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which, again, contains the variables q and r only, without the parameter t This relation is defined only for positive real-number values of q 4 The second answer to Problem 3 is an expression of the relation in which q is the independent variable and r is the dependent variable We can t apply a line test directly to Fig 16-7, but we can graph the relation in a Cartesian coordinate plane with q on the horizontal axis and r on the vertical axis When we do that, we get the curve shown in Fig 16-6, with q in place of x and r in place of y This graph passes the vertical-line test, indicating that r is a function of q 5 For reference, the parametric equations are x = a cos t and y = a sin t We ve been assured that a 0, so we can divide the equations both through by a to obtain x /a = cos t and y /a = sin t Squaring both sides of both equations gives us (x /a)2 = cos2 t and (y /a)2 = sin2 t When we add these two equations, left-to-left and right-to-right, we have (x /a)2 + (y /a)2 = cos2 t + sin2 t The Pythagorean trigonometric identity for the sine and the cosine tells us that cos2 t + sin2 t = 1 for all real numbers t Therefore, the preceding equation can be rewritten as (x /a)2 + ( y /a)2 = 1 Expanding the squared ratios on the left-hand side gives us x 2 /a2 + y 2 /a2 = 1
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Multiplying through by a2, we get x 2 + y 2 = a2 which is the equation of a circle of radius a, centered at the origin 6 When we subtract x 2 from both sides of the solution to Problem 5, we obtain y 2 = a2 x 2 Taking the positive-or-negative square root of both sides gives us y = (a2 x 2)1/2 This relation is not a function of x, as we can see when we apply the vertical-line test to the graph of Fig 16-10 Whenever we input any value of the independent variable x that lies within the open interval ( a,a), our relation produces two values of the dependent variable y 7 For reference, the parametric equations are x = sec t and y = tan t When we square both sides of both equations, we obtain x 2 = sec2 t and y 2 = tan2 t Subtracting the second equation from the first, left-to-left and right-to-right, we get x 2 y 2 = sec2 t tan2 t From trigonometry, the Pythagorean identity for the secant and the tangent tells us that sec2 t tan2 t = 1 for all real numbers t except odd-integer multiples of p /2 The preceding equation can therefore be rewritten as x2 y2 = 1 which represents the unit hyperbola in the Cartesian xy plane
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8 For reference, the parametric equations are x = a csc t and y = b cot t We ve been assured that a 0, so we can divide the first equation through by a to obtain x /a = csc t We ve also been told that b 0, so we can divide the second equation through by b, getting y /b = cot t Squaring both sides of both equations gives us (x /a)2 = csc2 t and (y /b)2 = cot2 t When we subtract the second equation, left-to-left and right-to-right, from the first one, we obtain (x /a)2 (y /b)2 = csc2 t cot2 t The Pythagorean identity for the cosecant and the cotangent tells us that csc2 t cot2 t = 1 for all real numbers t except integer multiples of p Knowing this, we can rewrite the preceding equation as (x /a)2 (y /b)2 = 1 Expanding the squared ratios on the left-hand side gives us x 2/a2 y 2 /b2 = 1 which represents a hyperbola centered at the origin in the Cartesian xy plane The width of the horizontal (x-coordinate) semi-axis is a units, and the height of the vertical ( y-coordinate) semi-axis is b units
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9 We start with the relation x = sin (cos y) Suppose we assign our parameter t such that cos y = t Taking the arccosine of both sides, we get arccos (cos y) = arccos t which simplifies to y = arccos t That s one of our parametric equations We can substitute t for cos y in the original equation to get x = sin t That s the other parametric equation 10 As in Problem 9, we start with the relation x = sin (cos y) Taking the arcsine of both sides gives us arcsin x = arcsin [sin (cos y)] which simplifies to arcsin x = cos y When we take the arccosine of both sides, we obtain arccos (arcsin x) = arccos (cos y) which simplifies to arccos (arcsin x) = y Transposing the left- and right-hand sides, we get the sought-after equation y = arccos (arcsin x)
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