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

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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.

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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

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