Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


c. The parametric equations x=t, y=t², for t≥0, describe the complete parabola y=x².

10–12. Parametric curves

a. Eliminate the parameter to obtain an equation in x and y.

x = ln t, y = 8ln t², for 1 ≤ t ≤ e²; (1, 16)

37–52. Curves to parametric equations Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique.

The left half of the parabola y=x ² +1, originating at (0, 1)

37–52. Curves to parametric equations Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique.

The upper half of the parabola x=y ², originating at (0, 0)

53–56. Circular motion Find parametric equations that describe the circular path of the following objects. For Exercises 53–55, assume (x, y) denotes the position of the object relative to the origin at the center of the circle. Use the units of time specified in the problem. There are many ways to describe any circle.

The tip of the 15-inch second hand of a clock completes one revolution in 60 seconds.

14–18. Parametric descriptions Write parametric equations for the following curves. Solutions are not unique.

The segment of the curve x=y ³ +y+1 that starts at (1, 0) and ends at (11, 2).

14–18. Parametric descriptions Write parametric equations for the following curves. Solutions are not unique.

The circle x ² + y ² =9, generated clockwise

Air drop A plane traveling horizontally at 80 m/s over flat ground at an elevation of 3000 m releases an emergency packet. The trajectory of the packet is given by

x = 80t, y = −4.9t² + 3000, t ≥ 0

where the origin is the point on the ground directly beneath the plane at the moment of the release (see figure). Graph the trajectory of the packet and find the coordinates of the point where the packet lands.

Air drop—inverse problem A plane traveling horizontally at 100 m/s over flat ground at an elevation of 4000 m must drop an emergency packet on a target on the ground. The trajectory of the packet is given by

x = 100t, y = −4.9t² + 4000, t ≥ 0

where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the target?