Functions+Notes+_updated_.pdf

For example the value of an investment bearing

Info icon This preview shows pages 30–35. Sign up to view the full content.

View Full Document Right Arrow Icon
proportional to their size. For example, the value of an investment bearing interest that is continuously compounding increases at a rate proportional to that value. The mass of undecayed radioactive material in a sample decreases at a rate proportional to that mass. All of these phenomena, and others exhibiting similar behavior, can be modelled mathematically in the same way. It Q = Q ( t ) denotes the value of a quantity Q at time t , and if Q changes at a rate proportional to its size, then dQ dt = kQ, where k is the constant of proportionality. The equation above is called the differential equation of exponential growth or decay because, for any value of the constant C , the function Q ( t ) = Ce kt satisfies the equation. Moreover, the initial-value problem dQ dt = kQ, Q (0) = Q 0 has unique solution Q ( t ) = Q 0 e kt . We say that the quantity Q exhibits exponential growth if k > 0 and exponential decay if k < 0. Doubling Time. Half-life Exponential growth is characterized by a fixed doubling time . If t d is the time at which Q has double from its size at t = 0, then 2 Q (0) = Q ( t d ) = Q 0 e kt d = Q (0) e kt d , since Q (0) = Q 0 . But then e kt d = 2, which implies that kt d = ln 2, or t d = ln 2 /k . Similarly, exponential decay involves a fixed halving time, usually called the half-life .
Image of page 30

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2. Functions 29 Application. How long, to the nearest year, will it take an invest- ment to double its value if invested at 6% annually compounded continuously? Solution: Let us assume that the amount invested is P . Then recall that the future value A ( t ) of P invested at 6% p. a. compounded continuously is given by the following exponential growth function: A ( t ) = Pe 0 . 06 t We then have to find T such that A ( T ) = 2 A (0) = 2 P . Since A ( T ) = Pe 0 . 06 T we obtain the equation Pe 0 . 06 T = 2 P e 0 . 06 T = 2 ln e 0 . 06 T = ln 2 0 . 06 T = ln 2 Thus T = ln 2 / 0 . 06 11 . 55 An alternative way to solve this problem is to simply apply the formula for doubling time. Notice that in this case, k = 0 . 06 and Q 0 = A (0) = P . The doubling time is t d = ln 2 / 0 . 06 11 . 55 Therefore, we can conclude that it will take 12 years in order to double the value of the given investment.
Image of page 31
30 J. S´ anchez-Ortega 2.9 New Functions from Old Functions In this section, we will learn how to obtain new functions by shifting, stretch- ing, and reflecting their graphs. As we will see, it will allow us to sketch the graphs of many functions quickly by hand. Translations Let c be a positive number . We can translate the graph of a function f horizontally and vertically as indicated in the table below.
Image of page 32

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2. Functions 31 Example 2.19. The graph of y = x 2 + 1 is the parabola y = x 2 shifted upward 1 unit. The graph of y = x 2 - 3 is the parabola y = x 2 shifted downward 3 units. 0 1 -1 x y y = x 2 y = x 2 + 1 y = x 2 - 1 Example 2.20. The graph of y = ( x - 3) 2 is the parabola y = x 2 shifted 3 units to the right. The graph of y = ( x + 1) 2 is the parabola y = x 2 shifted 1 unit to the left. -1 3 0 x y y = x 2 y = ( x + 1) 2 y = ( x - 3) 2
Image of page 33
32 J. S´ anchez-Ortega Example 2.21. Sketch the graph of y = x 2 - 4 x +3 shifting the graph of y = x 2 .
Image of page 34

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 35
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern