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Now let s derive this equation from the parametric equations in the solution to Problem 9 Those equations are x = sin t and y = arccos t We can take the arcsine of both sides of the first equation, getting arcsin x = arcsin (sin t) This simplifies to arcsin x = t Substituting arcsin x for t in the right-hand side of the second parametric equation, we obtain y = arccos (arcsin x) which is the same equation we got from the original relation in terms of x and y without the parameter t
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1 The standard-form vector 4i + 4j 4k can be described by the ordered triple (a,b,c) = ( 4,4, 4) We ve been told that one of the points in our plane is (x0,y0,z0) = (0,0,0) The general formula for a plane in Cartesian xyz space is a(x x0) + b(y y0) + c(z z0) = 0 Plugging in the known values, we get 4(x 0) + 4[y 0] + ( 4)(z 0) = 0
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which simplifies to 4x + 4y 4z = 0 2 The standard-form vector 2i + 0j + 0k can be described by the ordered triple (a,b,c) = ( 2,0,0) One of the points in the plane is (x0,y0,z0) = (4,5,6) The general formula for a plane in Cartesian xyz space is a(x x0) + b(y y0) + c(z z0) = 0 Plugging in the known values, we get 2(x 4) + 0[y 5] + 0(z 6) = 0 which simplifies to 2x + 8 = 0 3 We ve been told that the equation of a certain sphere is x 2 + 2x + 1 + y 2 2y + 1 + z2 + 8z + 16 = 64 Grouping the addends by threes, we get (x 2 + 2x + 1) + (y 2 2y + 1) + (z2 + 8z + 16) = 64 Factoring each of the trinomials enclosed by parentheses, we obtain (x + 1)2 + (y 1)2 + (z + 4)2 = 64 The general equation for a sphere in Cartesian xyz space is (x x0)2 + (y y0)2 + (z z0)2 = r2 where (x0,y0,z0) are the coordinates of the center, and r is the radius Based on this information, we can deduce that the coordinates of this sphere s center are (x0,y0,z0) = ( 1,1, 4) and the radius r is the positive square root of 64, which is 8
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4 We ve been told that the coordinates of the center of a certain sphere are (x0,y0,z0) = (5,7, 3) and the radius is r = 231/2 Once again, the general equation for a sphere in Cartesian xyz space is (x x0)2 + (y y0)2 + (z z0)2 = r2 where (x0,y0,z0) are the coordinates of the center, and r is the radius Plugging in the known values directly, we conclude that the equation for this particular sphere is (x 5)2 + (y 7)2 + (z + 3)2 = 23 5 Stated again for reference, the equation of our object is 8(x 1)2 + 8(y + 2)2 + 6(z + 7)2 = 24 Dividing through by 24, we obtain (x 1)2 /3 + (y + 2)2 /3 + (z + 7)2 /4 = 1 This is the equation for a distorted sphere centered at (x0,y0,z0) = (1, 2, 7) The length of the axial radius in the x direction is the positive square root of 3 The length of the axial radius in the y direction is also the positive square root of 3 The length of the axial radius in the z direction is the positive square root of 4, or 2, which is a little longer than the other two axes Therefore, our object is an ellipsoid 6 Stated again for reference, the equation for the object under scrutiny is 400(x + 2)2 + 225(y 4)2 + 144z2 3600 = 0 When we add 3600 to each side, we get 400(x + 2)2 + 225(y 4)2 + 144z2 = 3600 Dividing through by 3600 yields (x + 2)2 /9 + (y 4)2 /16 + z2 /25 = 1 This is the equation for a distorted sphere centered at (x0,y0,z0) = ( 2,4,0)
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The length of the axial radius in the x direction is 91/2, which is 3 The length of the axial radius in the y direction is 161/2, which is 4 The length of the axial radius in the z direction is 251/2, which is 5 Because no two of the axial radii are the same, our object is an oblate ellipsoid 7 We ve been told that the equation of a certain object is x 2 + 2x + 1 + y 2 2y + 1 z2 + 6z 9 = 36 Grouping the terms by threes, we get (x 2 + 2x + 1) + (y 2 2y + 1) + ( z2 + 6z 9) = 36 which can be rewritten as (x 2 + 2x + 1) + (y 2 2y + 1) (z2 6z + 9) = 36 Factoring each of the trinomials enclosed by parentheses, we obtain (x + 1)2 + (y 1)2 (z 3)2 = 36 Dividing through by 36 gives us (x + 1)2 /36 + (y 1)2 /36 (z 3)2 /36 = 1 This equation represents a hyperboloid of one sheet whose center is at (x0,y0,z0) = ( 1,1,3) and whose axis is a line parallel to the coordinate z axis 8 We ve been told that the coordinates of the vertex of an elliptic cone are ( 2,3,4), and that the cone s axis is parallel to the coordinate y axis The equation must therefore be of the form (x + 2)2 /a2 (y 3)2 /b2 + (z 4)2 /c2 = 0 where a, b, and c determine the eccentricity and orientation of the cross-sectional ellipses that we get when we slice through the cone with planes perpendicular to its axis On the basis of the information given, all we can say about these constants is that they re positive real numbers 9 Here s the generalized equation for the elliptic cone described in Problem 8: (x + 2)2 /a2 (y 3)2 /b2 + (z 4)2 /c2 = 0 This cone intersects the xz plane in a curve where the y value is always equal to 0 If we set y = 0 in the above equation, we get (x + 2)2 /a2 (0 3)2 /b2 + (z 4)2 /c2 = 0
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