final review MA 141 Calc

product rule

(f*g)' = f' g + f g' (f g h)' = f'gh + fg'h + fgh'

quotient rule

(f/g)' = (f' g - f g') / g^2

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integral is larger on top than bottom, and how to do that

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long way of finding definite integral

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the equation of the tangent line to the curve at the point, use definition of derivative

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derivative of expotenial

1. lim n->infinity (1+1/n)^n 2. e is the unique positive number for which lim x-> 0 e^h-1/h = 1 f(x) = e^x f'(x) = e^x f(x) = e^x f'(x) = e^x f(x) = a^x f'(x) = a^x ln(a)

evaluate using fundamental theorem of calc

1. plug top number on S into the R. 2. Simplify and for the dr part, take the derivative, for final answer

inverse trig

1/x^2+1 dx = tan^-1x + c 1/sqtr(1-x^2) dx = sin^-1x + c

d(y^3) / dx

3y^2 * dy/dx

find equation of tangent line (using definition of derivative)

A tangent line to the function f(x) at the point is a line that just touches the graph of the function at the point in question and is "parallel" (in some way) to the graph at that point. So, in the first point above the graph and the line are moving in the same direction and so we will say they are parallel at that point. We want the tangent line to f(x) at a point x=z, First, we know that the point P=(a,f(a)) will be on the tangent line. Next we'll take a second point that is on the graph of the function call it Q(x,f(x)) and compute the slope of the line connecting P and Q as follows, mpq = secant line slope f(x)-f(a)/ (x-a) We than take values of x that get closer and closer to x=a and use this list of values to estimate the slope of the tangent line, m The tangent line will be y=f(a)+m(x-a) exp: f(x) 15/4x+7 at the point (2,1) 1. put in definition of limit form 15/4(x+h) + 7 - 15/4x+7 / h 2. get common demoninator on top half and pull out bottom h so its 1/h (15/4(x+h) + 7 * (4x+7)/(4x+7) - 15/4x+7 * (4(x+h)+7/ (4(x+h)+7 ) * 1/h 3. put together like dominator into one equation 15(4x+7) - 15(4x+4h+7) / 4(x+h)+7 (4x+7) 4. solve it out and get rid of like terms 60x + 105 - 60x - 60h - 105 / 4(x+h)+7 (4x+7) * 1/h 5. Get rid of stand alone h and h on top -60 / 4(x+h)+7 (4x+7) 6. Replace h with 0 and factor back in -60 / (4x+7)^2 7. plug in x point into new equation for slope -60/(4(2) + 7)^2 = -60/225 = -4/15 8. plug slope into x y form y - (y1) = m(x-x1) y - 1 = -4/15(x-2)

power rule

If f(x) = x^n then f'(x) = nx^n-1 Simple exp: 12x^4 = 48x^3

chain rule

If we define F(x) = (f of g) (x) then the derivative of F9x) is F'(x) = f'(g(x)) g'(x) exp: (2x-5)^3 = 2(2x-5)^2 * 2

newtons method

If xn is an approximation a solution of and if the next approximation is given by, Xn+1 = Xn - f(Xn)/f'(Xn) exp: sinx - x^2 = 0 x1 = .8 find x2 and x3 1. first set equation to 0 if it isn't all ready 2. than take derivative of equation 3. use equation. Xn+1 = Xn -f(Xn)/f'(Xn) 4. Plug in number given for x1 to get an answer. 5. Use that answer as the next plug in for the same repeated equation

limits - x approaching infinity if you get infinity over infinity - l'hopatal rule: 0/0 , infinity/infinity

If you plug in the limit and it goes to 0/0 or infinity/infinity, called indeterminate forms Suppose that we have one of the following cases, limx->a f(x)/g(x) = 0/0 limx->a f(x)/g(x) = infinity / infinity Where a can be any real number, infinity or negative infinity. In these cases we have, limx->a f(x) / g(x) = limx->a f'(x)/g'(x) exp: lim x->0 e^4x-1-4x /x^2 1. plug in limit 2. if 0/0, take derivative of the problem 3. plug back in again, if 0/0, take derive again, repeat until no longer 0 you may have to revert to standard identical form a*b = b/(1/a)

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-2 First, 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) = 0 The problem however, the derivative of the function is, f'(x) = 20x^4 + 3x^2 + 7 Because 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) = 0 We 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.

mean value theorem

Suppose f(x) is a function that satisfies both 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) Then there is a number c such that a<c<b and f'(c) = f(b) - f(a) / b-a or f(b) - f(a) = f'(c)(b-a) Tells ut is that these two slops must be equal or in other words the secant line connecting a and b and the tangent line at x=c must be parallel. exp: Determine all the numbers c which satisfy the conclusions of the Mean Value Theorem for the following function. f(x) = x^3+2x^2 - x on (-1,2) f'(x) 3x^2 +4x - 1 Now, to find the numbers that satisfy the conclusions of the Mean Value Theorem all we need to do is plug this into the formula given by the Mean Value Theorem. f'(c) = f(2) - f(-1) / 2-(-1) 3c^2 + 4c -1 = 14-2/3 = 12/3 = 4 Now this just a quadratic equation 3c^2 +4c -1 = 4 3c^2 +4c - 5 = 0 c=-4+-Sqtr(16-4(3)(-5) / 6 = -4+-sqtr76 / 6 Facts proved by mean value 1. if f'(x) = 0 for all x in an interval (a,b) then f(x) is constant on (a,b) 2. if f'(x) = g'(x) for all x in an interval (a,b) then in this interval we have f(x) = g(x) + c where c is some constant

definition of derivative f(x+h) - f(x) / h

The derivative of f(x) with respect to x is the function f'(x) and is defined as f'(x) lim(x->0) f(x+h) - f(x) / h exp: for f(x)=5x^2+3x-2 find f(x+h) - f(x) / h 1. replace all x with x+h. than - normal equation (5(x+h)^2 + 3(x+h) -2) - 5x^2+3x-2 / h 2. factor the equation out 5(x^2+2xh+h^2) + 3x+3h -2) - (5x^2+3x-2) / h 3. eliminate like terms 10xh+5h^2+3h/ h 4. Pull out the h h(10x+5h+3) / h 5. Cancel out the numerator and denominator h 6. answer 10x+5h+3

definition of a limit

We say that the limit of f(x) is L as x approaches a and write this as lim(x->a) f(x) = L provided we can make f(x) as close to L as we want for all x sufficiently close to a, from both sides, without actually letting x be a as x gets closer to x=a (from both sides) then f(x) must be getting closer to L factoring both the numerator and denominator. multiply out terms, cancel out terms, get rid of the denominator.

integrate and evaluate with higher numerator than nominator

When fraction is Cubed sqtr(x^2) - itmeans x^2/3 exp: x-1/x^2/3 1. Separate it into two different fractions 2. get them into simpler forms and take integral of equation. 3. Simplify and plug in the two integral numbers and solve.

Definition of derivative f(x)-f(a) / x-a

a function is called differentiable at x=a if f'(a) exists and f (x) is called differentiable on an interval if the derivative exists for each point in that interval if f(x) is differentiable at x=a then f(x) is continuous at x=a exp: f(x) = f/x^2 (f(x) - f(a) / x-a) 1. fill in the top x with the equation and fill in the top a with the equation with a 4/x^2 - 4/a^2 / x-a 2. Get common denominator for top half 4a^2/x^2a^2 - 4x^2/a^2*x^2 / x-a 3. factor out bottom half so its 1 / x-a 4. factor out all equation 4(a^2-x^2)/a^2x^2 (x-a) 5. get rid of like terms -4(a+x) / a^2x^2

12-16 questions

average of 3-4 for each test than 1 trig substitution 1 table of integrals

definite integral

b f(x) dx = lim n -> inf f(xi*) Delta X a Delta X = endpoint - endpoint / n xi* = endpoint - endpoint / n * i

partial fractions

break each dominator part into its own fraction

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

graph the split domain function and use it to answer, limit and continuous

continuous at x=a if lim (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>/0 1. make three different charts, and solve out x and y values for each one 2. 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 forward 4. continuous at point, check if f(x) exists check if there is a limit for both sides if they are the same and exist, than yes, if not on

trig stuff I should probably know

cos(x) sin(x) tan(x) = sinx/cos/ cot(x)= cosx/sinx sec(x)= 1/cosx csc(x)= 1/sin(x) cos = adjacent/hypotenuse sin = opposite/hypotenuse tan = opp/adj cot = adj/opp sec = hyp/adj csc = hyp/opp degress = radian 0 = 0 30 = pie/6 45 = pie/4 60 = pie/3 90 = pie/2 180 = pie 270 = 3pie/2 360 = 2pie cos^-1 = arccos sin^-1 = arcsin tan^-1 = arctan f(x)=b^x natural exponential function =e^x y=logbX = x=b^y exp: log2,16=2^x=16 logbB=1 logb1=0 logbB^x=x b^logbX=X logbXY=logbX+logbY logb(x/y)=logbX-logbY Logb(X^r)=rlogbX sin = y/1 cos = x/1 tan = y/x csc = 1/y sec = 1/x cot = x/y csc = 1/sin sin = 1/csc sec = 1/cos cos = 1/sec cot = 1/tan tan = 1/cot sin^2+cos^2=1 tan^2+1=sec^2 1+cot^2=csc^2 sin (a ± b ) = sin a cos b ± cos a sin b cos (a ± b ) = cos a cos b -+ sin a sin b tan(a±b)= tana±tanb / 1-+ tana tan b sin 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)

related rates

cube = a^3 recantular prism = a*B*c cylinder = pi r^2 * h pyramid = (1/3) b*h cone = (1/3) pi r^2 h sphere = (4/3) pi r^3 f'(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 zero exp 2: determine where the following function is increasing and decreasing A(t) = 27t^5 - 45t^4 - 130t^3 + 150 A'(t) = 135t^4 - 180t^3 - 390t^2 = 15t^2 (9t^2 -12t -26) Find where the function isn't changing. aka find the zero The 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 car x= 500 - (35(3) = 395 y= 50(3) = 150 Use the pythagorean theorem to find z z^2 = 395^2 + 150^2 = 178525 Take square root of that = 422.5222 Now, 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 mph 1. draw a diagram 2. 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 change 4.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 equation 6. substitute all information given exp: 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 deep 1. draw picture 2. find rate of change given dv/dt =2cm 3. find the rate of change needed to find - dh/dt 4. write out other information h = 10 r = 3 5. 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.

derivatives of log

d/dx (a^x) = a^x ln(a) d/dx (ln(x)) = 1/x d/dx (logaX) = a/X ln(a) log rules - log(A*X) = log(a) + log(x) log (a/x) = log(a) - log (x) log (a^n) = n* log(m)

implicit differentiation

differentiate x normally and everytime there is a y, you differentiate it as dy/dx Than 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...

parametric equation

dx/dt & dy/dt find dy/dx = dy/dt / dx/dt chain rule: dy/du * du/dx = dy/dx

x in terms of t y in terms of t

dx/dt dy/dt

integration with exponential and logarithm functions

e^x dx = e^x + c a^x dx = a^x / lna + c 1/xdx = x^-1 dx = ln abs(x) + c

decompose into partial fractions and integrate

exp: 11x^2 -28x -23 / (x-2)^2 (x+5) hint: (when bottom is (x-2)^2 - it counts as single terms. (x-2) and (x-2)^2 if x^2-2 it is Ax+B 1. distribute out the denominator 2. fill out each demonimator part as own fraction, giving corresponding letters on top. 3. now multiply to get same terms on detonator. (hint) if one is squared (x-2)^2, when you multiply with the (x-2) make sure you add in a (x-2)plus the other term 4. solve the top out with corresponding letters 5. set equation = to the original equation 6. group like terms and pull out the x^2 and x appropriately 7. substitute the x^2 and x correctly 8. solve the for the easiest one, making it relate to another letter. 9. Plug that letter into both equations and solve out. 10. subtract both equations from each other, eliminating one variable and finding the solution to the other 11. substitute that solution into the other equations to get answers 12. Plug those into letters and take integrals of the new equation

find f^-1(x) What the the domain and range of f(x) or f^-1(x)

exp: 2x-5 /1+3x 1. To find inverse, switch the y and x x=2y-5/1+3y 2.solve back to get y alone x(1+3y) = 2y-5 X+3xy=2y-5 3xy-2y=-x-5 3. factor out the ys y(3x-2) = -x-5 y= -x-5/(3x-2) 4. Find domain and range for both f(x) = domain equals range for f-(x) f-(x) = domian equals range for f(x) f(x) domain equals -1/3, range f-(x) -1/3 f-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^2 1. f of g means you take f(g) and than plug in equation for every x f(-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

distance, height, position function (derivative or anti derivative)

exp: s(t) = -16t^2+64t + 100 Def of derivative Find avg velocity 1. Write out equation - equation / t2 - t1 2. Plug in the given time for t2 and t1 and solve out For instantaneous velocity 1. set up in f(x+h) - f(x) / h form 2. distribute out and get rid of like terms 3. Factor out top h and cancel bottom h 4. plug in instantaneous number to get slope

direction curve is traced as the parameter increases; eliminate the parameter to find a cartesian equation of the curve

exp: x=sint, y=csct 0<t<pie/2 1. Make a chart filling out the range with x and y values 2. Graph those value. 3. Change y or x equation to equal each other using trig identityx y=csc t -> csc t = 1/sin y=1/sin -> x=sin Therefore y = 1/x

given points on the curve, where he tangent line is (are) horizontal

exp: y = 4x/ x^2 + 1 1. find the derivative use quotient rule for this problem 2. Find where f'(x) = equals 0 3. Take these point(s) and plug them back into the regular equation for answer. so f'x=0 # and that output of that number into first equation

logarithmic differentiation use if x is in the base and expontenet

exp: y=(sinx)^x lny = ln(sinx)^x ln y = x ln(sinx) take derivative of both sides 1/y * dy/dx = x(1/sinx*cosx) + ln(sinx) * 1 solve for dy/dx x^x = x^x (1+lnx) add an ln to both side of the equation than differnatie both sides with respect to x. y'/y Than multiply the y to the right, and then plug the original equation back into y. exp: y = (sinx)^lnx 1. if there is an x in the base and the exponent, need to ln both sides 2.the complete exponent comes down 3. take derivatives of both sides, for every y, make it (*dy/dx) 4. solve out the problem with whatever rule 5. get y back on other side, and plug original y back into the problem

find dy/dx

exp: y^4+3x^2y^3 = 5x^2 + xe^y^2 1. every derivative you take of y add (*dy/dx) next to it 4y^3+(3x^2*3y^2*dy/dx) + y^3(6x) = 10x + (x*e^y^2(2y dy/dx) + e^y^2(1) 2. distribute out 3. put all dy/dx connected terms to one side 4. factor out dy/dx 5. get dy/dx by itself, thats answer

integration by parts

f g'dx = fg - inte f'g dx u = f(x) du= f'(x) dx v= g(x) dv = g'(x) dx integral of udv = uv - inte vdu 1. let blank(x) equal u 2. find du by taking derivative of u, so du= f'(x)* dx 3. dv = the other part of equation 4. v = the integral of dv 5. plug numbers into equation. S udv = (u)(v) - S vdu 6. solve out and add c

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)

graph increasing, decrease, flat, steep 1st derivative understanding what it tells us second derivative understanding what it says

f'(x) = 0 flat f'(x) > 0 increaseing f'(x) < 0 decreasing f'(x) undefined - vertical line point of inflection - 2nd derivative at zero. Change in concavity change in concavity - the signs change from pos to neg / neg to pos if concave up - 2nd derive is positive if concave down - 2nd derive is neg critical points - We say that x=c is a critical point of the function f(x) if f(x) exists and if either of the following are true. f'(c) = 0 or f'(c) = doesn't exist exp 1: Determine all the critical points for the function f(x) = 6x^5 + 33x^4 - 30x^3 + 100 First need the derivative of the function f'(x) = 30x^4 + 132x^3 - 90x^2 = 6x^2(5x-3(x+5) The three critical points are, x= -5, 0, 3/5 We say that f(x) has an absolute maximum at x=c if f(x) < f(c) for every x in the domain we are working on. We say that f(x) has a relative maximum at x=c if f(x) < f(c) for every x in some open interval around x=c We say that f(x) has an absolute minimum at x=c if f(x) > f(c) for every x in the domain we are working on We say that f(x) has a relative minimum at x=c if f(x) > f(c) for every x in some open interval around x=c open interval around x=c means that we can find some interval (a,b) not including the end point, such that a<c<b. extreme value theorem - suppose that f(x) is continuous on the interval (a,b) then there are two numbers a<c,d<b, so that f(c) is an absolute maximum for the function and f(d) is an absolute minimum for the function. fermat's theorem - If f(x) has a relatve extrema at x=c and f'(c) exists then x=c is a critical point of f(x). In fact, it will be a critical point such that f'(c) = 0 1. Verify that the function is continuous on the interval 2. Find all critical points of f(x) that are in the interval. 3. Evaluate the function at the critical points and end points 4. identify the absolute extrema exp 1: Determine the absolute extrema for the following function g(t) = 2t^3 +3t^2 -12t + 4 (-4,2) derivative g'(t) = 6t^2 + 6t - 12 6(t+2)(t-1) = critical points -2,1 now evaluate critical points and end points with original equation g(-2)=24 abs max g(-4) = -28 abs min g(1) = -3 g(2) = 8 full exp: g(x) = 3x^4 + 4x^3 - 12x^2 on (-3,2), find critical points, points of inflection, where increasing, decreasing, concave up, down; all relative/max/min gals 1. take derivative of equation 2. find zeros for that equation, ( make sure hey are on the points) 3. solve those zeros back into original equation. (rel mix and max) 4. Then plug the end points in as well 5. make a line and put the three critical points on it, plug numbers into the derivative that are in between these numbers. If negative, g(x) is decrease. If positive g(x) is increase, the critical points are zeros 6. take second derivative 7. Find zero (may have to use quadratic formula) 8. plug points back into original equation. (inflection points) 9. Make line with both inflection points on it. Plug in numbers between these inflection points. if g''(x) is positive, g(x) is concave up, if g''(x) is negative, g(x) is concave down

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 is L(x) = f(a) + f'(a) (x-a) exp 1: determine the linear approximation for f(x) 3sqtr(x) at x=8 Use 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) 1/12 The linear approxmation is then L(x) = 2+1/12(x-8) = 1/12x + 4/3 L(8.05) = 2.00416667 L(25) = 3.4166667 (3)sqtr(8.05) = 2.00415802 (3)sqtr(25) = 2.92401774 exp: sqtr(1+x) at a=0 and use it to approximate sqtr(1.2) 1. find the derivative of equation 2. 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 = x 4. plug number into equation to solve for the answer

limits

finding limit with finite number. factor out or in xs until you have common demoninator and numerator to cancel out exp: x^2-4x / x^2 -3x -4 1. x(x-4) / (x-4) (x+1) 2. cancel out so x/x+1 = 4/5 If 0 ends on bottom, than D.N.E. If the limit is infinity. Than whoever has the highest exponent is the limit if the dominator has it - its 0 If has same exponent, than its the leading coefficients over each other If the numerator is the larger, its infinity

integration using substitution

integral f(g(x)) g'(x) dx = integral f(u) du u=g(x) exp: 1S0 - x^2(1+5x^3)^4 dx 1. find u, whatever is in the paranethesis usually is it 2. find du - add dx - take reciprocal of leading coefficient and place it in front of the Integral 3. Take the integral of U 4. Plug back in for U 5. Plug in the top and bottom of the integral and solve out f(top) - f(bot)

trig substitution

integral sqt(a^2-u^2) let u = asin(theta) find du= a * cos(theta) d(theta) integral sqt(a^2+u^2) let u = atan(theta) du= a*sec^2(theta) d(theta) Integral sqt(u^2+a^2) u=asec(theta) du= a*sec(theta)tan(theta) d(theta) Make a right triangle somehow exp: sqtr (x^2 -1) / x^4 1. let x = sec(theta) dx = sec theta tan theta d theta 2. plug sec for x and sec tan for dx sqtr(sec^2t)-1) / sec^(t) * sec(t)tan(t)d(t) 3. plug in trig identity if can sqtr(tan^2t) / sec^4(t) * sec(t)tan(t)d(t) 4.simplify tan^2(t) / sec^3(t) * d(t) 5. put into terms of sin and cos sin^2(t) / cos^2(t) / 1/cos^3(t) * d(t) 6. multiply by reciprocal to get rid of uneven fraction and simplify sin^2(t) * cos (t) * d(t) 7. substitute u in for the problem u^2*du 8. Find the integral u^3/3 + c 9. Subs sin back in Sin^3(t)/3 + c 10. Make a right angle triangle graph for theta. Find trig identity for original trig function. to get two parts of triangle if Sec its X as long line and 1 as bottom 11. add other part of triangle by pygarithan theorem 12. take trig function you have now and use right triangle trig identity to get correct identity 13. Plug that back into the problem if tan theta = x/2, for right triangle its opp / adj and long part is sqtr(4+x^2) if sin theta = x/2, x is short length and 2 is the long length

derivative of natural log

ln(x) = 1/x exp: ln(x^3-7) at point (2,0 1. take 1/over equation * f'(equation) plug in number to get slope. Use x1 and y1 to get tangent line

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 = xy Constraint 500 = x+2y Okay, 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=500-2y Substitute this into the area function A(y) = (500-2y)y = 500y -2y^2 So, 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^2 x=500-2(125) = 250 Method 1: method in finding absolute extrema Method 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 + 48wh Constraint : 50=lwh=3w^2 h h=50/3w^2 Plugging into the cost gives, C(w) = 60w^2 + 48w (50/3w^2) = 60w^2 + 800/w Now, get the first and second derivatives C'(w) = 120w - 800w^-2 C''(w) = 120 + 1600w^-3 120w^3-800=0 = 1.8821 w=1.8821 l = 3w = 3(1.8821) = 5.6463 h= 50/3w^2 = 50/3(1.8821)^2 = 4.7050 1. Draw a diagram 2. assign variables to the quantity to be optimized and all other unknown quantities given 3. write an equation that associates the optimal quantity to the other variables 4. if expressed in terms of more than one variable, we must eliminate the extra variables. 5. Should be in terms of one variable 6. find the absolute max or min my own words 0. draw diagram 1. Find S equation depending on setup 2. Find V equation depending on set up 3. plug in numbers for V 4.Since S has two numbers, arrange V equation to make S only have one variable. Plug it into S 5. 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 pos Prove max by concave down using sec drive if neg

using a riemann sum and sigma notation formula

riemann sum A lim as n approach infinity n sigma i=1 f (xi*) delta x sigma notation A = n sigma i = 1 f(xi*) delta X 1. Find delta X = b-a/h 2. plug into equation f(a+i delta x) * delta x a is usually 0 3. now for every #i/h - plug that in for x in original equation) * #/h 4. Solve out, multiply #/h through 5. Pull out the number/h on both sides and leave i and i^2 with the integral #/h^2 Sigma i 6. Plug in the given n equations for i and i^2 7. swamp h^2 for demonator and h^3 for other demonator and cancel out 8. solve the rest of the problem 9. check by tking integral of original problem

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^2 cot-1 = -1/1+x^2 sec-1 = 1/ abs(x) sqtr(x^2-1) csc-1 = -1/ abs(x) sqtr(x^2-1)

integration with trig functions

sinx dx = -cosx +c cosx dx = sinx + c sec^2x dx = tanx + c secx tanx dx = sec x + c csc^2x dx = -cotx + c cscx cotx dx = -cscx + c exp: S sin^4x cos^3x dx 1. take the odd trig function and take it a part to an even an odd. 2. replace trig function so they are both the same trig identitys 3. solve out if have too 4. let u = trig identiy (x) let du = the left over function you have. (if need be, change sign) 5. Take integral with U and than plug back in and add c. You do not need to add back in du after shown in the new equation.

intermediate value theoreom

suppose that f(x) is continuous on (a,b) and let M be any number between f(a) and f(b) Then there exists a number c such that, 1. a<c<b 2. f(c) = M exp e^x+2x-3 has a root on the interval (0,1) 1.Take a number lower and higher than number given and plug it into equation 2. If you get an output that is one is negative and one is positive, than there has to be a number in between thats 0. Therefore IVT would prove yes (also prove that it is continuous) polynomials are continuous

squeeze theorem

suppose that for all x on (a,b) we have, f(x) is less than h(x) is less than g(x) Also suppose that lim (x->c) f(x) lim g (x) = L for some a<c<b then, lim (x->c) h (x) = L