Questions: A curve is defined by the parametric equations x(t)=-4 tan t+3 and y(t)=3 cos t-3. Draw this graph by using the draggable points.

Solution Steps

We can use _t_ = 0 and _t_ = 3π/4 because these values are already shown in the graph in the original image. We can also select other values near these to get a better idea of the shape of the graph. Since the period of tan(_t_) is π and the period of cos(_t_) is 2π, we know that repeating values of _t_ shifted by multiples of 2π will create the same _x_ and _y_ outputs, which won't help us to understand the graph's shape. So, we will select _t_ values within the interval [0, 2π).

• If _t_ = 0, _x_ = 4tan(0) + 3 = 3, and _y_ = 3cos(0) - 3 = 0. So, we plot the point (3, 0).
• If _t_ = π/4, _x_ = 4tan(π/4) + 3 = 7, and _y_ = 3cos(π/4) - 3 ≈ -0.88. So, we plot the point (7, -0.88).
• If _t_ = π/2, _x_ is undefined because tan(π/2) is undefined.
• If _t_ = 3π/4, _x_ = 4tan(3π/4) + 3 = -1, and _y_ = 3cos(3π/4) - 3 ≈ -5.12. So, we plot the point (-1, -5.12).
• If _t_ = π, _x_ = 4tan(π) + 3 = 3, and _y_ = 3cos(π) - 3 = -6. So, we plot the point (3, -6).
• If _t_ = 5π/4, _x_ = 4tan(5π/4) + 3 = 7, and _y_ = 3cos(5π/4) - 3 ≈ -5.12. So, we plot the point (7, -5.12).
• If _t_ = 3π/2, _x_ is undefined because tan(3π/2) is undefined.
• If _t_ = 7π/4, _x_ = 4tan(7π/4) + 3 = -1, and _y_ = 3cos(7π/4) - 3 ≈ -0.88. So, we plot the point (-1, -0.88).

Draw a hyperbolic curve passing through the plotted points. The curve is oriented such that _t_ increases from right to left across the top portion of the curve, and _t_ increases from left to right across the bottom portion of the curve. Vertical asymptotes appear at _t_ = π/2 and _t_ = 3π/2.

Final Answer: