1. Here are the plots for k = 1, 2, and 3.

k=1                                                  k=2                                                    k=3

1. The k= 1 case is easy: y(t) = x(t) is the strainght line of slope 1 passing through the origin. For k = 2, use the trigonometric identity cos(2t) = 2cos²(t) -1 to obtain y(t) = 2x(t)² – 1, a parabola opening upward with minimum at (0, -1). For k = 3, use cos(a + b) = cos(a)cos(b) – sin(a)sin(b) and sin(2a) = 2sin(a)cos(a) to obtain y(t) = cos(3t) = cos(t + 2t) = cos(t)cos(2t) – sin(t)sin(2t) = cos(t)(2cos²(t) – 1) – sin(t)2sin(t)cos(t) = 2cos³(t) – cos(t) – 2cos(t)(1 – cos²(t)) = 4cos³(t) – 3 cos(t) = 4x(t)³ – 3x(t)

1. The curve passes through (x, y)= (1, 0) twice, and has two tangent lines at that point: y= -(x – 1) and y= x – 1

1. The slope of the tangent line is: dy/dx = (t³ – t)’/(t²)’= (3t² – 1)/(2t) To find the slope of the tangent line at the point (x, y) = (1, 0), we must find the value of t corresponding to that point. First, y= 0 means t³ – t= 1; that is, t= -1, t= 0, and t= 1. Second, x= 1 means t²= 1; that is, t= -1 and t= 1. Consequently, this curve passes through the point (x, y) = (1, 0) at t = -1 and at t = 1. At t = -1 the slope of the tangent line is -1 and so the equation of the tangent line is y = -(x – 1) At t = 1 the slope of the tangent line is 1 and so the equation of the tangent line is y = (x – 1)

1. Please note the following image:

1. First note that x² + y² = t², so x² + y² + z²= 1. That is, the curve lies on the unit sphere. In fact, on the upper hemisphere because the z-coordinate is positive. Projected into the xy-plane, the curve is a spiral converging to (0,0). The spiral makes a complete turn between 1/2π and 1/4π, between 1/4π and 1/6π, between 1/6π and 1/8π, and so on, spinning ever faster. Moreover, the distance to (0, 0) decreases as t decreases toward 0. Combining these observations, we see the curve is a spiral on the upper hemisphere of the unit sphere, spiraling in to the north pole.

1. Please note the following image:

1. First, note that x² + z²= t², so the curve lies on the cone y = (x² + z²)^1/2, having the positive y-axis as its axis. Projected into the xz-plane, the curve is a spiral converging to (0,0), so in three dimensions, the curve spirals to the apex of the cone.

1. The curve has vertical tangents at (0,0) and at (2π,0), and has a horizontal tangent at (π,2). The curve is concave down everywhere.

1. If the curve has a horizontal tangent at parameter t, then 0 = (1 – cos(t))’ = sin(t), so t = 0, π, and 2π are the possible parameters of horizontal tangents. If the curve has a vertical tangent at parameter t, then 0 = (t – sin(t))’ = 1 – cos(t), so t = 0 and 2π are the possible parameters of vertical tangents. First, we see that the curve has a horizontal tangent at t = π, at the point (x, y) = π,2). To check of the curve has a horizontal or vertical tangent, or neither, at t = 0 and t = 2π, compute the limit lim_{t → 0}dy/dx = lim_{t → 0}(dy/dt)/(dx/dt) = lim_{t → 0}(sin(t))/(1 – cos(t)) = lim_{t → 0}(cos(t))/(sin(t)) by l’Hopital’s rule = ∞ So the curve has a vertical tangent at (0,0). A similar calculation shows the curve has a vertical tangent at (2π, 0). To determine concavity, we compute d²y/dx², a bit tricky because we do not know y as an explicit function of x. So instead we use the chain rule, twice. d²y/dx² = (d/dx)(dy/dx) = (d/dx)((dy/dt)/(dx/dt)) = (d/dx)(sin(t)/(1 – cos(t))) = (d/dt)(sin(t)/(1 – cos(t))) / (dx/dt) = -1/(1 – cos(t)) The curve is concave down throughout its domain.