Hyperbolic Sine

Hyperbolic Sine In this problem we study the hyperbolic sine function: ex − e−x 2 reviewing techniques from several parts of the course. sinh x = a) Sketch the graph of y = sinh x by finding its critical points, points of inflec­ tion, symmetries, and limits as x → ∞ and −∞. b) Give a suitable definition for sinh−1 x (the inverse hyperbolic sine) and sketch its graph, indicating the domain of definition. c) Find d sinh−1 x. dx � d) Use your work to evaluate √ dx . + x2 a2 Solution This problem provides a nice review of a variety of material covered in the course and also explains why electronic integration tools frequently give answers that involve the hyperbolic sine function. a) Sketch the graph of y = sinh x by finding its critical points, points of inflec­ tion, symmetries, and limits as x → ∞ and −∞. We need to test for symmetry, take derivatives, and take limits. We can perform these tasks in any order; here they’re presented in the order they are listed in the problem. To find the critical points of sinh x we take its first derivative and set it equal to zero. d d 1 x sinh x = (e − e−x ) dx dx 2 1 x = (e + e−x ) 2 = cosh x Because ex is always positive, the derivative of sinh x is never 0 and so sinh x has no critical points. To find points of inflection we set the second derivative equal to 0. d2 sinh x dx2 d 1 x (e + e−x ) dx 2 1 x = (e − e−x ) 2 = sinh x = 1 The second derivative of sinh x equals 0 when ex = e−x , or when x = 0. To confirm that this is an inflection point, we test the value of the second derivative of sinh x to the left and right of this point: 1 1 (e − e−1 ) = 2 1 −1 (e − e1 ) = 2 y �� (1) = sinh(1) = y � (−1) = sinh(−1) = 1 1 (e − ) > 1 2 e 1 1 ( − e) < 1 2 e We conclude that the graph of y = sinh x has no critical points and an inflection point at x = 0. To find the symmetries of a function, we find the output of the function when −x is input. sinh(−x) 1 −x (e − ex ) 2 1 = − (ex − e−x ) 2 = − sinh x = If f (−x) = −f (x) we say that f (x) is an odd function; the graph of y = sinh x is symmetric about the origin. Recall that sinh x = 12 (ex − e−x ). As x approaches positive infinity, the value of ex approaches positive infinity and the value of e−x approaches zero. lim sinh x = ∞ x→∞ Similarly, as x approaches −∞ the value of e−x dominates and ex vanishes. lim sinh x = −∞ x→−∞ We now have a good deal of information about the graph, including the values of sinh(1) and sinh(−1). In addition, we can easily determine that sinh(0) = 0. Our graph will look something like: 2 b) Give a suitable definition for sinh−1 x (the inverse hyperbolic sine) and sketch its graph, indicating the domain of definition. While it’s possible to find an algebraic expression for the inverse of sinh x, it’s not necessary for this problem. We can tell from the graph of sinh x that the function is invertible, so we can apply the definition of the inverse of a function: y = sinh−1 x is the unique value of y for which x = sinh y. The graph of y = sinh1 x is the reflection of the graph of y = sinh x across the line y = x. 3 The domain of sinh−1 x is all real numbers. d sinh−1 x. dx To answer this we must review material from the first unit of the semester! dy of We learned how to use implicit differentiation to find the derivative dx the inverse of a function. c) Find y sinh y d sinh y dx dy (cosh y) dx dy dx = sinh−1 x = x d x = dx = 1 = 1 cosh y At this point we wish to rewrite cosh y in terms of x. When dealing with trigonometric functions we could use a right triangle and the Pythagorean theorem for this task; here we must rely on hyperbolic trig identities. (If you need these identities on the final exam, they will be given to you.) Using the following identity, we can transform an expression in terms of cosh y to one in terms of sinh y = x. cosh2 t − sinh2 t = 4 1 cosh2 t = cosh t = 1 + sinh2 t � 1 + sinh2 t Thus: dy dx dy dx = 1 cosh y = � = 1 1 + sinh2 y 1 √ . 1 + x2 We can get an idea of whether our work is correct by noting that this equation should describe the slope of the tangent line to the graph of sinh−1 x. � d) Use your work to evaluate √ dx . a2 + x2 We can reinterpret our results from the previous problem to say that: � dx √ = sinh−1 x + c. 1 + x2 This gives us an alternative to the trig substitution x = a tan θ when dealing dx with integrands of the form √ . 2 a + x2 � √ dx a2 + x2 � = a du � a2 + (au)2 � du √ = 1 + u2 = sinh−1 u + c �x� = sinh−1 +c a (Substitute x = au) Note that this is much simpler than the calculation involving the substitution x = a tan θ. 5