×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

## Math Lecture Notes 1/19/16 and 2/2/16

by: Grant Pitarys

83

4

10

# Math Lecture Notes 1/19/16 and 2/2/16 MATH 114

Marketplace > George Mason University > Mathematics (M) > MATH 114 > Math Lecture Notes 1 19 16 and 2 2 16
Grant Pitarys
Mason
GPA 3.94

Enter your email below and we will instantly email you these Notes for Analytic Geometry & Calculus 2

(Limited time offer)

Unlock FREE Class Notes

### Enter your email below to receive Analytic Geometry & Calculus 2 notes

Everyone needs better class notes. Enter your email and we will send you notes for this class for free.

Includes neatly organized layout of all the notes gone over in class as well as step by step solutions for in class example problems
COURSE
Analytic Geometry & Calculus 2
PROF.
Matthew Holzer
TYPE
Class Notes
PAGES
10
WORDS
CONCEPTS
Math, Calculus
KARMA
Free

## Popular in Mathematics (M)

This 10 page Class Notes was uploaded by Grant Pitarys on Thursday January 28, 2016. The Class Notes belongs to MATH 114 at George Mason University taught by Matthew Holzer in Winter 2016. Since its upload, it has received 83 views. For similar materials see Analytic Geometry & Calculus 2 in Mathematics (M) at George Mason University.

×

## Reviews for Math Lecture Notes 1/19/16 and 2/2/16

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 01/28/16
Math notes for CALC 114 (Holzer) 1/19/16 Chapter 6: Applications of Integration Review: What is a derivative? Mathematical de▯nition for f(x): 0 f(a)▯f(0) f (0) = lim a!0 a The derivative is the instantaneous rate of change for a point in a function. Review: What is an integral? #1: Net area under a curve If there’s a function from point starting at point a and ending at point b and part A is above the x-axis, part B is below the x-axis, and part C is above the x-axis, then the net area under that curve is: Rb a f(x)dx A ▯ B + C #2: Anti Derivative 0 f(x) ! f (x) To get to f (x) from f(x) one takes the derivative. Vice versa for the anti derivative. R 0 f(x) = f (x)dx 0 f(x) is the anti derivative of f (x) G(x) = R g(x)dx G(x) is the anti derivative of g(x) The anti derivative can be seen as the inverse of the derivative with the exception for inde▯nite integrals because of the constant. R 0 f (x)dx = f(x) + C #3: Riemann sums As the number of partitions of a function increases the area under a curve will become closer to the integral of that curve 1 R b partition size! 1 = a f(x)dx FTOC: Fundamental Theorem of Calculus: #1. Z b f(x)dx = F(b) ▯ F(a) a The anti derivative/area from b to a is the anti derivative for f(x) at b minus the anti derivative for f(x) at a #2. Z d x = f(t)dt = f(x) dx a Example: Z 1 2 ▯t [t ▯ 4sin(t) + e ]dt 0 t3 ▯t F(t) = + 4cos(t) ▯ e 3 Taking the anti derivative c3n be done by breaking the body up into three parts t ’s anti power rule is t , the anti derivative of sin(x) = ▯cos(x) and ▯t ▯t 3 e = ▯e Z 1 [t ▯ 4sin(t) + e▯t]dt = F(1) ▯ F(0)dt 0 = 1 + 4cos(1) ▯ e▯1 ▯ 4 ▯ 1 3 Using the fundamental theorem of calculus we can solve the de▯nite integral. Review: u-substitution u-substitution is used as the anti chain rule for integrals. chain rule: 0 0 f(g(x)) = f (g(x)) ▯ g (x) 2 Example: Z p ( x + 1)4 p dx x we set u equal to an operation or multiple opepations of x to make the anti derivative easier to solve. In this example we use u = The goal of u-substitution is to make sure the anti derivative is only in terms of u after substitution. p We take the derivative of whatever u equals. fox + 1 we get 1 du = p dx 2 x p Get dx by itself for 2 = dx and plug that back into the integral: Z p 2(u) x p du x p x cancels and we are left with: Z 2(u) du Now we take the anti derivative to get: 2 5 u + C 5 Finally we plug what u equals back into the equation. 2 p ( x + 1) + C 5 Section 6.1: Velocity and Net Change De▯ne: s(t) = Position of an object at time t v(t) = Velocity of an object at time t The displacement between t = a and t = b is: 3 Z b s(b) ▯ s(a) = v(t)dt a The distance traveled between t = a and t = b is: Z b s(b) ▯ s(a) = jv(t)jdt a Example: for 0 ▯ t ▯ 3 v(t) = t ▯ 4t + 3 Displacement Z 3 = t ▯ 4t + 3dt 0 t 2 3 = [ ▯ 2t + 3t] 0 3 = 27 ▯ 18 + 9 ▯ 0 3 = 0 Distance traveled Z 3 = jt ▯ 4t + 3jdt 0 In order to ▯nd distance traveled we need to break this integral into parts where y is positive for v(t) and another part y is negative. In this equation y is positive from 0 to 1 and negative from 1 to 3. We take the integrals for when y is positive and subtract the integrals for when y is negative from then like so: Distance traveled Z Z 1 3 = t ▯ 4t + 3dt ▯ t ▯ 4t + 3dt 0 1 t t3 = [ ▯ 2t + 3t] ▯ [ ▯ 2t + 3t] 3 3 0 3 1 1 1 8 = ▯ 2 + 3 ▯ 0 + ( ▯ 2 + 3) = 3 3 3 4 Math notes for CALC 114 (Holzer) 2/2/16 Section 6.1 continued Previously: Position/Velocity/Acceleration Example: Car starting from rest, ▯nd time necessary to reach 60 mph given: a(t) = 30 f=s 2 1 + t Find velocity by taking the anti derivative of acceleration. Z t v(t) = a(s)ds + v(0) 0 v(0) = 0 so: Z t 30 v(t) = ds 0 1 + s We can factor out the 30 to outside the integral and Z t 1 30 ds = 30ln(1 + s) 0 1 + s = [30ln(1 + s)]0 = 30ln(1 + t) ▯ 30ln(1) 30ln(1) = 0 so: v(t) = 30ln(1 + t) The question asks to ▯nd the time to reach 60 mph, but our units are in ft=s , rather than change it at the start it’s easier to wait until we have the velocity function solved ▯rst and then convert. 60m 5280ft 1h 5280ft ▯ ▯ = = 88ft=s 1h 1m 3600s 60s 88 = 30ln(1 + t) Divide by 30 88 = ln(1 + t) 30 Raise both sides by e to remove the natural log 88 e30 = 1 + t Subtract 1 88 e30 ▯ 1 = t 1 17:8 ▯ t Example: -At t = 0 snow is on the ground and falling at a constant rate -A plow is plowing with a velocity inversely proportional to the depth of snow -The plow travels twice as fast in the ▯rst hour than the second Q:When did it start snowing To begin this problem we start by de▯ning simply functions: s(t) = Position of plow r = Rate of snow fall v(t) = Velocity of plow d(t) = Depth of snow We know that d(t) will equal the rate of snow fall times time plus a constant so: d(t) = C + rt We also know that v(t) equals a constant value divided by d(t) so: v(t) = k = k d(t) C+rt Using this we can take the integral of the position of plow to ▯nd how much snow is on the ground Z t s(t) = s(0) + v(▯)ds 0 plug in the velocity, note we’re using ▯au instead of t Z t k = ds 0 C + r▯ like the 30 from the previous example we can factor k to outside the integral. Z t 1 = k ds 0 C + r▯ Because of the tau in the denominator we divide k by r. k t = [ln(C + r▯)]0 r k = r [ln(C + rt) ▯ ln(C)] Distance traveled in the ▯rst hour: 2 k k S(1) ▯ S(0) = ln(C + r) ▯ ln(C) r r Total distance traveled over two hours: k k s(2) ▯ s(0) = ln(C + 2r) ▯ ln(C + r) r r Because the distance traveled in the ▯rst hour was twice as much as the second hour: 2(s(2) ▯ s(1) = s(1) ▯ s(0) plugging in for all the variables we get: 2k 2k k k ln(C + 2r) ▯ ln(C + r) = ln(C + r) ▯ ln(C) r r r r All the k cancel out we move all like terms to one side and we’re left with: r 2ln(C + 2r) ▯ 3ln(C + r) + ln(C) = 0 Using logarithmic rules we can manipulate that to become: 2 3 ln(C + 2r) ▯ ln(C + r) + ln(C) = 0 Using another logarithmic identity we can manipulate that to become: C ▯ (C + 2r) 2 ln( 3 ) = 0 (C + r) Because ln(x) = 0 is the same as x = 1 we can manipulate that further to become: C ▯ (C + 2r) 2 (C + r)3 3 Multiply bother sides by (C + 2r) 2 3 C ▯ (C + 2r) = (C + r) Expand the quantities C(C + 4Cr + 4r ) = (C + 3C r + 3Cr + r ) 2 3 3 Multiple the "C" on the left side in the quantity 3 2 2 3 2 2 3 C + 4C r + 4Cr = C + 3C r + 3Cr + r Combine like terms on one side 0 = r ▯ Cr ▯ C 2 factor out an "r" = r(r ▯ Cr ▯ C ) Use the quadratic formula to get C 1p r = ▯ 5C 2 2 2 rate has to be p 1 5 = C( 2+ 2 ) Now we can solve for d(t) = 0, recall that d(t) = c + rt p 1 5 0 = C + C( + )t 2 2 divide both sides by "C" p 1 5 0 = 1 + ( + )t 2 2 Use algebra to get "t" on one side and our answer for time is ▯1 t = 1 p5 2+ 2 Marginal Cost Q(x) - Cost to produce x units Q (x) - Cost ot produce 1 extra unit given that you have already produced x units Q (x) = Q(x+h)▯Q(f) h 4 Example: 2 Q(x) = 1000 + x ▯ x 0 1 x 2 100 Q (x) = 2 ▯ 50 Suppose Q (x) ▯ 0, decreasing True or False: The cost to increase the production from 2A units is greater than the cost to Increase from 2A to 3A units. True R 2A 0 Q(2A) ▯ Q(A) = A Q (x)dx R 3A 0 Q(3A) ▯ Q(2A) = 2A Q (x)dx Loans: Exponential Growth/Decay Q (t) 0 #1. Q(t) = r or Q (t) = rQ(t) Example: Loan r - Interest Rate Q - Principle Integrate #1. Z 0 Z Q (t) dt = rdt Q(t) use u-substitution for u = Q(t) Z 0 Q (t) 0 = rt + C uQ (t) The Q (t) cancel leaving just 1 u ln(Q(t)) = rt + C Raise both sides by e to get: Q(t) = e rt+C = e rt▯ C Example: Co▯ee that is currently 120 degrees is sitting in a room that is 70 degrees and it’s temperature is changing at the rate dT ▯ 5 = ▯10e dt 5 At what time is T = 80? This written out as an equation is: Z t s T(t) ▯ T(0) = ▯ 10e ▯5ds 0 ▯rst T(0) = 120 and you can take the 10 and factor it outside the integral so Z t ▯ s T(t) = 120 ▯ 10 e 5ds 0 s ds You can use u-substitution so that u = ▯ 5and du = ▯ 5 so that Z t ▯ 5 eu T(t) = 120 ▯ 10 du 0 ▯5 Don’t forget to change the bounds of the integral in terms of u so t is changed t to ▯ 5 Factor out the ▯5 like we did with 10 to get Z t ▯ 5 T(t) = 120 + 50 e du 0 Now we can solve for what the integral might equal ▯ t 0 ▯ 5 T(t) = 120 + 50[e 5 ▯ e ]0 Remember e = 1 ▯t 80 = 120 + 50e 5 ▯ 50 ▯ t 10 = 50e 5 1 = e ▯5 5 Take the natural log of both sides 1 t ln( ) = ▯ 5 5 1 t = ▯5ln( ) 5 6

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Amaris Trozzo George Washington University

#### "I made \$350 in just two days after posting my first study guide."

Steve Martinelli UC Los Angeles

Forbes

#### "Their 'Elite Notetakers' are making over \$1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com