Calculus B Workshop #4 (March 23, 2012)

Transcription

1 Calculus B Workshop #4 (March 3, 0) Name: Circle one: -4pm 4-6pm Rules: () A minimum of TWO and a maimum of FOUR students in each group. () Work together and do the problems in order! (3) Your answers must be justified, eact, and reasonably simplified. (4) Do all of your work on the answer page. Do NOT attach etra pages. you will need your calculator. etra credit (but all other previous problems must be completed first).. This problem is a continuation of problem in the last workshop. You will derive the other formula for the derivative of sec. 3 Consider the function f () = sec with domain 0,,. (a) Draw the graph of f. (b) Why is f invertible? (c) Draw a graph of f ( ) sec. d (d) Show that, with this definition, sec d.. If one side of a rectangle,, is increasing at a rate of 3 inches per minute while the other side, y, is decreasing at a rate of 3 inches per minute, which of the following must be true about the area A of the rectangle? (Be sure to justify your answer.) (A) A is always increasing. (B) A is always decreasing. (C) A is decreasing only when < y. (D) A is increasing only when < y. (E) A is constant. 3. Let f ( ) e e and g( ) e e. (a) Determine the following : (i) f and (ii) g. Write your answers in terms of f and g. (b) The Gateway Arch in St. Louis was designed by Eero Saarinen and was constructed using the equation y g( ) for the central curve of the arch, where and y are measured in meters. (For each of the following questions, write your answers correct to two decimal places.) Assuming the correctness of the previous formula, answer each of the following questions. (i) How wide is the arch at the base? (ii) What is the height of the arch at its center? (c) The function f is one-to-one on its domain (,) (you do not need to prove this). Find the inverse of f. [Instructor s Note: Because f is defined by eponential functions, then the inverse will involve logarithms (and a square root if everything is done correctly).]

2 Calculus B Solutions to Workshop #4. (a) (b) (c) Because it passes the Horizontal Line Test. (d) 3 Note that in the interval 0,,, tan y is always positive. y sec IFP sec y = (Differentiate both sides) d d sec y d d (Use implicit differentiation) sec y tan y y = (Solve for y) y sec ytan y (tan y is always positive) tan y (Use the identity z z ) (Use the identity tan sec ) tan y y sec Thus, d sec d.. Start with the area equation given by A = y. Differentiating both sides with respect to time t, then A y (differentiate w.r.t. time) d d A y (use product rule)

3 dy d y (substitute d/ = 3 and dy/ = 3) 3 3y (simplify) 3( y ) (Equation ) This implies the following: A is increasing when 0. By Equation, y > 0 y >. A is constant when 0. By Equation, y = 0 y =. A is decreasing when 0. By Equation, y < 0 y <. Thus, the answer is (D). 3. These functions, f and g, are two of the hyperbolic functions. More specifically, f is called the hyperbolic sine, denoted sinh, and g is called the hyperbolic cosine, denoted, cosh (there are four other functions, all defined in terms of sinh and cosh in a similar manner as the trigonometric formulas). (a) f ( ) e e g( ) (i) (ii) g( ) e e f( ) (b) More information about the St. Louis Arch and the mathematics behind it can be found at the Wikipedia page: y g( ) for the central curve of the arch, where and y are measured in meters. (For each of the following questions, write your answers correct to two decimal places.) Assuming the correctness of the previous formula, answer each of the following questions. (i) In the provided equation set y = 0 and solve for. y g( ), Solution #: The easiest way to answer is just to program it into your calculator. Solution #: Solving by hand is much more difficult. The steps are very similar to those given in part (c). However, I m not going to provide all of the steps because I wasn t epecting any of the groups to do this. To avoid having to carry through the provided constants in the equations, we will generalize the problem a bit. The solution to the equation 0 a bg( c) where a, b, and c are positive constants with a > b is a a b ln. c b Using the specific values associated to this problem a =.49, b = 0.96, and c = gives the solutions = 9.0 and = 9.0. The difference of these two solutions gives the wih of the base: 8.4 m.

4 (ii) The highest point occurs when = 0 y g(0) (c) Let y = f (), then e e m y y y e e e e To solve for y, first make the substitution z = e y. Note that z > 0 from properties of the eponential function. Thus, solving for z, using the quadratic equation, we have y y y e e e y e z z z z z z0 4 4 z z Since for any real, then we must take the positive root of z in order for z to be positive. Rewriting in terms of y and solving gives Thus, f ( ) ln. y z e ln( e ) ln y ln y

5 Number of groups: 8 Calculus B Workshop #4 March 3, 0 The ranks of the top groups were provided. P # A(%) M(%) O a b c d a b Total A(%) M(%) O