2 + 800/wNow, get the first and second derivativesC'(w) = 120w - 800w^-2C''(w) = 120 + 1600w^-3120w^3-800=0 = 1.8821w=1.8821l = 3w = 3(1.8821) = 5.6463h= 50/3w^2 = 50/3(1.8821)^2 = 4.70501. Draw a diagram2. assign variables to the quantity to be optimized and all other unknown quantities given3. write an equation that associates the optimal quantity to the other variables4. if expressed in terms of more than one variable, we must eliminate the extra variables.5. Should be in terms of one variable6. find the absolute max or minmy own words0. draw diagram1. Find S equation depending on setup2. Find V equation depending on set up3. plug in numbers for V4.Since S has two numbers, arrange V equation to make S only have one variable. Plug it into S5. Take derivative of S equation and than set it to 0.6. Solve out and sub that number back into for H equation. Plug those numbers into the regular surface area equation.7. Prove minimization by concave up using sec derive if posProve max by concave down using sec drive if neg
fundamental theorem of calculus : f is continuous on [a,b] then, g(x) - inte (x,a) f(t) dt is continuous on [a,b] and it is differentiable on and that, g'(x) = f(x)
long way of finding definite integral : ...
integrate and evaluate with higher numerator than nominator : When fraction is Cubed sqtr(x^2) - itmeans x^2/3exp: x-1/x^2/31. Separate it into two different fractions2. get them into simpler forms and take integral of equation.3. Simplify and plug in the two integral numbers and solve.
rolle's theorem : Suppose f(x) is a function that satisfies all of the following.1. f(x) is continuous on the closed interval (a,b)2. f(x) is differentiable on the open interval (a,b)3. f(a) = f(b)Then there is a number c such that a<c<b and f'(c) =0. Or, in other words f(x) has a critical point in (a,b)exp:show that f(x)=4x^5+x^3 +7x-2First, we should show that it does have at least one root. Note that f(0) = -2 and that f(1) = 10 and so that we can see that f(0) < 0< f(1)We now need to show that this is in fact the only real root. To do this we'll use an argument that is called contradiction proof. What we'll do is assume that f(x) has at least two real roots. This means that we can find real numbers a and b (there might be more, but all we need for this particular argument is two) such that f(a) = f(b) = 0 But if we do this then we know from Rolle's Theorem that there must then be another number c such that f'(c) = 0The problem however, the derivative of the function is, f'(x) = 20x^4 + 3x^2 + 7Because the exponents on the first two terms are even we know that the first two terms will always be greater than or equal to zero and we are then going to add a positive number onto that and so we can see that the smallest the derivative will ever be is 7 and this contradicts the statement above that says we MUST have a number c such that f'(c) = 0We reached these contradictory statements by assuming that F(x) has at least two roots. Since this assumption leads to a contradiction the assumption must be false and so we can only have a single real root.
table of integrals : compare problem given with that subset. Make it match exactly, so same u^2 is same u^2 wherever appears in table of integrals
