171, Calculus 1. Summer 1, CRN 50248, Section 001. Time: MTWR, 6:30 p.m. 8:30 p.m. Room: BR-43. CRN 50248, Section 002

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1 171, Calculus 1 Summer 1, 018 CRN 5048, Section 001 Time: MTWR, 6:0 p.m. 8:0 p.m. Room: BR-4 CRN 5048, Section 00 Time: MTWR, 11:0 a.m. 1:0 p.m. Room: BR-4 CONTENTS Syllabus Reviews for tests 1 Review for the final eam

2 171 Calculus 1 Syllabus Catalog Description Functions, graphs, limits, continuity, derivatives and antiderivatives of algebraic and transcendental functions; techniques of differentiation; applications of derivatives, polynomial approimation; indeterminate forms; maima and minima and applications; curve sketching; the definite integral; the fundamental theorem of calculus; integration by substitution. Prerequisite: MATH 16 with a grade of C or better. Learning outcomes Upon successful completion of this course, students will be able to: 1. Evaluate limits of functions. Differentiate algebraic and transcendental functions. Solve problems involving rates of change and optimization problems 4. Graph functions and determine features of graphs such as intervals of increase and decrease, concavity, inflection points, asymptotes, holes, etc. 5. Find anti-derivatives of functions and evaluate definite integrals using the definition of the integral and the Fundamental Theorem of Calculus 6. Evaluate definite and indefinite integrals using substitution Book: James Stuart. Calculus. Early Transcendentals. Edition , Special edition for the community college of Philadelphia Cengage Learning ISBN: Instructor: Dr. Arkady Kitover Office: Main B - 5J, NE - 7

3 Office hours: Main S :0 pm 5 pm, NE MR 5 pm 6:5 pm, T 5 pm 6 pm. (the best way to contact me) akitover@ccp.edu or akitover@hotmail.com Web Page: The web page contains the syllabus, the reviews with detailed solutions, and MAPLE attachements. Contents: Chapters 1-5 Limits, Derivatives, Transcendental functions, Applications of derivative. Antiderivatives and definite integrals. Appendi (provided): Applications of Taylor and Maclaurin polynomials. Tests: Test 1. Limits. 40 points with possibilities for etra credit. Test. Rules of differentiation. 40 points with possibilities for etra credit. Test. Graphing problems. Applied minimum and maimum problems, Newton s method, L Hopital s Rule. Linear and quadratic approimation, Taylor polynomials. 60 points with possibilities for etra credit. You will need a scientific calculator for tests 1 and. Cumulative Final 100 points. Grading: A B C D F Average of all tests. 90%-100% (16 40 points) 80%-89% (19 15 points) 70%-79% ( points) 60%-69% ( points) Less than 60% (less than 144 points).

4 No matter what is your average, you will not get an A or a B if you get less than 50% or 40%, respectively, on the final. You can make up the regular class tests, one make up for each test. No make ups for the final. Only students with satisfactory attendance record (not more than two unecused absences) will be entitled to take make up tests (if necessary). If you miss a test I will give you a possibility to take it only if you have a valid and documented ecuse. No food is allowed in the classroom. You may not use any electronic devices for the purposes not related to the class work (teting, surfing, etc.) Put your cell phones in the vibration mode before the class starts. No cell phones during a test (unless you use your cell phone as a calculator). What you have to repeat from precalculus. (Chapters numbers are given for the book of Larsen and Hostetler Algebra and Trigonometry, the fifth edition. You can use any other precalculus tetbook). Chapter. Functions and their graphs (functions, graphs, translations and combinations, compositions, inverse functions). Chapter 5. Eponential and logarithmic functions and their graphs, properties of logarithms. Chapter 6. Angles and their measure. Graphs of basic trigonometric functions. Chapter 7. Trigonometric identities. Sum and difference formulas. Trigonometric equations.

5 Recommended Homework The numbers of Sections, Pages, and Problems are given from Stewart, Early Transcendentals, Edition 8. Section Page (Pages) Problems , 50 1., 4 1,, 8, , 9, ,, , 5, , 17,, 9, 51, ,, 9,, 5, , 18, 0, , 15,, 9, 1, 49, 65, , 5, 1, 9, , 9, 15, , 17, 5, , 11, 19, 9, 7, , 11, 9, 4, , 1,, , 9, 10, 6, , 5, 57, , 1 17, 1, 5, 47, 57, , 6, 15, 7, 7, , 19, 5, 9, , , 19,, 9, , 401 7, 7, 41, 57, 61, , 419, 11,, 55, 69

6 171 Calculus 1 Review for Test 1 Find the following limits (if the limit is positive or negative infinity, or does not eist state it eplicitly) lim. This is an indeterminate form 0/0. To solve the problem we 4 64 factor the numerator as difference of two squares 16 = 4 = ( + 4)( 4) and the denominator as difference of two cubes (according to the formula a b = ( a b)( a + ab+ b )), 64 = 4 = ( 4)( ). Therefore, 16 ( 4)( + 4) lim = lim = lim = = ( 4)( ) lim( ). Here we have an indeterminate form. To solve the problem we will first convert it to an indeterminate form in the following way. lim( 1 1) lim ( )( ) = = ( + 1) ( + + 1) 6 = lim = lim Because it is now an indeterminate form the value of the limit will not change if we leave only the leading terms in the denominator, i.e lim = lim = lim = lim 0 1. Again we have an indeterminate form 0/0. We will solve the u e 1 problem with the help of one of our basic eponential limits lim = 1. To u 0 u ln ln (ln) do it notice that = e and = e whence = e and = e (ln). Net,

7 (ln) (ln) 1 e 1 e 1 ln ln lim = lim = lim. The limits of the first 0 0 (ln) 0 (ln) 1 e 1 ln e 1 ln and the second factor are equal to one (put in the first case u = ln, in the u e 1 second u = ln and apply the basic lim = 1 ). Therefore the answer in u 0 u this problem is ln ln. cos( s) 1 4. lim s 0 sin. We deal with this indeterminate form 0/0 in the following (5 s ) cos( s) 1 [ cos( s) 1][ cos( s) + 1] cos ( s) 1 lim = lim = lim = s 0 0 sin (5 s) s sin (5 s) [ cos( s) + 1] s 0 sin (5 s) [ cos( s) + 1] way sin ( s) = lim. s 0 sin (5 s) cos( s) + 1 Because s 0 [ ] lim[cos( s) + 1] = cos0 + 1= we have to compute 1 sin ( s) lim s 0 sin (5 s ). We will do it sin u with the help of our basic trigonometric limit lim = 1 as follows. u 0 u 1 sin ( s) 1 sin ( s) (5 s) 9 9 lim = lim =. (The second and the s 0 0 sin (5 s) s ( s) sin (5 s) 5 50 third factors in the computation above have limits equal to 1 because sin u lim = 1 = 1.) u 0 u Review of problems similar to etra credit problems. 5. Consider the following function f () = tan, 0 Compute the values f (0.1), f (0.01), f (0.001), and f (0.0001). Based on your calculations what seems to be the value of tan lim? 0 Remember that your calculator must be in the radian mode. We compute

8 f (0.1) f (0.01).4667 Now we can guess that lim 0 tan = 1 f (0.001) 0.47 f (0.0001) Consider the function g() = ln, > 1. Compute g(10), g(10 ), g(10 6 ), g(10 9 ). Based on your calculations what seems to be the value of lim ln? We compute Now we can guess that lim 0 ln = g(10) 1.4 g(10 ) 4.6 g(10 6 ) 7.4 g(10 9 ) 155

9 171 Calculus 1 Review In the following problems find the derivative. Simplify if possible. 1. f( ) = csc(5 ). We combine the product and the chain rules and use the formulas da ( ) d(csc ) = a ln a and = csc cot. Thus we obtain d d df ln csc(5 ) csc(5 ) cot(5 ) 5 csc(5 )[ln 5cot(5 )] d = =.. f( ) = We use the quotient rule f fg fg =. g g (4 )( + + 1) ( + 1)(4 + ) f ( ) = = 4 ( + + 1) = = 4 ( + + 1) ( 1) = = ( + + 1) ( + + 1) 4 4. sin. f( ) cot( e ) =. The problem is on the application of the chain rule. df d 1 sin sin = ( csc ( e )) e cos. sin cot( e ) ln + sin 4. f( ) =. Like in Problem we apply the quotient rule. ln + cos df (1/ + cos )(ln + cos ) (ln + sin )(1/ sin ) = = d (ln + cos ) (1/ ) ln + (1/ ) cos + cos ln + cos (1/ ) ln + sin ln sin (1/ ) + sin = = (ln + cos ) (1/ )(cos sin ) + ln (cos + sin ) + 1 = (ln + cos )

10 5. Use implicit differentiation to find the slope-intercept equation of the tangent line to the ellipse 4 + y = 1 at the point (1, / ). 9 After differentiating both parts of the relation we get + y dy 9 d = 0. Plugging in = 1 and y = / we obtain 1 + dy d = 0 whence dy (1, / ) =. An equation of the tangent line at the point d (1, / ) in the point-slope form is y = ( 1) whence the slope-intercept form is y = +. 6 ( 1) 8 6. f () =. We use the formula for logarithmic differentiation ( ) 10 1 ( ) df d(ln( f ( ))) = f( ). We have ln( f ()) = 6ln + 8ln( 1) 10ln( ) 1ln( ) d d whence d(ln( f ())) d Finally, df d = 45 ( 1) 7 ( ) ( ) 11 ( ) 1. = = 4( ) ( 1)( )( )

11 Calculus Review In Problems (1) (4) consider the function f( ) = ( + ) e. 1. Find the critical (stationary) points; establish their character (relative minimum, relative maimum, or neither); find intervals where the function is increasing or decreasing. Solution. By product rule we have f ( ) = (+ 1) e + ( + ) e = ( + + 1) e. The stationary points will be the solutions of quadratic equation + + 1= 0. By the ± ± 5 quadratic formula = =. These two stationary points 1 divide the real ais into three intervals. The net table shows the behavior of the function f on these intervals Interval ,,, Sign of f ( ) Behavior of f( ) Increasing Decreasing Increasing Therefore the point minimum. 5 is a relative maimum and the point + 5 is a relative. Find the inflection points and the intervals where the function is concave up or concave down. Solution. Applying again the product rule we see that f ( ) = (+ ) e + ( + + 1) e = ( ) e = ( + 1)( + 4) e. There are two inflection points at -1 and at -4. The table below shows the intervals of concavity up and down. Interval (, 4) ( 4, 1) ( 1, ) Sign of f ( ) Concavity Up Down Up

12 . Graph the function (your graph should show all the essential details (The - intercepts, critical points, and inflection points). To find the -intercepts we have to solve the equation f( ) = ( + ) e = ( + 1) e = 0 whence the -intercepts are at = 1 and = 0. Moreover we see that the function is positive on (, 1) (0, ) and negative on ( 1,0). A computer generated graph is shown below. 4. Use Maclaurin polynomial of degree to estimate the value of f (0.1). Estimate the error of your approimation. Solution. Recall the formula for Maclaurin polynomial of degree ; f (0) f (0) M( ) = f(0) + f (0) + +. We already know that!! f ( ) = ( + + 1) e and f ( ) = ( ) e. Therefore by the product rule f ( ) = (+ 5) e + ( ) e = ( ) e and we have M( ) = + +. Therefore (0.1) M = + + =. Recall that the absolute value of the error of our approimation is not greater than (4) f ( t) 4 ma 0.1. The fourth derivative of f is 0 t 0.1 4! (4) f ( ) = (+ 7) e + ( ) e = ( ) e.

13 The function f (4) ( ) obviously is positive and increasing on the interval[0,0.1]. Therefore the error is not greater than (4) f ( t) ma 0.1 = e t 0.1 4! 4 For comparison, we can compute directly that f (0.1) and therefore the error of our approimation is about which is smaller than our estimate above. 5. Find the limit Solution. Because lim (1 cos ) π / tan +. π cos = 0, lim tan =, and lim tan = we have here an π / π /+ indeterminate form1. As always with this kind of problem we will try first to find the limit of natural logarithm of our epression. tan ln (1 + cos ) = tan ln(1 + cos ). The limit lim tan ln(1 + cos ) is an indeterminate form 0. We will rewrite it as π / ln(1 + cos ) lim which is an cot π / indeterminate form 0 and now we can apply the L Hospital s rule. According to this 0 rule we differentiate the numerator and the denominator and look at the limit 1/(1 + cos )( sin ) lim = 1. π / csc tan 1 Finally, lim (1 + cos ) = e = e. π / 6. Find the smallest value of the function f( ) = sec+ 4cscon the interval (0, π / ). Solution. Because the function is differentiable on (0, π / ) it takes its smallest value df at a stationary point. Thus we have to solve the equation 0 d =. df sin 4cos sin 4cos tan sec 4cot csc = = =, and therefore at d cos sin sin cos a stationary point we have sin = 4cos whence tan = 4 / and = arctan( 4 / ) 0.8. To find the smallest value itself let us notice that at the stationary point sec 1 tan 1 16/ 9 = + = + and min f( ) 1 16 / / (0, π /) = csc 1 cot 1 9/16 = + = + whence

14 The two computer generated graphs below (made with MAPLE and GRAPHMATICA, respectively) are included to illustrate our calculations. On the test you are not required to graph the function in such a problem.

15 7. Use Newton s method to approimate the positive solution of the equation P ( ) = + 1= 0. Solution. Because the sign of the coefficients of the polynomial P changes only once the equation has only one positive real root by the Descartes rule of signs. Because P(0) = 1 and P(1) = 1this root is located between 0 and 1 and we will take as the initial approimation 0 = 0.5. Now it remains to use successively the formula f( n) n + n 1 n+ 1 = n = n f ( n) n + n. Below it is shown how the calculations can be organized for TI calculators (TI-8, TI-85, et cetera). If you have a different type of calculator you have to organize your calculations yourself or consult the manual. STO 0.5 X (This command gives the initial value of 0.5) STO ( ^ 1) /( ) X X + X X + X X (This command computes 1 ) nd Enter Enter (This command repeats the previous one and therefore computes ). We repeat the last command until the desired accuracy is achieved (e.g. two successive approimations coincide up to the accuracy of our calculator).

16 Below is shown how it looks if we use TI-85. = = = = = = = Here we stop, the accuracy of our approimation is at least 10 or one over one hundred billion. 8. An open cup is in the shape of a right circular cone and must have the volume of V cm. Find the radius and the height of the cone that would minimize its surface area and also find the minimum value of the surface area. Solution. The volume and the surface area of a right circular cone with radius r and height h are given by the formulas 1 V = π r h S = π r r + h Minimizing S is equivalent to minimizing S. Therefore our problem can be written as follows. V Minimize r ( r + h ) under the condition that rh=. We will use implicit π differentiation assuming that r is a function ofh. Differentiating both parts of the V relation rh= by r we obtain π Whence The derivative of the function At a critical point we have dh r dr rh + = 0 dh h = dr r 4 r ( r + h ) = r + r h by r is dh 4r + rh + r h dr

17 Whence r rh r h dh dr = 0 r h rh dh dr dh h dr r h= r V r π + + = 0 By plugging into the last equation = we get From the last formula we obtain Then 6 h r v =, and r r = h or = V. 6 = = and S r V π π = π = π.

18 Math 171 Review for the Final Eam 1. Find the limits (4 points each) (a) lim 4 ; (b) lim ( 1 ) ; (c) lim( 1 1 ); 1 ln 1 sin ( ) (d) lim. 4 Solutions (a) The limit lim 4 represents an indeterminate form. Therefore we can keep only the leading terms in the numerator and in the denominator of the fraction. lim 4 = lim 4. Because tends to positive infinity we have 4 = whence lim 4 = lim =. (b) The limit lim ( 1 ) represents an indeterminate form 1. As always when we deal with such an indeterminate form we will first find the limit of the natural logarithm of our epression. We have ln ( 1 ) = ln, and the limit lim ln is an indeterminate 1 1 form 0 because the first factor tends to and the second one - to ln 1 = 0. To apply the L Hospital s rule we have to write the last limit as an indeterminate form 0 or. In our case it is easier to write 0 it in the form 0, namely, lim ln = lim ln. According to the 0 1 L Hospital s rule we will now differentiate the numerator and the denominator and look at the following limit (when differentiating the numerator we combine the chain rule and the quotient rule). 1 1 lim 1 1( 1) ( )1 ( 1) 1 = lim Therefore lim ( 1 ) = e 1 = 1 e. 1 ( )( 1) 1 1 = lim ( )( 1) = 1.

19 (c) We have here an indeterminate form. First we bring it to the form 0; lim( 1 1 ) = lim ( 1) ln. 0 1 ln 1 1 ( 1) ln Net we use the L Hospital s rule to obtain the following limit. lim ln + 1 It is still an indeterminate form 0 so we apply the L Hospital s rule 0 once again and get the limit (d) lim = 1.. This is an indeterminate form 0 0 sin ( ) lim 4 cos ( ) lim and we apply the L Hospital s rule. = Find the derivatives of y = f() with respect to (4 points each) (a) f() = cos (sin ( + 1)); (b) f() = 1+ ; (c) f() = sin ; (d) 5y + sin y =. Solutions. (a) f() = cos (sin ( + 1)). We apply the chain rule to get f () = sin (sin ( + 1)) cos ( + 1).

20 (b)f() = 1+. By the quotient rule combined with the chain rule we have f () = We simplify the last epression by multiplying both numerator and denominator by 1 +. f () = (1 + ) (1 + ) 1 + = 1 (1 + ) /. (c) f() = sin. By the product rule combined with the chain rule we have f () = 1 sin + cos 1 = 1 ( 1 sin + cos ). If we recall the formula A cos θ + B sin θ = A + B sin (θ + arctan B A ) then we can also write our answer as f () = 1 + sin ( + arctan ). (d) 5y + sin y =. We apply implicit differentiation to get Solving it for dy d we get 10y dy dy + cos y d d =. dy d =. Find the integrals (5 points each) 10y + cos y.

21 (a) d; ( +1) (b) tan sec d; (c) (d) Solutions (a). ( + 1) d; 6t (t + 1) 10 dt. ( + 1) d If we notice that is proportional to the derivative of + 1 we can use the following substitution; u = + 1. Then du = 6, du = 6d, d and therefore d = 1 du. The integral becomes 6 Applying the power rule we get (b) u du u du = 1 6u + C = 1 6( + 1) + C. tan sec d Notice that sec equals to the derivative of tan and therefore it is convenient to use the substitution u = tan. Then du = sec d and our integral becomes u du = u + C = tan + C. (c) We apply the power rule and the Newton - Leibnitz formula to get 4

22 1 0 ( ( 4 ) 1 + 1) d = 4 + = (d) 0 6t (t + 1) 10 dt. 1 We use the substitution u = t + 1. Then du = t dt whence 6t dt = du. If t = 1 then u = 0, and if t = 0 then u = 1; we will change the limits of integration accordingly t (t + 1) 10 dt = 1 0 u 10 du = u = Use a linear approimation to estimate the following values (5 points each) (a) 6; (b) sin 1. Solutions In both cases we use the formula for linear approimation 0 L() = f(a) + f (a)( a) where point a should satisfy two (informal) conditions a should be close to the values of f(a) and f (a) should be easy to compute. (a) In this case we take f(t) = t whence f (t) = 1. We also take t a = 64 and = 6. Then f(a) = 4, f (a) = 1, and a = 1. The 48 corresponding linear approimation to 6 is

23 (b) Remember that we do not use the degree measure in calculus. In our case we take Then f(t) = sin t, a = 0 = π 6 and = 1 = π 6 + π 180. f(a) = sin π 6 = 1, f (a) = cos a = cos π 6 =, and a = π 180. Therefore the value of the linear approimation is 1 + π Find the values of a and b, if the tangent line to y = a b at (, 5) has slope m =. (5 points) Solution. Because y( ) = 5 we have the following equation for a and b. 4a + b = 5 The derivative dy dy is a b. Because ( ) = we have the second d d equation for a and b. 4a b =. Adding the equations we get b = 7 and plugging this value of b into the first equation we get 4a + 14 = 5 whence a = If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is h(t) = 80t 16t. What is the velocity of the ball when it is 96 ft above the ground on its way up? On its way down? (5 points) Solution The velocity of the ball equals to the derivative of its height. v(t) = dh dt 6 = 80 t.

24 We can find the moments of time when the ball is 96 ft high by solving the quadratic equation 80t 16t = 96. Dividing both parts by 16 and moving all terms to the left we get t 5t + 6 = 0 whence t = or t =. At the moment t = the velocity of the ball is 16 ft/sec and the ball is moving up whilst at the moment t = the velocity is -16 ft/sec and the ball is moving down. 7. A rectangular bo with open top must have the volume of V cm and its length must be twice greater than its height. Find the dimensions of the bo that will minimize its surface area S..(10 points) Solution Let, y, z be the length, the width, and the height of the bo, respectively. Then = h, V = h y, and S = hy + (4h + y)h = 4h + 4hy. We will consider y as a function of h. Then dv dh = 4hy + h dy dh = 0. Therefore dy dh = y h. At a critical point of S we have ds dh = 8h + 4y + 4hdy dh = 8h + 4y + 4h( h y = 8h 4y = 0. Thus, y = h =. Now it becomes easy ti find the dimensions of the bo. Indeed, 4h = V whence h = V 4 and y = = V. 7

25 8. Sketch the graph of y = +. Find - and y- intercepts; plot the stationary points and the inflection points and determine the intervals where y is increasing and decreasing, concave up and concave down.(10 points) Solution. To find the -intercepts we have to solve the equation + = 0. Notice that = 1 is a solution of this equation and therefore 1 is a factor of +. We can factor + using long division, grouping, or synthetic division. The table below shows synthetic division Therefore + = ( 1)( + ) = ( 1) ( + ) and the -intercepts are (, 0) and (1, 0). The y-intercept is (0, ). The first derivative of y is dy d = = ( 1)( + 1). The function has two stationary points, 1 and 1 which divide the -ais into the intervals (, 1), ( 1, 1), and (1, ). The net table shows the behavior of the function on these intervals. Interval (, 1) ( 1, 1) (1, ) Sign of dy + + d Behavior of y Increases Decreases Increases To find the inflection points of the cubic function y we compute its second derivative d y d = 6. 8

26 We see that y has only one inflection point 0. The information about concavity of y is contained in the net table. Interval (, 0) (0, ) Sign of d y - + d Concavity Down Up A computer generated graph of the function y = + is shown below. 9

27 10

28 9. Use the fundamental theorem of calculus:(4 points each) (a) Find the derivative of the function G() = sin sin t dt; Solution. The fundamental theorem of calculus in its generalized form tells us that d d b() a() f(t) dt = f(b()) db d f(a())da d. Applying this formula when a() =, b() = sin, and f(t) = sin t we get dg d = cos sin (sin ) sin ( 4 ). (b) Prove that the function F () = 4 t dt is a constant on the interval (, 0). Solution. Look at two ways to solve the problem. First way. We apply the fundamental theorem of calculus to see that df d = 4 4 = 0. Because the derivative of F is identically 0 the function F is a constant function. Second way. F () = 4 4 dt = ln t t = ln 4 ln = ln 4 + ln ln ln = = ln 4 ln = ln. 11 =