Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Homework 6: Derivatives of logarithmic, inverse and trigonometric functions 1 Calculus problems [75 points] 1.1 Derivatives of logarithms and Logarithmic derivatives • f (x) = ln(x2 + 1) 1 • f (x) = ln √2x+3 • f (x) = ln √xx3 +1 • f (x) = log10 (sin(x)) √ • f (x) = log10 ( 1 + x) 1 • f (x) = ln cos(x) • f (x) = • f (x) = 2x x2 (x6 −3)(x2 +1) (x+1)(ex −1) √ 3x x2 +1 3 2 (1+2x ) (10x+3x2 +5)1/3 • f (x) = (2x)1/x • f (x) = xx 1.2 x Derivatives of inverse functions Using the formula for the derivative of the inverse of a function, find the derivatives of : • f (x) = sinh−1 (x) (Note: sinh−1 is the inverse of the sinh function, see Wolfram alpha for detail.). SIMPLIFY your answer using the fact that cosh2 (x) − sinh2 (x) = 1. • f (x) = cosh−1 (x) (Note: cosh−1 is the inverse of the cosh function, see Wolfram alpha for detail.). SIMPLIFY your answer using the fact that cosh2 (x) − sinh2 (x) = 1. • f (x) = tanh−1 (x) (Note: tanh−1 (x) is the inverse of the tanh function, see Wolfram alpha for detail.). SIMPLIFY your answer using a very similar method to the one used in class for tan−1 (x) 1.3 Trigonometric equations Find all the solutions of the following equations: • sin(2x) = 1 2 • cos2 (x) = 3/4 • cos(x) = sin(x) • cos(2x) = − sin(2x) 1 1.4 Trigonometric functions and tangent lines • What is the equation of the tangent line to the function y = tan(x) at the point x = 0? • Using the formula for the derivative of inverse functions, calculate the derivative of tan−1 (x), the inverse function of tan(x). • Simplify your expression for [tan−1 (x)]0 using the trigonometric formula 1 + tan2 (a) = sec2 (a) = 1 cos2 (a) • On some graphing paper, sketch the function tan(x) in the interval (−π/2, π/2), as well as its tangent at x = 0. Indicate clearly the vertical asymptotes and their position (i.e. what is the value of x on the asymptotes). • By symmetry across the relevant line, draw on the same diagram the function tan−1 (x). Indicate clearly the horizontal asymptotes and their position (i.e. what is the value of y on the asymptotes). • What is the slope of the tangent to the curve y = tan−1 (x) at the point x = 0? 2 Applied Problem [25 points] This problem starts with a very simple question: is there a mathematical function we can use to model the shapes of nature, such as spirals, flowers, plant leaves, etc? The basic difficulty is that most of these shapes don’t pass the vertical line test, so that they can’t be represented as the graph of a function y = f (x). However, they can be represented as a polar graph, of the kind r = f (θ). In a polar graph r = f (θ), each point on the graph is drawn in the following way: • Draw the half-line passing through the origin and at an angle θ from the y = 0 line. • On that half-line, put the point at a distance r from the origin. √ This is illustrated (see next page), for the function r = θ. Other shapes can be drawn using this method, as for instance the 3-leaf clover. A possible equation for it is: r = 1 + cos(3θ) (1) Question 1: Using standard graphing paper, draw the shape corresponding to the polar graph r = 1 + cos(3θ). You can check your answer on Wolfram using the command ” polar plot r = 1 + cos(3theta)”. Try to plot it for yourselves first, however, before checking the answer. Question 2: The leaves of the clover are oriented at angles that look like 2π/3 and 4π/3. Show this using Calculus, by considering the fact that the leaves are also points where r is maximum. Question 3: What function f (θ) would you choose so that the graph r = f (θ) describes a 4-leaf clover?. Check your answer on Wolfram, and draw it on another graphing paper. Question 4: Using the same method as above (i.e. using Calculus, rather than just looking at the plot) , find the position of the leaves. 2 1.5 θ= 3π ,r = 4 3π = 2.35 4 θ= π π ,r = = 1.25 2 2 1 € € θ= 0.5 θ = 2π ,r = 2π = 2.5 θ = π ,r = π = 1.77 0 π π ,r = = 0.88 4 4 € € € 0.5 1 θ= 1.5 5π 5π ,r = = 1.98 4 4 θ= 2 3π θ= ,r = 2 € 3π = 2.17 2 7π ,r = 4 7π = 2.34 4 2 2.5 € 2.5 2 1.5 1 0.5 € 0 0.5 3 1 1.5 3