Additional Exercises for Chapter 10

Transcription

1 Aitional Eercises for Chapter 0 About the Eponential an Logarithm Functions 6. Compute the area uner the graphs of i. f() =e over the interval [ 3, ]. ii. f() =e over the interval [, 4]. iii. f() = over the interval [, ]. 63. Compute the volume of the soli obtaine by rotating the graph of f() =e,, one revolution about the -ais. 64. Consier the function f() =e 3. Let a>0 an show that the point (ln a, a 3 )isonits graph. Epress the length of the graph of f() from (0, ) to (ln a, a 3 )asaefinite integral. (Don t evaluate). 65. Epress the length of the graph of f() =ln from (, 0) to (a, ln a) with a>0asaefinite integral. (Don t evaluate). 66. Recall that F () =ln is an antierivative of f() = for >0. Suppose that <0 an compute ln( ). Conclue that the function F () =ln is an antierivative of the function f() = with omain all with 0.Sketch the graph of F () =ln. More About the Tractri 67. We saw in Section 0.4 that the tractri, the pocket watch curve that Leibniz ha stuie in Paris, is the graph of the function f() =a ln( a+(a ) ) (a ). This curve can be generate as follows: You are given a circular watch an a chain to which it is attache. Place the watch on the -ais so that its center rests over the point (a, 0). Keeping the watch fie, stretch the chain buy pulling on its open en, an arrange things so that this open en is on the y-ais. Now release the watch an move the open en of the chain along the y-ais in the positive irection. The curve that the center of the moving watch traces out in the process is the graph of the function f(). This is the vertical format of the tractri. Now o this once more, but place the center of the watch at the point (0,a)ofthe y-ais, an rag the open en of the chain along the positive -ais. i. Refer to the figure below an convince yourself that the curve that the center of the watch traces out this time, the tractri in horizontal format, is the graph of the inverse function f () ofthe function f() above.

2 y f() = trac() a f () = trac () a ii. Show that f () = f (), first by analyzing the geometry of the curve, an a f () again by using the formula for the erivative of an inverse function. Unlike the situation of the function f(), it turns out that it is not possible to epress the function f () in a close way as a combination of log functions, an trig functions, an their inverses. Eercising Inverse Trig Functions 68. Fin the average value of the function f() = from 0 to. 69. Fin the area uner the graph of the function f() = + from to4. About a Hyperbola 70. Consier the function f() = c for a positive constant c. i. Show that the an y aes are horizontal an vertical asymptotes respectively of the graph of f(). ii. Show that f () =f() an therefore that the graph of f oes not change when it is revolve aroun the line y =. iii. Compute f (). Show that the graph of y = f() isecreasing over both (, 0) an (0, ). iv. Compute f (). Show that the graph of y = f() isconcave own over (, 0) an concave up over (0, ). v. Show that the two points on the graph at which the slope is equal to are ( c, c) an ( c, c).

3 vi. Sketch the graph of the function. vii. Review the efinition of the hyperbola in Section 3.. It can be shown that the graph of f() isahyperbola with focal points ( c, c) an ( c, c) an k = c. Verify this in the case c =. 7. Consier the function f() = c for a positive c. i. How oes its graph iffer from that of the function in Eercise 67. ii. Let a an b be positive constants. Show that the graph obtaine by shifting the graph of y = f() tothe left by a units an up by b units is the graph of the function g() = b+ab c. +a 7. Consier the function f() = c for positive constant c an. Show that its graph can be + obtaine by shifting the graph of f() = c by units to the left an c units up. Hyperbolic Functions The two eponential epressions e e an e + e occur frequently in applie mathematics an engineering. We will see that they have a relationship to each other that is very similar to the relationship between the functions sin an cos. We will also see that they are relate to the hyperbola y =in the same way that sin an cos are relate to the circle + y =. Because of the frequency of their occurrence, their connections with sin an cos, an their relationship to the hyperbola, the two epressions are calle respectively the hyperbolic sine an the hyperbolic cosine of. More precisely, the hyperbolic sine enote by sinh an the hyperbolic cosine enote by cosh are the functions efine by for any real number. sinh = e e an cosh = e + e 73. Use the graphs of the functions y = e an y = e to sketch the graphs of the hyperbolic functions sinh an cosh. Sketch them sie by sie on the same coorinate system. We saw in Chapter 9 that the curve of the main cable of a suspension brige is a parabola. The unerlying reason is the fact (or reasonable assumption) that such a cable is subject to the uniform vertical loa generate by the eck that it supports. But what about a freely hanging fleible cable or chain that is not subject to such a loa? Say, a telephone wire or a power line? In this case, the shape of the hanging cable is escribe by the graph of the hyperbolic cosine. Many ientities similar to those for trigonometric functions hol for the hyperbolic sine an cosine functions. For eample, 3

4 74. Verify the ientities (notice that each is analogous to a trigonometric ientity) i. cosh sinh =(where cosh an sinh enote (cosh ) an (sinh ), respectively. ii. sinh( + y) =(sinh )(cosh y)+(cosh )(sinh y) iii. cosh( + y) =(cosh )(cosh y)+(sinh )(sinh y) Let s return to the sine an cosine for a moment. What curve oes the point (cos u, sin u) lie on for any real number u? Nowlet u vary from 0 to. How oes the point (cos u, sin u) move in the process? 75. Sketch the graph of the equation y =an show that it is a hyperbola with asymptotes the lines y = ±. Show that the point (cosh u, sinh u) lies on the right branch of this hyperbola for any real number u. Let u vary from an an show that the point (cos u, sin u) traces out the entire right branch from the bottom to the top. The striking similarity between the trigonometric sine an cosine an the hyperbolic sine an cosine suggests that aitional hyperbolic functions shoul be efine analogously to their trigonometric counterparts. Specifically, the hyperbolic tangent, secant, contangent, an cosecant functions, enote by tanh, sech, coth, an csch, respectively, are efine by tanh = sinh cosh, sech = cosh, cosh coth =, an csch = sinh sinh. 76. Show that tanh = sech. The erivatives of the hyperbolic functions are easy to compute. For eample, since e = e an e = e, sinh = ( ) e e = e + e = cosh. 77. Compute the erivatives of the functions cosh, tanh, an sech. Let s turn to the stuy of the graph of the function tanh = sinh. Observe that cosh for cosh any, an that sinh 0 for 0 an sinh <0 for <0. Notice also that cosh sinh = e > 0 when 0, an cosh ( sinh ) =e > 0 when <0. It follows that cosh > sinh for all, an hence that tanh < for all. 4

5 78. Epan on the observations just mae to show that the lines y =an y = are both horizontal asymptotes of the graph of f() =tanh an that the graph falls between these two lines. Determine the intervals over which f() = tanh is increasing an ecreasing an concave up an concave own. Then sketch the graph of f() =tanh. 79. Sketch the graphs of the functions sech, coth, an csch by analyzing the flow an pattern of the graphs of cosh, tanh, an sinh. Inverse Hyperbolic Functions Refer to the graphs of the hyperbolic sine, hyperbolic cosine, an hyperbolic tangent functions an consier the efinition of their inverses. For any real number, efine sinh to be that number y with the property that sinh y =. For any real number, efine cosh to be that number y with y 0 such that cosh y =. <<, efine Finally, for any real number with tanh to be that number y with the property that tanh y =. All these inverse functions can be epresse in terms of the natural log as follows. 80. i. sinh = ln( + +) for all real numbers. ii. cosh = ln( + ) for all. iii. tanh = ln ( + ) for <<. We will verify (ii) an leave the other two equalities to the reaer. Because y = cosh means that cosh y =, wenee to set cosh y = ey +e y =, an solve for y. Rewrite the last equality as e y + e y =0,then multiply through by e y to get e y e y + =0. Because e y =(e y ),we can apply the quaratic formula, to get e y = ( ± 4 4) = ± ). Suppose that were possible. The fact that y 0, woul then give us, an hence. So ( ), an therefore, +. Since this cannot be so, we now know that e y = + ). After taking ln of both sies, we get y = ln( + ). 8. Sketch the graphs of the functions sinh, cosh, an tanh. 5

6 We turn to the computation of the erivative of y = sinh. Note that = sinh y. Since both sies of this equation are functions of, wecan use the chain rule to obtain = cosh y y. So y =.Byusing cosh y cosh y = sinh y+, we get y = (sinh. After substituting y = y) + sinh, we see that y = (sinh(sinh =. Therefore, )) + + sinh = +. Eercise 77 (i) provies another way to o this, namely sinh = + + (+( +) ) = +( +) +( +) ( +) = ( +). 8. Show in two ifferent ways that cosh =. 83. Show that tanh =. The iscussion an the conclusions of the eercises above provie the following integration formulas: + = sinh + C = cosh + C = tanh + C We will leave the efinition an eploration of the functions sech, coth, an csch the reaer. to 6