Math 1120 Class 1 Dan Barbasch Aug. 23, 2012 Course Website Math - PowerPoint PPT Presentation

Math 1120 Class 1 Dan Barbasch Math 1120 Class 1 Dan Barbasch Aug. 23, 2012

Course Website Math 1120 Class 1 Dan Barbasch http://www.math.cornell.edu/ ˜ web1120/index.html Dan Barbasch (dmb14) Office Hours: Mo, We 1:30-2:30 in Malott 543 or by appointment Other lecture sections will announce their own office hours. Regular Office Hour Schedule will be at website. First Assignment due 8/30 or 8/31 depending on the lecture section. Frequent quizes and assignments due during the week. Evening Prelims Tuesday Spetember 25, and Thursday October 25.

Is 1120 the right course for you? Math 1120 Math 1120 is a second semester course, so you need to Class 1 remember material from first semester calculus. Dan Barbasch If you are interested in a more theoretical course that covers somewhat more material, you might want to take Math 1220 - Honors Calculus II. If you need to take a mathematics course but don’t need to take calculus, you might want to take Math 1105, 1300, 1340, 1350, or 1710. Math 1340: Mathematics and Politics Math 1710: Statistics Math 1105: Finite Math. for the Life and Social Sciences (Fall) Math 1350: Art of Secret Writing (Fall) If you have an AP score of 4 or 5 in BC Calculus, we do not recommend you repeat the course.

Grading Math 1120 Class 1 Dan Barbasch At the end of the term we will assign grades based on each prelim 20%, final 30%, 40% in fall 2015 homework 20%, and section grade in fall 2015 section grade 10%. The section grade will be based on participation, quizzes, warmup questions, interaction with your lecturer, and the quality (not just correctness) of your homework writeups. Typically half the students receive a grade of B or higher.

Math 1012 Math 1120 Class 1 Dan Barbasch Math 1012 is a course that parallels Math 1120. www. lsc. sas. cornell. edu/ suppcourses/ Supplemental_ courses_ descriptions/ Math1011_ 1012. html Wednesday 2:55-4:10pm in Malott 251. Wednesday 7:30-8:45pm in Malott 251.

Review of Calculus 1 Math 1120 1 Derivative Class 1 2 Rules for Sums, Products, Quotients, Trigonometric Dan Barbasch Functions 3 Chain Rule 4 Linear Approximation 5 Powers, Logs 6 Monotonicity, Critical Points, Maximum/Minimum, 2nd Derivative Test 7 Inverse Functions 8 Riemann Sums, Antiderivative Check review Sessions of MSC, August 28 and Auguast 30 First new topic of the course: Fundamental theorem of Calculus

Definition of Derivative Math 1120 Class 1 Dan Barbasch f ( x + ∆ x ) − f ( x ) f ′ ( x ) = lim ∆ x ∆ x → 0 The interpretation of the derivative is the (instantaneous) rate of change of the quantity f with respect to the quantity x .

Chain Rule � d Math 1120 (3 x 2 + 1) 4 . � Example: Compute Class 1 dx Dan Barbasch For differentiating compositions of functions, many ways to write the formula: e.g. f = g ◦ h f ′ ( x ) = g ′ ( h ( x )) · h ′ ( x ) . or z = g ( y ) , y = h ( x ) dz dx = dz dy dy dx or z = g ( y ) , y = h ( x ) , z = f ( x ) f ′ ( x ) = g ′ ( y ) h ′ ( x )

Powers and Logs Math 1120 Class 1 y = e x is the unique function which satisfies Dan Barbasch y ′ = y and y (0) = 1 . This function is very important in applications. The equation y ′ = ky where k is a constant is interpreted as encoding the fact that the rate of change of y with respect to x is proportional to y ; radioactive decay, population growth, continuously compounded interest rate. The general solution is y = Ce kx .

Powers and Logs Math 1120 The inverse function of e x , y = ln x (defined for x > 0 only!) is Class 1 the unique function which satisfies Dan Barbasch y ′ ( x ) = 1 x and y (1) = 0 . It is also the unique function satisfying e ln x = x , ln ( e x ) = x , Arithmetic rules for logarithms and exponentials e a + b = e a · e b ( e a ) b = e ab ln( a b ) = b ln a ln( ab ) = ln a + ln b ln( a + b ) � = ln a + ln b .

Powers and Logs Math 1120 Calculus of logs and powers with other bases can be reduced to Class 1 base e using Dan Barbasch e ln b = b ln e b = b . dx (10 x ) =? d Example: 10 x dx =? � Example: d Example: dx (log 2 x ) =? All logarithms are proportional: log b x = ln x ln b ⇒ b y = x ⇐ ⇒ ln( b y ) = ln x ⇐ y = log b x ⇐ ⇒ y ln b = ln x .

Inverse functions Math 1120 Class 1 Inverse functions are about solving equations. Dan Barbasch If y = f ( x ) , express x as a function of y . x is a function of y only if the equation has a UNIQUE solution x for each y . We also write y = f − 1 ( x ) ⇐ ⇒ x = f ( y ) . Because we like the argument of a function to be called x , we write y = f ( x ) , has inverse function y = f − 1 ( x ) , satisfying f ( f − 1 ( x )) = x , f − 1 ( f ( x )) = x .

Inverse functions Math 1120 Class 1 Dan Barbasch Example: y = e x has inverse x = ln y ; we write the new function as y = ln x , interchanging x and y . Note that e x is increasing, has domain all real x , but has range y > 0 . So y = ln x is also increasing, but has domain x > 0 and range all y . Thus the inverse function ln x is only defined for x > 0 . The equations e ln x = x , ln e x = x encode the fact that these functions are inverse to each other.

Inverse functions Math 1120 Class 1 Dan Barbasch Example: y = sin x DOES NOT have an inverse because it is not 1-1. It is 1-1 and onto if x is restricted to the interval − π/ 2 ≤ x ≤ π/ 2. The inverse function is called y = Arcsin x ; it has domain − 1 ≤ x ≤ 1 and range − π/ 2 ≤ y ≤ π/ 2 . In words: y = Arcsin x is the unique value − π/ 2 ≤ y ≤ π/ 2 which satisfies sin y = x . In equation form: sin (Arcsin x ) = x , Arcsin (sin x ) = x ONLY IF − π/ 2 ≤ x ≤ π/ 2 .

Inverse functions Math 1120 Class 1 Dan Barbasch REFERENCE section 3.9. y = Arccos x , y = Arctan x , y = Arccot x , y = Arcsec x , y = Arccsc x are in section 3.9. The notation in the text is sin − 1 x rather than Arcsin x , but I think this leads to the possible confusion that sin − 1 x might be 1 / sin x .

Inverse functions Math 1120 1 Class 1 The derivative of y = f − 1 ( x ) is y ′ = f ′ ( f − 1 ( x )) also written as Dan Barbasch dy dx = 1 dx dy This follows from the chain rule: x = f ( f − 1 ( x )) ⇔ � ′ ⇔ 1 = f ′ ( f − 1 ( x )) · f − 1 ( x ) � f − 1 � ′ ( x ) = 1 � f ′ ( f − 1 ( x )) . Exercise: d dx (Arccsc x ) =?

Solution to Exercise The function y = Arccsc x is defined as: Math 1120 Class 1 Given x , Arccsc x is the unique value − π/ 2 ≤ y ≤ π/ 2 which Dan Barbasch satisfies csc y = x . Rewrite the equation as 1 / sin y = x , or sin y = 1 / x , and differentiate both sides (using the chain rule) 1 cos y · y ′ = − 1 / x 2 ⇔ y ′ = − x 2 cos y We need to express y ′ in terms of x alone. For this note that cos 2 y + sin 2 y = 1 , so solving for cos y we get � � 1 − sin 2 y = ± 1 − 1 / x 2 . cos y = ± However − π/ 2 ≤ y ≤ π/ 2 so cos y ≥ 0 . Since square roots are always positive quantities, the final answer is 1 y ′ = − x 2 � 1 − 1 / x 2

Solution to Exercise continued Math 1120 Class 1 Dan Barbasch You can use a right triangle with one angle equal to y , the hypotenuse of size 1 , and the opposite side equal to 1 / x to compute cos y . You still have the problem of deciding between ± .

Montonicity Math 1120 Class 1 1 st Derivative Test Dan Barbasch Suppose f ( x ) is continuous on [ a , b ] and f ′ ( x ) > 0 for x ∈ ( a , b ) . Then f ( x ) is (strictly) increasing on [ a , b ]. This means x < y ⇒ f ( x ) < f ( y ) . Suppose f ( x ) is continuous on [ a , b ] and f ′ ( x ) < 0 for x ∈ ( a , b ) . Then f ( x ) is (strictly) decreasing on [ a , b ]. This means x < y ⇒ f ( x ) > f ( y ) . f ′ ( x ) ≤ 0 would give f nonincreasing: x < y ⇒ f ( x ) ≥ f ( y ) .

Montonicity Math 1120 Class 1 Dan Barbasch 1 Exercise: Is 1 + x always bigger than 1 − x for x > 0? 1 e.g. 1 + . 25 = . 8 > 1 − . 25 = . 75 .

Solution to Exercise The question is the same as same as Math 1120 Class 1 1 Is f ( x ) := (1+ x ) − (1 − x ) ≥ 0 for x ≥ 0? Dan Barbasch We answer it by sketching the graph of f using the 1 st derivative test: 1 1 f ′ ( x ) = − (1 + x ) 2 − ( − 1) = 1 − (1 + x ) 2 . So we need to decide when this is ≥ 0 , and when it is ≤ 0 . This is why you need to be able to solve inequalities. f ′ ( x ) ≥ 0 for x ≥ 0 . Indeed, x ≥ 0 ⇒ 1 + x ≥ 1 ⇒ (1 + x ) 2 ≥ 1 2 ≥ 1 1 1 ⇔ 1 ≥ (1 + x ) 2 ⇔ 1 − (1 + x ) 2 ≥ 0 . Since f ′ ( x ) ≥ 0 , f ( x ) is increasing (nondecreasing) for x ≥ 0 . But f (0) = 0 , so f ( x ) ≥ f (0) = 0 for x ≥ 0 .

MaxMin and the 2nd Derivative Test Math 1120 Class 1 Dan Barbasch Exercise: Find and classify all local and global maxima and minima of f ( x ) = x 3 + ax for any value of a > 0? Exercise: Find and classify all local and global maxima and minima of f ( x ) = x 3 + ax for a any value of a < 0? Exercise: Find and classify all local and global maxima and minima of f ( x ) = x 3 . REFERENCE: Sections 4.1 - 4.4 cover this material.