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More Challenging Problems
1. For the curve x(t) = t2 and y(t) = 3t – 1, write y as an explicit function of x
Answer
1. y= 3x1/2 – 1
Solution
1. Solve x(t) = t2 for t, obtaining t = x1/2. Then substitute this in for t in y(t) = 3t – 1: y = 3x1/2 -1
2. Plot x(t) = sin(at), y(t) = sin(bt) for (a,b) = (2,1), (2,2), and (2,3).
Answer
1. (a,b) = (2,1) 2. (a,b) = (2,2) 3. (a,b) = (2,3)
Solution
1. The solutions are the following: For (a,b) = (2,1), the x-coordinate makes two complete circuits, the y-coordinate one. The curve passes through the origin so is a figure 8. For (a,b) = (2,2), the x- and y-coordinates are the same, so the curve is the line of slope 1, -1 ≤ x ≤ 1, through the origin. For (a,b)= (2,3) the x-coordinate makes two complete circuits, the y-coordinate three. Again, the curve passes through the origin.
3. For the curve x(t) = t2 + 2t – 1, y(t) = t3 + t2 – 5t, find the points (x,y) at which the tangent is horizontal; find the points at which the tangent is vertical.
Answer
1. The horizontal tangents occur at (x, y) = (175/27, 4/9) and (x, y) = (-3, 4). The vertical tangents occur at (x, y) = (5, 0).
Solution
1. The slope of the tangent line is dy/dx = (t3 + t2 – 5t)’/(t2 + 2t + 1)’ = (3t2 + 2t – 5)/(2t + 2). The tangent line is horizontal when 3t2 + 2t – 5 = 0. That is, t = -5/3 and t = 1. The corresponding points are (x, y) = (175/27, 4/9) and (x, y) = (-3, 4). The tangent line is vertical when 2t + 2 = 0. That is, t = -1. The corresponding point is (x, y) = (5, 0).
4. Describe the curve x(t) = 3 cos(t), y(t) = 3 sin(t), z(t) = 3 cos(t), for 0 ≤ t ≤ 2π.
Answer
1. The curve is an ellipse lying in the plane z = x, with semi-minor axis length 3 and semi-major axis length 3√2.
Solution
1. The curve is the graph of the cosine function in the plane y = x.
5. Describe the curve x(t) = t, y(t) = t, z(t) = cos(t), for 0 ≤ t ≤ 2π.
Answer
1. The curve is the graph of the cosine function in the plane y = x.
Solution
1. Because x = t and y = t, the curve lies in the plane y = x. Projected into the xz plane, the curve is z = cos(t) = cos(x). Similarly, in the yz plane the curve projects to z = cos(y). Then the curve is the graph of the cosine function, but stretched horizontally, because successive maxima of the curve are (0,0,1) and (2π,2π,1). That is, the distance between successive maxima is 2π√2.