4.10E: Exercises for Section 4.10 - Mathematics

1) What is the linear approximation for any generic linear function (y=mx+b)?

2) Determine the necessary conditions such that the linear approximation function is constant. Use a graph to prove your result.

(f'(a) = 0)

3) Explain why the linear approximation becomes less accurate as you increase the distance between (x) and (a). Use a graph to prove your argument.

4) When is the linear approximation exact?

The linear approximation exact when (y=f(x)) is linear or constant.

In exercises 5 - 10, find the linear approximation (L(x)) to (y=f(x)) near (x=a) for the function.

(L(x)=frac{1}{2}−frac{1}{4}(x−2))

7) [T] (f(x)= an x, quad a=frac{π}{4})

8) [T] (f(x)=sin x, quad a=frac{π}{2})

(L(x)=1)

(L(x)=0)

In exercises 11 - 16, compute the values given within (0.01) by deciding on the appropriate (f(x)) and (a), and evaluating (L(x)=f(a)+f′(a)(x−a).) Check your answer using a calculator.

11) [T] ((2.001)^6)

12) [T] (sin(0.02))

(sin(0.02)approx 0.02)

13) [T] (cos(0.03))

14) [T] ((15.99)^{1/4})

((15.99)^{1/4}approx 1.9996875)

15) [T] (dfrac{1}{0.98})

16) [T] (sin(3.14))

(sin(3.14)approx 0.001593)

In exercises 17 - 22, determine the appropriate (f(x)) and (a), and evaluate (L(x)=f(a)+f′(a)(x−a).) Calculate the numerical error in the linear approximations that follow.

17) ((1.01)^3)

18) (cos(0.01))

(cos(0.01) approx L(0.01) = f(0) + f'(0)(0-0.01) = 1;) error, (~0.00005)

19) ((sin(0.01))^2)

20) ((1.01)^{−3})

((1.01)^{−3}approx L(1.01) = f(1) + f'(1)(1.01 - 1) = 0.97;) error, (~0.0006)

21) (left(1+frac{1}{10} ight)^{10})

22) (sqrt{8.99})

(sqrt{8.99} approx L(8.99) = f(9) + f'(9)(8.99 - 9)= 3−frac{1}{600};) error, (~4.632×10^{−7})

In exercises 23 - 26, find the differential of the function.

23) (y=3x^4+x^2−2x+1)

24) (y=xcos x)

(dy=(cos x−xsin x),dx)

25) (y=sqrt{1+x})

26) (y=dfrac{x^2+2}{x−1})

(dy=(dfrac{x^2−2x−2}{(x−1)^2})dx)

In exercises 27 - 32, find the differential and evaluate for the given (x) and (dx).

27) (y=3x^2−x+6, ;x=2, ;dx=0.1)

28) (y=dfrac{1}{x+1}, ;x=1, ;dx=0.25)

29) (y= an x, ;x=0, ;dx=frac{π}{10})

30) (y=dfrac{3x^2+2}{sqrt{x+1}}, ;x=0, ;dx=0.1)

31) (y=dfrac{sin(2x)}{x}, ;x=π, ;dx=0.25)

32) (y=x^3+2x+dfrac{1}{x}, ;x=1, ;dx=0.05)

(dy=left(3x^2+2−dfrac{1}{x^2} ight)dx, quad dy = 0.2)

In exercises 33 - 38, find the change in volume (dV) or in surface area (dA.)

33) (dV) if the sides of a cube change from 10 to 10.1.

34) (dA) if the sides of a cube change from (x) to (x+dx).

(dA = 12x,dx)

35) (dA) if the radius of a sphere changes from (r) by (dr.)

36) (dV) if the radius of a sphere changes from (r) by (dr).

(dV=4πr^2dr)

37) (dV) if a circular cylinder with (r=2) changes height from (3) cm to (3.05cm.)

38) (dV) if a circular cylinder of height 3 changes from (r=2) to (r=1.9) cm.

(dV = −1.2π, ext{cm}^3)

In exercises 39 - 41, use differentials to estimate the maximum and relative error when computing the surface area or volume.

39) A spherical golf ball is measured to have a radius of (5) mm, with a possible measurement error of (0.1) mm. What is the possible change in volume?

40) A pool has a rectangular base of 10 ft by 20 ft and a depth of 6 ft. What is the change in volume if you only fill it up to 5.5 ft?

(−100 , ext{ft}^3)

41) An ice cream cone has height 4 in. and radius 1 in. If the cone is 0.1 in. thick, what is the difference between the volume of the cone, including the shell, and the volume of the ice cream you can fit inside the shell?

In exercises 42 - 44, confirm the approximations by using the linear approximation at (x=0.)

42) (sqrt{1−x}≈1−frac{1}{2}x)

43) (dfrac{1}{sqrt{1−x^2}}≈1)

44) (sqrt{c^2+x^2}≈c)

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4 (Class 10 Ex. 4.4) Quadratic Equations (Dwighat Samikaran) in PDF format to free download. All the Solutions are available to study online or in PDF format to free download for session 2021-22. UP Board Students are now using NCERT Books for their course study. They can use these solutions for solving their doubts. They can download UP Board Solution for Class 10 Maths Exercise 4.4 in Hindi Medium free of cost. NCERT Solutions 2021-2022 are available in Video Format as well as Hindi Medium and English Medium free for CBSE board as well as UP Board students.

All the solutions are based on NCERT Books for the academic years 2021-22. Vedic Maths is very effective to improve calculation. Use Vedic Maths to make your calculation easier and faster than ever.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

10 Maths Chapter 4 Exercise 4.4 Solutions

NCERT CBSE Solutions for Class 10 Maths Chapter 4 Exercise 4.4 Quadratic Equations in English & Hindi Medium for online study is given below. Visit to Class 10 Maths Chapter 4 main page to get all exercises in PDF form. Download Class 10 Maths App for offline use. Students should go for Vedic Maths to improve calculations. Download (Exercise 4.4) in PDF form to use it offline otherwise online solutions are given below.

Class 10 Maths Chapter 4 Exercise 4.4 Solution in Videos

Class 10 Maths Exercise 4.4 Solution in Hindi Class 10 Maths Chapter 4 Exercise 4.4 Solution

Word Problems with Answers for Practice

1. If the list price of the book is reduced by ₹5, a person can buy 5 more books for ₹300. Find the original cost price of the book. [Answer: ₹20]
2. A Piece of cloth costs ₹200. If the piece was 5m longer and each metre of cloth costs ₹2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre? [Answer: Length = 20m, rate = ₹10/meter]
3. A trader bought a number of articles for ₹900. Five articles were found damaged. He sold each of the remaining articles at ₹2 more than what he paid for it. He got a profit of ₹80 on the whole transaction. Find the number of articles he bought. [Answer: 75]
4. A plane left 30 minutes later than the schedule time and in order to reach the destination 1500 km away in time it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed. [Answer: 750 km/h]
5. If a student had walked 1 km/h faster, he should have taken 15 minutes less to walk 3 km. Find the rate at which he walking. [Answer: 3 km/h]
EXTRA QUESTIONS FROM CBSE BOARD PAPERS
• If the sum of first n even natural numbers is 420. Find the value of n. [Answer: n = 20]
• While boarding an aeroplane a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed /hour of the plane. [Answer: 500 km/h]
• A person wishes to fix three rods together in the shape of a right angled triangle. The hypotenuse is 4 cm longer than the base and 8 cm longer than the altitude. Find the length of each rod. [Answer: 12 cm, 16 cm, 20 cm]
• A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work alone. What are the moral values reflected in this question which are to be adopted in our life? [Answer: 30 days]
Feedback for 10th Maths Chapter 4 Solutions

We have updated the solutions of Chapter 4 Class 10 Maths according to latest CBSE NCERT Syllabus 2021-2022. Please provide your feedback about the contents. Ask your doubts related to NIOS or CBSE Board and share your knowledge with your friends and other users through Discussion Forum.

4.10E: Exercises for Section 4.10 - Mathematics

Back in the chapter on Limits we saw methods for dealing with the following limits.

In the first limit if we plugged in (x = 4) we would get 0/0 and in the second limit if we “plugged” in infinity we would get (/<-infty >) (recall that as (x) goes to infinity a polynomial will behave in the same fashion that its largest power behaves). Both of these are called indeterminate forms. In both of these cases there are competing interests or rules and it’s not clear which will win out.

In the case of 0/0 we typically think of a fraction that has a numerator of zero as being zero. However, we also tend to think of fractions in which the denominator is going to zero, in the limit, as infinity or might not exist at all. Likewise, we tend to think of a fraction in which the numerator and denominator are the same as one. So, which will win out? Or will neither win out and they all “cancel out” and the limit will reach some other value?

In the case of (/<-infty >) we have a similar set of problems. If the numerator of a fraction is going to infinity we tend to think of the whole fraction going to infinity. Also, if the denominator is going to infinity, in the limit, we tend to think of the fraction as going to zero. We also have the case of a fraction in which the numerator and denominator are the same (ignoring the minus sign) and so we might get -1. Again, it’s not clear which of these will win out, if any of them will win out.

With the second limit there is the further problem that infinity isn’t really a number and so we really shouldn’t even treat it like a number. Much of the time it simply won’t behave as we would expect it to if it was a number. To look a little more into this, check out the Types of Infinity section in the Extras chapter at the end of this document.

This is the problem with indeterminate forms. It’s just not clear what is happening in the limit. There are other types of indeterminate forms as well. Some other types are,

These all have competing interests or rules that tell us what should happen and it’s just not clear which, if any, of the interests or rules will win out. The topic of this section is how to deal with these kinds of limits.

As already pointed out we do know how to deal with some kinds of indeterminate forms already. For the two limits above we work them as follows.

In the first case we simply factored, canceled and took the limit and in the second case we factored out an ()from both the numerator and the denominator and took the limit. Notice as well that none of the competing interests or rules in these cases won out! That is often the case.

So, we can deal with some of these. However, what about the following two limits.

This first is a 0/0 indeterminate form, but we can’t factor this one. The second is an (/) indeterminate form, but we can’t just factor an ()out of the numerator. So, nothing that we’ve got in our bag of tricks will work with these two limits.

This is where the subject of this section comes into play.

L’Hospital’s Rule

Suppose that we have one of the following cases,

where (a) can be any real number, infinity or negative infinity. In these cases we have,

So, L’Hospital’s Rule tells us that if we have an indeterminate form 0/0 or (/) all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

Before proceeding with examples let me address the spelling of “L’Hospital”. The more modern spelling is “L’Hôpital”. However, when I first learned Calculus my teacher used the spelling that I use in these notes and the first text book that I taught Calculus out of also used the spelling that I use here.

“In the 17th and 18th centuries, the name was commonly spelled "l'Hospital", and he himself spelled his name that way. However, French spellings have been altered: the silent 's' has been removed and replaced with the circumflex over the preceding vowel. The former spelling is still used in English where there is no circumflex.”

So, the spelling that I’ve used here is an acceptable spelling of his name, albeit not the modern spelling, and because I’m used to spelling it as “L’Hospital” that is the spelling that I’m going to use in these notes.

1. (displaystyle mathop limits_ frac<>)
2. (displaystyle mathop limits_ frac <<5- 4 - 1>><<10 - t - 9>>)
3. (displaystyle mathop limits_ frac<<<<f>^x>>><<>>)

So, we have already established that this is a 0/0 indeterminate form so let’s just apply L’Hospital’s Rule.

In this case we also have a 0/0 indeterminate form and if we were really good at factoring we could factor the numerator and denominator, simplify and take the limit. However, that’s going to be more work than just using L’Hospital’s Rule.

This was the other limit that we started off looking at and we know that it’s the indeterminate form (/) so let’s apply L’Hospital’s Rule.

Now we have a small problem. This new limit is also a (/) indeterminate form. However, it’s not really a problem. We know how to deal with these kinds of limits. Just apply L’Hospital’s Rule.

Sometimes we will need to apply L’Hospital’s Rule more than once.

L’Hospital’s Rule works great on the two indeterminate forms 0/0 and (<< pm ,infty >>/<< pm ,infty >>). However, there are many more indeterminate forms out there as we saw earlier. Let’s take a look at some of those and see how we deal with those kinds of indeterminate forms.

We’ll start with the indeterminate form (left( 0 ight)left( < pm ,infty > ight)).

Note that we really do need to do the right-hand limit here. We know that the natural logarithm is only defined for positive (x) and so this is the only limit that makes any sense.

Now, in the limit, we get the indeterminate form (left( 0 ight)left( < - infty > ight)). L’Hospital’s Rule won’t work on products, it only works on quotients. However, we can turn this into a fraction if we rewrite things a little.

The function is the same, just rewritten, and the limit is now in the form ( - /) and we can now use L’Hospital’s Rule.

Now, this is a mess, but it cleans up nicely.

In the previous example we used the fact that we can always write a product of functions as a quotient by doing one of the following.

Using these two facts will allow us to turn any limit in the form (left( 0 ight)left( < pm ,infty > ight)) into a limit in the form 0/0 or (<< pm ,infty >>/<< pm ,infty >>). Which one of these two we get after doing the rewrite will depend upon which fact we used to do the rewrite. One of the rewrites will give 0/0 and the other will give (<< pm ,infty >>/<< pm ,infty >>). It all depends on which function stays in the numerator and which gets moved down to the denominator.

Let’s take a look at another example.

So, it’s in the form (left( infty ight)left( 0 ight)). This means that we’ll need to write it as a quotient. Moving the (x) to the denominator worked in the previous example so let’s try that with this problem as well.

Writing the product in this way gives us a product that has the form 0/0 in the limit. So, let’s use L’Hospital’s Rule on the quotient.

Hummmm…. This doesn’t seem to be getting us anywhere. With each application of L’Hospital’s Rule we just end up with another 0/0 indeterminate form and in fact the derivatives seem to be getting worse and worse. Also note that if we simplified the quotient back into a product we would just end up with either (left( infty ight)left( 0 ight)) or (left( < - infty > ight)left( 0 ight)) and so that won’t do us any good.

This does not mean however that the limit can’t be done. It just means that we moved the wrong function to the denominator. Let’s move the exponential function instead.

Note that we used the fact that,

to simplify the quotient up a little. This will help us when it comes time to take some derivatives. The quotient is now an indeterminate form of ( - /) and using L’Hospital’s Rule gives,

So, when faced with a product (left( 0 ight)left( < pm ,infty > ight)) we can turn it into a quotient that will allow us to use L’Hospital’s Rule. However, as we saw in the last example we need to be careful with how we do that on occasion. Sometimes we can use either quotient and in other cases only one will work.

Let’s now take a look at the indeterminate forms,

These can all be dealt with in the following way so we’ll just work one example.

In the limit this is the indeterminate form []. We’re actually going to spend most of this problem on a different limit. Let’s first define the following.

Now, if we take the natural log of both sides we get,

Let’s now take a look at the following limit.

This limit was just a L’Hospital’s Rule problem and we know how to do those. So, what did this have to do with our limit? Well first notice that,

and so our limit could be written as,

We can now use the limit above to finish this problem.

With L’Hospital’s Rule we are now able to take the limit of a wide variety of indeterminate forms that we were unable to deal with prior to this section.

Below is a schedule for the course. This schedule is subject to change, and therefore you should check this webpage fregquently.

Webassign HW

and Matlab HW

Exam I sample and the solutions

Review session for Chapter 5

Click here for a detailed proof and some remarks on matrix of inner product

Click  here  for the solution to Spring 2015 final

﻿Click here for the solution to Fall 2012 final

Click  here  for the solution to Fall 2011 final

Students are encouraged to bring suggestions and to discuss with the instructor about any concerns they may have, including anything they think is not handled properly in the course. But if you do not feel comfortable about doing that, here you have the opportunity to send some anonymous feedback.  Click here to enter your feedback for this course.

West Lafayette, IN 47907 USA, 765-494-4600
An equal access/equal opportunity university

Yuanzhen Shao
Department of Mathematics, Purdue University
150 N. University Street, West Lafayette, IN 47907-2067
Phone: (765) 494-1901 - FAX: (765) 494-0548

4.10E: Exercises for Section 4.10 - Mathematics

We’re now going to take a brief detour and look at solutions to non-constant coefficient, second order differential equations of the form.

[pleft( t ight)y'' + qleft( t ight)y' + rleft( t ight)y = 0]

In general, finding solutions to these kinds of differential equations can be much more difficult than finding solutions to constant coefficient differential equations. However, if we already know one solution to the differential equation we can use the method that we used in the last section to find a second solution. This method is called reduction of order.

Let’s take a quick look at an example to see how this is done.

given that (y_<1>(t)=t^<-1>) is a solution.

Reduction of order requires that a solution already be known. Without this known solution we won’t be able to do reduction of order.

Once we have this first solution we will then assume that a second solution will have the form

for a proper choice of (v(t)). To determine the proper choice, we plug the guess into the differential equation and get a new differential equation that can be solved for (v(t)).

So, let’s do that for this problem. Here is the form of the second solution as well as the derivatives that we’ll need.

Plugging these into the differential equation gives

Rearranging and simplifying gives

Note that upon simplifying the only terms remaining are those involving the derivatives of (v). The term involving (v) drops out. If you’ve done all of your work correctly this should always happen. Sometimes, as in the repeated roots case, the first derivative term will also drop out.

So, in order for (eqref) to be a solution then (v) must satisfy

This appears to be a problem. In order to find a solution to a second order non-constant coefficient differential equation we need to solve a different second order non-constant coefficient differential equation.

However, this isn’t the problem that it appears to be. Because the term involving the (v) drops out we can actually solve (eqref) and we can do it with the knowledge that we already have at this point. We will solve this by making the following change of variable.

[w = v'hspace <0.25in>Rightarrow hspace<0.25in>w' = v'']

With this change of variable (eqref) becomes

and this is a linear, first order differential equation that we can solve. This also explains the name of this method. We’ve managed to reduce a second order differential equation down to a first order differential equation.

This is a fairly simple first order differential equation so I’ll leave the details of the solving to you. If you need a refresher on solving linear, first order differential equations go back to the second chapter and check out that section. The solution to this differential equation is

Now, this is not quite what we were after. We are after a solution to (eqref). However, we can now find this. Recall our change of variable.

With this we can easily solve for (v(t)).

This is the most general possible (v(t)) that we can use to get a second solution. So, just as we did in the repeated roots section, we can choose the constants to be anything we want so choose them to clear out all the extraneous constants. In this case we can use

Using these gives the following for (v(t)) and for the second solution.

Then general solution will then be,

If we had been given initial conditions we could then differentiate, apply the initial conditions and solve for the constants.

Reduction of order, the method used in the previous example can be used to find second solutions to differential equations. However, this does require that we already have a solution and often finding that first solution is a very difficult task and often in the process of finding the first solution you will also get the second solution without needing to resort to reduction of order. So, for those cases when we do have a first solution this is a nice method for getting a second solution.

given that (left( t ight) = t) is a solution.

The form for the second solution as well as its derivatives are,

[left( t ight) = tvhspace<0.25in>left( t ight) = v + tv'hspace<0.25in>left( t ight) = 2v' + tv'']

Plugging these into the differential equation gives,

[left( <2v' + tv''> ight) + 2tleft( ight) - 2left( ight) = 0 = 0]

Rearranging and simplifying gives the differential equation that we’ll need to solve in order to determine the correct (v) that we’ll need for the second solution.

Next use the variable transformation as we did in the previous example.

[w = v'hspace <0.25in>Rightarrow hspace<0.25in>w' = v'']

With this change of variable the differential equation becomes

and this is a linear, first order differential equation that we can solve. We’ll leave the details of the solution process to you.

As with the first example we’ll drop the constants and use the following (v(t))

[vleft( t ight) = >hspace <0.25in>Rightarrow hspace<0.25in>left( t ight) = tleft( <>> ight) = >]

Then general solution will then be,

On a side note, both of the differential equations in this section were of the form,

These are called Euler differential equations and are fairly simple to solve directly for both solutions. To see how to solve these directly take a look at the Euler Differential Equation section.

Teachers

Instructor Teaching Assistant
Name Miguel A. Lerma David Feng
Office Lunt 203 N/A
Phone 1-8020 N/A
E-mail mlerma at math dot northwestern dot edu [email protected]
Office Hours by appointment
in Lunt 203
TH 12-2pm
in Tech Express
Teaching in TECH M120 TECH M164
Code N.A. TBA

4.10E: Exercises for Section 4.10 - Mathematics

Welcome to the course webpage for Math 115A, Section 2. Here you will find some general info about the course. This is also the place to look for homework assignments, occasional course notes, and other things that might interest you.

Textbook   TA and Discussion   Homework Policy   Exams   Grading   Homework Assignments   Material by Day   Various Notes   Back to Main Page
ANNOUNCEMENT: The final for our class is on Thursday, 12/12, 3:30-5:30PM, in the usual classroom.

You will be allowed one side of a regular sheet of handwritten notes to bring to the final, but no calculators, books, other notes, phones, etc. Please do bring your student ID to the exam.

The exam will cover all of the material covered in class and on homework: it will be cumulative, with no intentional bias towards any particular part of the course.

Practice problems for the final are now available here. Solutions for these will be posted next week . We will have a review session for the final, time and place TBA.

ANNOUNCEMENT 2: There is no second midterm. However, I have put together a mock midterm 2 for you to test your knowledge of what you have been doing, from congruences through Fermat's Little Theorem. Treat it like you would an exam. Study for it, write up a half sheet cheat sheet just as for midterm 1, and then do the problems in 50 minutes with no books or notes or internet besides your cheat sheet. Solutions will be posted on Canvas. If you find that you really struggled on some topic, it may be good to come talk to me about it in office hours. The mock midterm is here.

Course Syllabus

Professor: Elena Fuchs
Office: 3109 Mathematical Sciences Bouilding
Email: efuchs at math dot ucdavis dot edu
Office hours: MW 10-11AM

The textbook this course is based on is Rosen's Elementary Number Theory and Its Applications, 6th Edition.

The prerequisite for this course is Math 21B. While the material in 21B itself is not a strict requirement, this course does expect that the students have some amount of mathematical maturity.

The TA for our section is Matt Litman (mclitman at ucdavis dot edu). Starting the first week of class, he will hold discussion on Thursdays, 4:10-5PM in Olson 106. It is highly recommended that you attend these sections, in which you will see review of homework and more examples of topics covered in class.

Your grade for the course is determined as follows: 20% for homework, 35% for the midterm, and 45% for the final. There will be no make up exams.

Homework (along with occasional supplementary notes) will be posted here and will be due every Friday at 4:30PM in the appropriate homework box. Homework boxes re located in the Math Sciences Building: enter from the front, walk past the elevators and calculus room on your left and take a left turn where you will see many slots in the wall: find your section and slip in your homework. Please do not hand in your homework in class or via email, unless you have checked with me beforehand that this is ok. It should only be done in very special circumstances. You are encouraged to work with your classmates on homework and discuss the homework problems amongst each other. However, solutions should be written up independently, and you should write down which classmates you worked with at the top of your homework. The lowest homework score will be dropped. Therefore the policy is that no late homework will be accepted, especially since we will sometimes discuss the solutions to the homework problems in class.

The midterm will be in class on Monday, October 21st. More details about the exam will be announced in class and posted here closer to the date.

Sometime in November, I will also post a mock second midterm for you to try at home to test how well you have retained the material up to that point. Solutions will also be posted. Think of this as a useful studying tool and a way not to have to cram before the final.

If you have a cold or flu, I ask you to please do yourself, your classmates, me, and all the people we live with a huge favor by staying home and resting, rather than coming to class! If this happens, Matt and I will help you catch up on notes and things you missed in class once you are healthy!

4.10E: Exercises for Section 4.10 - Mathematics

MATH 106 Calculus - FALL 2018

MIDTERM 1: NOVEMBER 2 (UP TO SECTION 4.10)

FINAL: DECEMBER 26 at 9:00am

ATTILA AŞKAR

Section 2 : TUE & THU 16:00-17:15

ALTAN ERDO ĞAN

Section 3 : TUE & THU 11:30-12:45

Section 4: TUE & THU 08:30-09:45

Office Hours: TUE & THU 10:00-11:00 or by appointment

Section 1: MON & WED 16:00-17:15

Section 5: MON & WED 10:00-11:15

Office Hours: MON & WED 14:00-15:00 or by appointment

PS Hours and Locations:

PS A - Fri 10:00-11:15 - - CASE B24

PS B - Fri 08:30-09:45 - - CASE B24

PS C - Fri 13:00-14:15 - - CASE B24

PS D - Fri 14:30-15:45 - - CASE B24

PS E - Fri 16:00-17:15 - - CASE B24

PS F - Fri 11:30-12:45 - - SCI Z24

Textbook: Calculus, A complete course, 9th Ed. by R. Adams and C. Essex.

Applications of Derivatives

Techniques and Applications of Integration

Homework and Problem Sessions: Homework problems will be posted on the web page. While these will not be collected, you should take them seriously and solve them. There will be regular problem sessions, centered on the homework.

Attendance: Attendance will be not be taken for lectures. But we STRONGLY recommend regular attendance to both lectures and PS sections. There will be a simple one question quizz during each PS section. The grades for these quizzes will be considered as bonus points: up to 10 BONUS points will be added to the final grade according to performance in PS quizzes. The grading will be as follows: 1 point (for writing your name), 2 points (for making an effort at a solution), 3 points (for a correct solution). The question will be from the previous PS sessions.

Study guideline: In order to benefit from the course and perform well on the exams, you need to spend three hours on average studying on your own for every class lecture. It is essential that you read your book and your notes, and work through the assigned exercises as a minimum. The more exercises you solve, the better your understanding of the material will be. Practice solving exercises quickly by timing yourself. Avoid as much as possible looking up the solution until after you have exerted significant effort in trying to solve the exercises.

Evaluation: Your progress will be evaluated according to your performance in two midterm exams and a final exam. Their contribution to your total grade will be as follows: Midterm exams 30% each and the final exam 40%.

The midterm exams will take place outside class hours. Exam dates will be determined by the Registrar Office. All the information regarding to the exam dates and places will be announced on the web page of the course. Calculators will not be allowed in the exams.

Passing Grade: The passing grade (D) is at least 40 out of 100.

Make-up Exam: There will be only 1 common make-up exam for a missed exam (midterm or final) with an excuse (health report, etc.) at the end of the semester, after the final exam. It will include all the material covered throughout the semester (even if it may be a make-up for the midterm).

Academic Integrity: Students are expected to maintain their academic integrity. Academic dishonesty in any form will not be tolerated and punished as described in the "academic integrity" section of the Koç University catalog p.36. A detailed definition of academic dishonesty is also provided in the catalog.

Prime and Composite Numbers Worksheets

Prime and composite numbers worksheets have a variety pdf exercises to understand recognize prime and composite numbers. Also amusing display charts which list the prime and composite numbers from 1 to 100 and extremely engaging activities like coloring, cutting, pasting and mazes are here for your children in grade 4 through grade 7. Free sample worksheets are included. The first two printable charts given below show the prime and composite numbers from 1-100. The third one has both prime and composite numbers listed till 100. Ideal for 4th grade and 5th grade children. Pick up your favorite color. Color all the prime numbers in chart 1 and all the composite numbers in chart 2. In chart 3, color all the prime and composite numbers green and yellow. An enjoyable Christmas theme is given here in these worksheets. Color all the prime stars yellow and the composite stars green. These pdf worksheets are tailor-made for children of grade 4 and grade 5. Colorful snails, tortoises and crabs carry a prime or a composite number on it. Write whether the number on each of them is prime or composite. Each printable worksheet has 18 standard questions for practice of 6th grade and 7th grade students. For the given numbers, recognize if they are prime or composite. Section A and B have a list of numbers given. Circle all the prime numbers in section A and composite numbers in section B. MCQs in section C. Carefully cut out the fishes and glue them on the prime and composite fish pots. Try out this activity and rejuvenate the little kittens who are expecting the fishes. Each pdf worksheet has 10 questions. List out the factors of the given number and also write if the numbers are prime or composite. Boost your mind! Try these interesting mazes on three distinctive themes. Find the path from top to bottom by coloring all the prime numbers with exactly two factors. Figure out the composite numbers: the numbers that have more than two factors color them all and help the animated characters find their way out of the maze. These printable prime and composite number worksheets have MCQs to find the odd prime, even prime, odd composite and even composite numbers. Recommended for grade 5 through grade 7 students.  Adding with grid support helps students who have trouble lining up place values themselves. Perhaps with a little practice, they might get a better understanding of not only lining up the place values, but why this is done. Pointing out that the 5 in 659 means 50, for example, is useful in helping students understand place value as it relates to addition. These adding worksheets also help students develop mental addition skills, but use a game context for familiarity and interest. For the adding with playing cards worksheets, a Jack is counted as 11, a Queen as 12, a King as 13 and an Ace as 1. Playing math games while enjoying some social time with their friends is a great way to develop strategic thinking and math fluency in children. Finding complements of numbers can help students a great deal in developing mental arithmetic skills and to further their understanding of number.  