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Each vertical division is 1 unit
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Illustration for the solution to Problem 6 in Chap 15 Each horizontal division represents p /2 units Each vertical division represents 1 unit The vertical dashed lines are asymptotes of h The positive dependent-variable axis is also an asymptote of h
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Each horizontal division is p /2 units Each vertical division is 1 unit
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Illustration for the solution to Problem 7 in Chap 15 Each horizontal division represents p /2 units Each vertical division represents 1 unit The vertical dashed lines are asymptotes of h The dependent-variable axis is also an asymptote of h
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7 In Fig B-28, the dashed gray curves are graphs of the tangent and cotangent functions The solid black curves compose the graph of h (q) = tan q cot q The domain of h is the set of all reals except the integer multiples of p /2 The range is the set of all real numbers 8 In Fig B-29, the dashed gray curves are graphs of the tangent-squared and cotangentsquared functions The solid black curves compose the graph of h (q) = tan2 q cot2 q The domain of h includes all reals except the integer multiples of p /2 The range is the set of all real numbers 9 In Fig B-30, the dashed gray curves represent the squares of the secant and tangent functions The solid black line with holes is a graph of f (q) = sec2 q tan2 q
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552 Worked-Out Solutions to Exercises: 11-19
Figure B-29 Illustration for the
solution to Problem 8 in Chap 15 Each horizontal division represents p /2 units Each vertical division represents 1 unit The vertical dashed lines are asymptotes of h The dependentvariable axis is also an asymptote of h
h(q )
Each horizontal division is p /2 units Each vertical division is 1 unit
The domain of f is the set of all reals except the odd-integer multiples of p /2 The range of f is the set containing the number 1 10 In Fig B-31, the dashed gray curves represent the squares of the cosecant and cotangent functions The solid black line with holes is a graph of f (q) = csc2 q cot2 q The domain of f is the set of all reals except the integer multiples of p The range of f is the set containing the number 1 Figure B-30
Illustration for the solution to Problem 9 in Chap 15 Each horizontal division represents p /2 units Each vertical division represents 1 unit
f (q )
Each horizontal division is p /2 units
Each vertical division is 1 unit
16
Figure B-31
Illustration for the solution to Problem 10 in Chap 15 Each horizontal division represents p /2 units Each vertical division represents 1 unit
f (q )
Each horizontal division is p /2 units
Each vertical division is 1 unit
16
1 For reference, the parametric equations are x = t2 and y = t3 We can take the 1/3 power of both sides of the second equation to get y1/3 = t Substituting y1/3 for t in the first equation yields x = (y1/3)2 which can be simplified to x = y 2 /3 This equation contains the variables x and y only, without the parameter t There s another way to approach this problem We can take the positive-or-negative 1/2 power of both sides of the first original equation, getting x1/2 = t
554 Worked-Out Solutions to Exercises: 11-19
Then we can substitute x1/2 for t in the second original equation to obtain y = ( x1/2)3 = x 3/2 which contains the variables x and y only, without the parameter t 2 The second answer to Problem 1 is an expression of the relation in which x is the independent variable and y is the dependent variable This relation is not a function of x We can see this by applying the vertical-line test to the graph of Fig 16-4 The graph fails the test because, for all positive values of x, there are two values of y 3 For reference, the parametric equations are q = t 1 and r = ln t We can take the natural exponential of both sides of the second equation to obtain e r = e(ln t) which simplifies to er = t provided that t > 0, so we re sure that ln t is defined Substituting e r for t in the first original parametric equation yields q = (e r) 1 which simplifies to q = e r This equation contains the variables q and r only, without the parameter t We can approach this problem another way If we take the reciprocal of both sides of the first original parametric equation, we get q 1 = (t 1) 1 as long as t 0, so we re sure that t 1 is defined This simplifies to q 1 = t When we substitute q 1 for t in the second parametric equation, we get r = ln (q 1) = ln q
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