Chapter 1 Review of Straight Lines We start with a brief review of properties of straight lines‚ since these properties are fundamentally important to our understanding of more advanced concepts (tangents‚ slopes‚ derivatives). Skills described in this introductory material will be required in many contexts. 1.1 Geometric ideas: lines‚ slopes‚ equations Straight lines have some important geometric properties‚ namely: The slope of a straight line is the same everywhere along its length.
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Collapse of Barings Bank Barings was the oldest investment bank in Britain‚ listing among its clients the Queen herself. Indeed‚ the bank’s pedigree was so distinguished that it did not have a logo‚ it had a crest. The firm traces its origins to John Baring of Bremen‚ who settled at Exeter in 1717 and set up in business as a merchant and manufacturer. He became one of the leading businessmen in the West Country. In 1762‚ his three sons established the London merchant house of John & Francis
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change between two variables‚ for example‚ x and y. This mathematical method can be used to reverse derivative back to its original form. For some one that is familiar with derivative‚ we know that d/dx (x2) = 2x or in mathematical notation we can write it as f ’(x2) = 2x. This is calculated simply by using the derivative formula nxn-1 where x2 will be 2* x2-1 = 2x. Now to reverse this derivative we have to use law of integral (power rule) that states for f(x)‚ x = xn+1n+1 (normally written
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algorithm using a hyperbolic tangent function for zero-memory nonlinear function has been proposed and applied to avoid this problem by Filho et al. But application of this algorithm includes the calculation of hyperbolic tangent function and its derivative or a look-up table which may need a large amount of memory due to channel variations. To reduce the computational and/or hardware complexity of Filho’s algorithm‚ in this paper‚ an improved method for the decision-directed algorithm is proposed
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End-of-Chapter Question Solutions 1 ____________________________________________________________ ________________________________ CHAPTER 5: FOREIGN CURRENCY DERIVATIVES 1. Options versus Futures. Explain the difference between foreign currency options and futures and when either might be most appropriately used. An option is a contract giving the buyer the right but not the obligation to buy or sell a given amount of foreign exchange at a fixed price for a specified time period. A future
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1. ht= -4.9t2+ 450‚ where t is the time elapsed in seconds and h is the height in metres. a) Table of Values t(s) | h(t) (m) | 0 | ht= -4.9(0)2+ 450= 450 | 1 | ht= -4.9(1)2+ 450= 445.1 | 2 | ht= -4.9(2)2+ 450= 430.4 | 3 | ht= -4.9(3)2+ 450= 405.9 | 4 | ht= -4.9(4)2+ 450=371.6 | 5 | ht= -4.9(5)2+ 450=327.5 | 6 | ht= -4.9(6)2+ 450= 273.6 | 7 | ht= -4.9(7)2+ 450= 209.9 | 8 | ht= -4.9(8)2+ 450= 136.4 | 9 | ht= -4.9(9)2+ 450=53.1 | 10 | ht= -4.9(10)2+ 450= -40 |
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Suppose the derivative of ( ) ( ) ) with ( ) exists. Assume that ( ) ( ) ( ) Find ( ) Let ( ) ( ) ( ) at a. Find an equation of the tangent line to ( ) at b. Find an equation of the tangent line to 3. Suppose tangent to at is and tangent to line tangent to the following curves at ( ) ( ) a. b. ( ) ( ) at is Find the Chain Rule using a table )) and ( ) ( 4. Let ( ) ( ( )) ‚ ( ) ( ( )) ‚ ( ) ( ( ( ) ) . Using the table to compute the following derivatives. a. ( ) b. (
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that uses differential calculus and integral calculus‚ linear algebra and multi linear algebra to study geometry problems. Partial differential equation is a differential equation that contains unknown multivariable functions and their partial derivatives. These are used to formulate problems involving functions of several variables. Mr. Nash Jr. used all of these skills and is known for developing the Nash embedding theorem. The Nash embedding theorem stated that every Riemannian manifold
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1. 7.5/8 The height in metres of a ball dropped from the top of the CN Tower is given by h(t)= -4.9t2+450‚ where t is time elapsed in seconds. (a) Draw the graph of h with respect to time (b) Find the average velocity for the first 2 seconds after the ball was dropped h(0)=(0‚450)‚ h(2)=(2‚430.4) = (430.4-450)/(2-0) = -9.8m/s √ (c) Find the average velocity for the following time intervals (1) 1 ≤ t ≤ 4 h(1)=(1‚445.1) h(4)=(4‚371.6) = (371.6-445.1)/(4-1) = -24.5m/s √ (2) 1 ≤ t ≤ 2 h(1)=(1‚445
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Find 1st derivative; > D(f); (3) > # Find turning/stationary point in the interval [1‚2]‚ 1st derivative expression = 0‚ 10 significant figures!; > evalf(fsolve(3*cos((1/4)*x^4)*x^3-3*sin((3/4)*x)^3*cos((3/4)*x)= 0‚x=1..2)‚10); 1.562756908 (4) > # Find 2nd derivative at x= 1.562756908; 10 significant figures!; > evalf(D[1$2](f)(1.562756908)‚10); (5) > # Remember to unassign variables/restart; > restart; > # Question 3; (same process as above) > f:=x-> 2^x + 2*cos(x); (6) > # Find 1st derivative; > D(f);
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