6.7 Hyperbolic Functions

Transcription

1 Hyperbolic Functions Even and Odd Parts of an Exponential Function We recall that a function f is called even if f( x) = f(x). f is called odd if f( x) = f(x). The sine function is odd while the squaring function (x ) is even. Please review the graphs of each. Also review the graphs of odd and even functions in general. Notice that every function defined on an interval centered at the origin can be decomposed into the sum of an odd and an even function. () f(x) = f(x)+f( x) } {{ } even part + f(x) f( x) }{{} odd part Now write the exponential function in this way. e x = ex +e x + ex e x }{{ }}{{ } even function odd function The even part of e x is call the hyperbolic cosine of x and the odd part is called hyperbolic sine of x. These functions are interesting on their own and will be the subject of this section.

2 6.7 Definition. Hyperbolic Functions () (3) sinhx = ex e x coshx = ex +e x (Hyperbolic sine) (Hyperbolic cosine) () (5) tanhx = sinhx coshx = ex e x e x +e x sechx = coshx = e x +e x (Hyperbolic tangent) (Hyperbolic secant) We sketch the graph of each of the four functions below (more details appear in the text). 3 y = coshx It turns out that a hanging cable lies along the graph of y = coshαx. Here α R is a constant α that depends on the weight of the cable and the tension at its lowest point. Also, see Figure on page 63 of the text.

3 y = sinhx 3 y = tanhx It is worth noting that, for example, lim tanhx = lim x x e x e x = lim e x +e x x e x 0 = +e x +0 = as suggested by the sketch above. y = sechx Similarly, one can show lim sechx = 0 x

4 6.7 It should come as no surprise that these functions satisfy some interesting identities. For example, using () and (3) we obtain ( ) e cosh t sinh t +e t e t = ( t e t ) = ( e t ++e t e t + e t) = () = In otherwords, (6) cosh t sinh t = And one can also show that tanh t+sech t = and cosh x = coshx+ A complete list of useful hyperbolic identities can be found in the text. Why hyperbolic? Recall that the graph of the equation y = /x is called a hyperbola. More generally, we know that the equations (x/a) (y/b) = ± generate a family of hyperbolas that are symmetric to either the x or y-axis. Now let (7) x = cosht and y = sinht Then (6) implies (8) x y = which we recognize as the equation of the familiar hyperbola shown. In this case, we call (7) a parametrization of the hyperbola given by (8) (cosht,sinht) For more on this, see example 6.7. and exercise See also the next example and the remarks that follow.

5 6.7 5 Example. Do you recognize the following equation? (9) x +xy y = We try completing the square. = x +xy y = x (y xy +x )+x (0) = = x (x y) Of course, we could now solve this for y as a function of x to obtain () y = x± x and graph both equations on a graphing utility. Instead, we try a different approach. Rearranging (0) we obtain () x (x y) = Now let x = cosht and x y = sinht. Then x (x y) = ( cosht ) (sinht) = cosh t sinh t = We have discovered the parametric equations for the curve given by the cartesian equation in (9). (3) x = x(t) = cosht and y = y(t) = cosht sinht We will have more to say about parametric equations in chapter 0.

6 6.7 6 Using the parametric plotting feature of any modern modern graphing calculator (or any one of a number of computer algebra systems), we obtain the graph shown below. To convince yourself that (9) and (3) are somehow equivalent, we suggest that you try plotting a few points with each. For example, from (9) it is clear that the hyperbola passes through the point (,0). On the other hand, it is not difficult to show that if t 0 = ln ( + ), then x(t 0 ) = cosht 0 = and y(t 0 ) = cosht 0 sinht 0 = 0 Now try it on your own by setting y = and solving for x in (9), etc. Continuing with the last example, notice that the graph appears to have tangent lines everywhere. Can we locate the vertical tangent lines shown in the sketch below? According to the Implicit Function theorem, we may differentiate both sides of (9) with respect to x to obtain Rearranging yields x+(y +xy ) yy = 0 dy dx = x+y y x y = x (, ) Substituting y = x into (9) (or equivalently into ()), we see that x / 0 = or x = ± as shown in the sketch. We will revisit this example in a later section.

7 6.7 7 We can say more about the previous example. Recall the trigonometric identity that resembles (6). () sec t tan t = This time let x = sect and x y = tant. Then x (x y) = ( sect ) (tant) = sec t tan t = In other words, we have obtained another parametrization of (9) of the hyperbola shown in Example. x = sect and We used this parametrization to generate the last sketch. y = sect tant Remark. The alert reader may soon discover that the parametrization x = cosht and y = sinht, say for t [ 6,6], generates only the right branch of the hyperbola x y = since cosht > 0 for all real t. This is indeed correct. In fact, we actually need to parameterize each branch separately. It is easy to verify that the parametrization ( cosh t, sinh t) generates the left branch. ( cosht,sinht) 3 (cosht,sinht) What about the hyperbola in Example?

8 6.7 8 Derivatives and Integrals of the Hyperbolic Functions Notice that, for example, the hyperbolic cosine function is differentiable (being the sum of two differentiable functions). In fact, d dx coshx = = = sinhx d ( e x +e x) dx ( e x e x) The other derivative formulas can be derived in a similar manner. We have (5) (6) (7) (8) d du (sinhu) = coshu dx dx d du (coshu) = sinhu dx dx d dx (tanhu) = sech u du dx d dx (sechu) = sechutanhu du dx

9 6.7 9 This leads immediately to the corresponding integral formulas. ˆ (9) coshudu = sinhu+c ˆ (0) sinhudu = coshu+c ˆ () sech udu = tanhu+c ˆ () sechu tanhudu = sechu+c