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1 Contents Limits. Basic Factoring Eample One-Sided Limit Squeeze Theorem Rationalizing Limits using Trig. Identities Limits involving Infinity, Part I Limits involving Infinity, Part II Continuity Precise definition of a Limit - Linear Case Derivatives. Limit Definition, Part I (i.e. +h definition) Limit Definition, Part II (i.e. -a defintion) Chain and Product Rule Quotient Rule E. involving all types of derivative techniques Derivatives w/ Fractional Eponents Integrals 7 3. Riemann sums U-substitution, basic Indefinite integral, u-substitution, basic manipulation of substitution Definite integral, u-substitution, basic manipulation of substitution Integrals w/ Symmetric Limits Fundamental Theorem of Calculus Applications of Derivatives 3 4. Related Rates Part I Related Rates Part II Differentials Linear Approimations Optimization (Standard) Optimization (Difficult) Curve Sketching Applications of Integrals 3 5. Area Between Curves Volume of Revolution, Disk Method, Part I Volume of Revolution, Disk Method, Part II Volume of Revolution, Shell Method Work, Chain Problem Work, Spring Problem Work, Involving Geometry Average Value of a Function

2 Limits. Basic Factoring Eample 3 Recognize the denominator as a difference of squares, since you want to cancel the ( ) factor on the bottom you should factor out an ( ) from the top either realizing it as a difference of cubes or simply doing long division: = ( )( + + ) ( )( + ) Cancel a factor of ( ) on top and bottom: = = = 3

3 . One-Sided Limit Recognize the numerator as the and rewrite: Have to rewrite the absolute value as a piecewise function: = { if < if Now consider the two one-sided its: and + It is now clear that the left-handed it goes to - and the right-handed it goes to, thus the it DNE DNE 3

4 .3 Squeeze Theorem ( ) sin Recognize the it as a times a function that is bounded which indicates that the Squeeze Thm may be useful. Setup the inequality that represents the fact that the sine function is bounded: sin ( ) Now manipulate the inequality until you arrive at the original problem; multiply everything by an : sin ( ) Now take the it of all three parts: sin ( ) ( ) = sin = sin ( ) 4

5 .4 Rationalizing + + a, where a is a positive constant case, so try rationalizing: + + a + a + a ( + a) + a a + a Now multiply top and bottom by the reciprocal of the highest power of in the denominator; here multiply by one over : a + a a + a = a 5

6 .5 Limits using Trig. Identities sin(a + u) sin(a) u u Epand using the addition formula for sine: sin(a) cos(u) + cos(a) sin(u) sin(a) u u sin(a)(cos(u) ) + sin(u) cos(a) = u u The key is to use the following identities: sin(u) cos(u) = and = u u u u Then, ( = sin(a) cos(u) + cos(a) sin(u) ) u u u = cos(a) 6

7 .6 Limits involving Infinity, Part I n n n Recall that: n = n(n + ) n n n i = i= n(n + ) Divide top and bottom by the highest power of n in the denominator: n + n n n + n = n n n = Recall the three it cases for rational functions where : When the degree of the numerator equals the degree of the denominator then you can read off the answer, in practice this amounts to dividing top and bottom by the highest power of in the denominator, e.g = 8 7 When the degree of the numerator is greater than the degree of the denominator then the answer will be infinite, e.g = And when the degree of the denominator eceeds that of the numerator then the it will go to zero, e.g = 7

8 .7 Limits involving Infinity, Part II m = m v c In the theory of relativity, the mass of a particle with velocity v is given by the equation above. Find the mass as v c. m v c v c As v c, v c Thus, v c And, m v c m v c v c = 8

9 .8 Continuity A function f is defined as follows: sin() if π f() = a + b if π < 5 + b > 5 where a,b are constants. Determine a,b such that the function f() is continuous everywhere. Geometrically a piecewise function is continuous if at the point where the functions switch (i.e. at π and, in our case) the two one-sided its equal each other and the it value equals the function values at that point. Because both the sine function and polynomials are continuous everywhere, we need only make sure the two one-sided its are equal (then the function values will automatically be the same). So we need: From the first equation: sin() = (a + b) and π π + sin(π) = = aπ + b (a + b) = ( + b) a = b π From the second equation: a(5) + b = (5) + b Plug in the epression for a in terms of b 5 b π + b = 5 + b b = 5π and a = 5π π = 5 5 y 5π 5 5( π) sin

10 .9 Precise definition of a Limit - Linear Case Prove that c (a + b) = ac + b Given ɛ > find δ > s.t. if < c < δ then (a + b) ac b < ɛ Scratch Work: a + b ac b < ɛ = a ac < ɛ = a( c) < ɛ = a c < ɛ = c < ɛ a = choose δ = ɛ a Formal: < c < δ < a c < a δ < a( c) < a δ < a ac + b b < a (a + b) ac b < ɛ ( ) ɛ a Note that if a =, then you have the constant function y = b, in this case an ɛ δ proof is unnecessary: b = b c Because b is a constant it is unaffected by the it and can be pulled outside: b c = b b = b

11 Derivatives. Limit Definition, Part I (i.e. +h definition) Using the it definition, find the derivative of f() = h + +h + h + It is easiest to first create a common denominator on the top and then rationalize the epression: + ( + + h) h h( + + h)( + ) h h + h h( + + h)( + ) ( + h) + + h + + h h( + + h)( + )( + + h) (Note that it is almost always easiest to leave the denominator unepanded) h h h h( + + h)( + )( + + h) ( + + h)( + )( + + h) Now you can plug h = in: = ( + ) ( )

12 . Limit Definition, Part II (i.e. -a defintion) Using the it definition, find the derivative of f() = a a a a Make a common denominator on the top and work towards canceling the problem factor of a on the bottom a a a (a ) a( ) ( )(a )( a) a a + a ( )(a )( ) a ( )(a )( a) Pull a negative one out of the a factor on the bottom a ( )(a ) = (a )

13 .3 Chain and Product Rule Calculate the derivative of f() = sin(cos ) cos(sin ) Using the product rule we get: f () = d [ ] sin(cos ()) d [ ] cos(sin ()) cos(sin ()) + sin(cos ()) d d = cos(cos ()) [ sin() cos() cos(sin ()) ] + sin(cos ()) sin(sin ()) [ sin() cos() ] ( ) = sin() cos() cos(cos ()) cos(sin ()) + sin(cos ()) sin(sin ()) If you want to simplify use the double angle formula for sine and the trig. addition formula for cosine = sin() cos(cos sin ) 3

14 .4 Quotient Rule Calculate the derivative of f() = sin sin( ) f () = sin( )( sin() cos()) sin () cos( )() sin ( ) ( sin( ) cos() s sin() cos( ) ) = sin() sin( ) ( cos() = sin() sin( ) sin() ) cos( ) sin ( ) ( ) = sin() csc( ) cos() sin() cot( ) 4

15 .5 E. involving all types of derivative techniques Calculate the derivative of f() = sin () ( + ) f () = ( + [ ) d d sin () ] sin [ () d d ( + ) ] [ ( + ) ] = ( + ) ( sin () + sin() cos()() ) sin () ( + )() ( + ) 4 = ( + ) ( sin () + 4 sin() cos() ) sin () 4( + ) ( + ) 4 = ( + ) ( sin () + sin(4) ) 4 sin ()( + ) ( + ) 4 To simplify, factor out a ( + ) term from the numerator = ( + ) ( sin () + sin(4) ) 4 sin () ( + ) 3 5

16 .6 Derivatives w/ Fractional Eponents Differentiate /5 + 4/5 ( ( + 4/5 d ) ) ( d /5 /5 d ) d + 4/5 ( + 4/5 ) = ( + 4/5 )( 5 4/5 ) /5 ( 4 5 9/5 ) ( + 4/5 ) = 4/5 + 8/ /5 5( + 4/5 ) = 4/ /5 5( + 4/5 ) = = + 5 4/5 5 4/5 ( + 4/5 ) 4/5 4/5 4/ /5 ( + 4/5 ) = 4/ ( 4/5 + ) 6

17 3 Integrals 3. Riemann sums Epress the integral as a it of Riemann sums, then evaluate. ( + + ) d = b a n = = n n i = a + i = + i( n ) = i n Note that if using right-hand sums then i=,,...,n. If using left-hand sums then i=,,...,n-. And if using midpoints then i=, 3 n,..., = n i= = n i= = n n ( i n n ( i 3 n 3 + i ) n + n ) + i n + n ( n i 3 n + i= ( 3 = n n 3 n n 3 i= i= i + n use the appropriate formulae for the summations: i n n + ) n i= n i + n ) i n i= i= n i = i= n(n + )(n + ) 6 and n i = i= n(n + ) and n i = n i= then, ( 3 n(n + )(n + ) = n 6n 3 + n(n + ) n + n ) n to simplify calculations simply notice that the degree of the numerators of the fractions are the same as the denominators, thus just look at the coefficients of the leading terms: =

18 3. U-substitution, basic 4 u = + 4 du = d ( + 4) d Changing the its of integration yields: u : = u u du = 8

19 3.3 Indefinite integral, u-substitution, basic manipulation of substitution 3 + d In general it is good to make u equal to whatever is beneath the square root. Then notice that you will want to separate the 3 into u = + u = du = d Plugging the substitution into the integral yields = = (u ) u du (u 3 u ) du ( 5 u 5 ) 3 u 3 + C After evaluating an indefinite integral you need to always substitute back to the original variable; in this case the answer needs to be epressed in terms of = ( + ) 5 5 ( + ) C 9

20 3.4 Definite integral, u-substitution, basic manipulation of substitution 4 + d Since the integral can t be evaluated directly, make the substitution equal to the radicand in hopes of simplifying the integrand u = + = u du = d Also change the its of integration for to its of integration for u by plugging the lower and upper -its into the substitution: Lower Limit : u = + () =, Upper Limit : u = + (4) = 9 9 u ( ) du u = u u du = ( ) u u du 4 = ( ) 4 3 u 3 9 u = ( ) ( 3 ) = ( ) 3 = 3

21 3.5 Integrals w/ Symmetric Limits I = sin() + d This integral cannot be evaluated using only Calc I techniques, instead notice that you are integrating an odd function over symmetric its = I =. First check that the integrand is indeed odd (i.e. f( ) = f()) f() = sin( ) + ( ) = sin() + = f() I = sin() + d = sin() + d + sin() + d =

22 3.6 Fundamental Theorem of Calculus Take the derivative of y() = 3+ = sin(t 4 ) dt = sin(t 4 ) dt + sin(t 4 ) dt sin(t 4 ) dt sin(t 4 ) dt Suppose the antiderivative of the integrand is F, then plugging in the its of integration yields y() = F () + F () + F (3 + ) F () = F (3 + ) F () Differentiating takes F back to the original integrand with the its of integration as the variable, DO NOT FORGET THE CHAIN RULE. dy d = sin((3 + )4 ) 3 sin(() 4 ) = 3 sin((3 + ) 4 ) sin(6 4 )

23 4 Applications of Derivatives 4. Related Rates Part I Two people start from the same point. One walks east at 3 mi/hr and the other walks northeast mi/hr. How fast is the distance between the people changing after 5 minutes? Start by drawing a distance triangle and a rate triangle. a c da dt = mi hr dc dt θ = π/4 b θ = π/4 db dt = 3 mi hr Translate the problem to mathematical language: Find dc dt at t = 4 hr Using Law of Cosines, write an epression that will contain (after differentiating implicitly) dc dt : c = a + b ab cos(θ) Differentiate implicitly with respect to t c dc ( dt = ada dt + bdb dt cos(θ) a db ) dt + bda dt () After 5 minutes: a = 4 = (mi.), b = 4 3 = 3 4 (mi.) ( ) ( ) 3 3 c = cos(π 4 ) 9 c = ( ) 4 3 = 6 3 = 8 6 (3 6 ) Now plug all the values into equation (): dc dt = ( ( 4 9 = ) = (3 6 ) 3 6 = 3 6 ( )) 3 3

24 4. Related Rates Part II A Ferris wheel with a radius of m is rotating at a rate of one revolution every minutes. How fast is a rider rising when his seat is 6 m above ground level? (, y) θ r = Since we are looking for dy dt, use ω = dθ dt = rev min = π rad min tan(θ) = y We have to einate the variable, so use the fact that the point (, y) is constrained to be on the circle: = y tan(θ) = y ( y ) / Differentiate implicity with respect to t sec (θ) dθ dt = dy [ dt ( y ) / + y ( y ) 3/ ( y dy ] dt ) sec (θ) dθ dt = dy [ ( y ) / y ] + dt ( y ) 3/ Since we fied the origin at the center of the circle, the height of the rider is y = 6 = 6, then we can solve for cos(θ) by looking at the triangle in the circle: We see that, cos(θ) = Plugging y and cos(θ) into our epression we get: 6 π = dy ( dt ) 8 3 Simplifying yields, dy dt = 8π 4

25 Problem Differentials Use differentials to estimate the sin( ). To use differentials we need to write a function; since the question involves the sine function, use y = sin() Now using the definition of the differential dy = f ()d, write dy = cos()d Since we are approimating which is very close to, take to be, and then d = = π 8. Note that we have converted degrees to radians because the normal formulas for the derivatives of trig. functions hold only when the argument is in radians. π dy = cos() 8 = π

26 Problem Linear Approimations Verify the linear approimation at for the sine function: sin() Recall that the linearization of f at a is L() = f(a) + f (a)( a) Where, in our case, a =. Thus, f() = sin() = f () = cos() = L() = + ( ) = This tells us that, for small enough (from to around ), the sine function can be approimated (to within % of the actual value) by the linear function y =. 6

27 Problem Optimization (Standard) Find the point on the curve y = that is closest to the point (3, ). Begin by drawing a picture: y (, y) y = (3, ) Choose a general point on the curve and then since we are trying to minimize distance, use this point to write out the distance formula: d = ( 3) + ( ) = = Now take the first derivative and find the critical point(s) d d (d) = = = 5 = 5 Therefore the point closest to (3, ) is the point ( 5, 5 ) 7

28 Problem Optimization (Difficult) Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral. a a h θ b θ The first thing to do with optimization problems is to draw a picture. Then the general procedure is to write out a constraint equation and the equation that you wish to optimize, obviously these equations depend on the geometry of the problem. In our case the triangle is constrained to have a given (i.e. constant) perimeter, say p: p = a + b Then the optimization equation is the area of the triangle: A = b h = ab sin(θ) Notice that to prove the triangle is isosceles we need to show that b = a. Before we can take the derivative (to find the critical points) we need to einate two of the variables. One eination follows directly from the constrain equation: b = p a Notice that we can einate sin(θ) because we can find cos(θ) entirely in terms of b and a: cos(θ) = b a cos (θ) = b 4a sin (θ) = cos (θ) = b 4a sin(θ) = b 4a Then, A = a(p a) (p a) 4a 8

29 Problem 6. Simplify a little before differentiating a(p a) 4a (p = 4ap + 4a ) 4a a(p a) = 4ap p 4a = p a 4ap p 4 Now differentiate and equate to zero: da da = ( 4 4ap p = p 4ap 4ap p 4ap p = p ap 6ap = p 6a = p 3a = p ) 4ap p 4p + (p a) = 4ap p Notice that this critical point must correspond to a maimum because physically a minimum would be A = which would imply the dimensions are zero. Plugging this value for a back into the equation for b b = 3a a = a The triangle is equilateral because all sides have the same length. 9

30 Problem Curve Sketching Sketch the curve y = 4 tan(), π < < π. Domain. In this case the domain is already specified.. Intercepts. To find the y-intercept plug in = : y() = 4() tan() = Thus the curve passes through the origin (i.e. (,)). 3. Symmetry. Check if y is even or odd: 4. Asymptotes. y( ) = 4( ) tan( ) = 4 sin( ) cos( ) = 4 + sin() = 4 + tan() cos() y( ) y() y is not even. y( ) = y() y is odd. (a) Horizontal. Since the domain is restricted there cannot be any horizontal asymptotes. (b) Vertical. (4 tan()) = ± a 5. First Derivative. 6. Second Derivative. y as π y as π dy d = 4 sec () = 4 = sec () ± = cos() cos() = ± = ± π 3 (critical points) d y d = sec () tan() = Notice that sec () never equals zero because of the domain restriction, so you can divide both sides by sec () tan() = Test concavity on each side of the inflection point: = (point of inflection) π < < : y > = = π 3 is a local min. < < π : y < = = π 3 is a local ma. 3

31 Problem 7. y = π π 3 π 3 = π 3

32 Problem 8. 5 Applications of Integrals 5. Area Between Curves Find the area of the region bounded by the parabola y =, the tangent line to this parabola at (, ), and the -ais. y y = After drawing a picture, the first step is to find the equation of the tangent line. y = = = Then using point-slope form, y = ( ) y = Notice that to find the area enclosed by the curves, two integrals are needed because the bottom curve changes. This point of change occurs when the two bottom curves intersect, i.e. when the tangent line hits the -ais y = = = 3

33 Problem 8. Then the total area (A) is, A = = 3 3 = ( ) d + + ( ) ( ( )) d 33

34 Problem Volume of Revolution, Disk Method, Part I Find the volume of the solid obtained by revolving the area enclosed by f() = 4 and g() = from 3 about the -ais. y ( 3, ) r inner r outer Volume = π = π = π b a 3 3 (r outer r inner) d (4 ) d (3 ) d ) = π ( ( = π ) 3 3 = π 3 34

35 Problem Volume of Revolution, Disk Method, Part II Using the same region enclosed by the functions from Problem 9, set up, but do not evaluate, the integral epression for volume if the ais of revolution is moved to y =. y ( 3, ) r inner r outer y = From the picture, it is easy to see that both radii have been lengthened by unit: r outer = 4 + r inner = + = Then plugging into the volume equation: Volume = π 3 [ ( 4 + ) 4 ] d 35

36 Problem Volume of Revolution, Shell Method Find the volume obtained by revolving the region bounded by the curves = (y 3), = 4 about y =. y (4, 5) h dy r y = (4, ) The epression for the height of a general shell will be, h = 4 (y 3) The radius, as seen from the picture, can be epressed as r = y And the its of integration will be y : 5 Now plugging into the equation for volume via shells we get: Volume = π = π = π = π = π (y )(4 (y 3) ) dy (y )(4 (y 6y + 9)) dy (y )( y + 6y 5) dy ( y 3 + 6y 5y + y 6y + 5) dy ( y 3 + 7y y + 5) dy ( ) = π y y3 3 y + 5y [( = π 65 ) [( ) ( )] 75 9 = π = π 56 = 8 3 π ( 4 )]

37 Problem Work, Chain Problem A cable 5 feet in length and weighing 4 pounds per foot (lb/ft) is hanging. Calculate the work done in winding up 5 feet of the cable. Neglect all forces ecept gravity. = d This problem can be broken into two parts; each little segment d in the first 5 feet of the cable will move a different length, namely a distance. Whereas each segment of the bottom 5 feet of the cable will move a constant distance of 5 feet. Top 5 feet: The force due to gravity on a small length, d, of the cable is the linear density times the length: 4 d. Fiing the origin of the number line at the top of the cable will prove to be the easiest configuration because then each small segment of the cable will move a length. Then because Work=Force Distance: Work = 5 4 d = 5 = 65 = 5 ft.-lb Bottom 5 feet: Because all segments move a constant distance, an integral is not needed: Work = (5 4) 5 = 5 ft.-lb Thus the total work is = 375 ft.-lb 37

38 Problem Work, Spring Problem A spring has a natural length of meter (m). A force of Newtons compresses it to.9 m. How much work is required to compress it to half of its natural length? What is the length of the spring when Joules of work have been epended? Here we have to use Hooke s Law, f() = k, then work = b k d. Keep in mind that all distances have a to be measured with respect to the natural length of the spring, e.g. stretching the spring to a length of.5 m means that you have to stretch it.5 m (w.r.t. natural length). The first thing to do is to solve for the constant k: = k(.) k = N m Compressing the spring to half of the natural length means that the its of our work integral will be from to.5 f = W =.5 = 5.5 d = 5 4 = 5 J In the second part of the problem we are given that W = J; W = = a = 5 a = 5a a = 5 = d 5 = 5 So if we are still compressing the spring, the length is now.8m 38

39 .. 3 m m m 3 m Problem Work, Involving Geometry 8 m If the tank is filled with water, how much work is required to pump all the water out of the top of the tank? Use the fact that the density of water is 6.5 lb./ft ft 4. ft 3 m frustum of a cone 3 ft 8 ft 6 ft ft 5. Suppose that for the tank in Eercise the pump breaks down after J of work has been done. What is the depth of the water remaining in the tank? 6. Solve Eercise if the tank is half Wfull ork = of ρgoil that V () has da density of 9 kg m 3 a. 7. When gas epands in a cylinder with radius r, the pressure at any given time is a function of the volume: P P V. The force eerted by the gas on the piston (see the figure) is the product of the pressure and the area: F r P. Show that the work done by the gas when the volume epands from volume to volume is y V V W y V PdV piston head in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). 3 perience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. V force = (mass) (gravity) = ρv g where ρ is the density of water and V is the volume. Then if we let = at the top of the tank, then each each slice of the tank is going to have to be lifted a distance of, so the total work becomes: b To come up with an epression for how the volume of a slice of the tank varies with height we can use similar triangles. Since the front face of the tank has a constant width, look at the tank in profile: Then the work becomes, 6 6 = = y 6 y = (6 ) V () = y d = (6 ) d Work = ρg 6 (6 ) d = ρg ) = ρg ( ( = ρg 8 6 ) = (6.5)(36) = 45 ft.-lb 6 (6 ) d 39

40 Problem Average Value of a Function Find the average value of f() = sin() from π. y sin sin π π Average value = sin() d π { sin() if < < π f() = sin() if π < < π = ( π π ) sin() d + sin() d π π = ( π π) cos() π + cos() π = 4 ( ) = π π = π 4