Parametric Equations: Graphs

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Sketch the graph of the parametric equations \(x(t)={t}^{2}+1,\) \(y(t)=2+t.\)

   जवाब दिखाएँ

   Construct a table of values for \(t,x(t),\) \(\) and \(y(t),\) as in , and plot the points in a plane.

   \(t\)	\(x(t)={t}^{2}+1\)	\(y(t)=2+t\)
   \(-5\)	\(26\)	\(-3\)
   \(-4\)	\(17\)	\(-2\)
   \(-3\)	\(10\)	\(-1\)
   \(-2\)	\(5\)	\(0\)
   \(-1\)	\(2\)	\(1\)
   \(0\)	\(1\)	\(2\)
   \(1\)	\(2\)	\(3\)
   \(2\)	\(5\)	\(4\)
   \(3\)	\(10\)	\(5\)
   \(4\)	\(17\)	\(6\)
   \(5\)	\(26\)	\(7\)

   The graph is a parabola with vertex at the point \((1,2),\) opening to the right. See .

2. Sketch the graph of the parametric equations \(x=\sqrt{t},\) \(y=2t+3,\) \(0\le t\le 3.\)

3. Construct a table of values for the given parametric equations and sketch the graph:

   \[\begin{array}{l}x=2\cos \ t \\ y=4\sin \ t\end{array}\]
   जवाब दिखाएँ

   Construct a table like that in using angle measure in radians as inputs for \(t,\) and evaluating \(x\) and \(y.\) Using angles with known sine and cosine values for \(t\) makes calculations easier.

   \(t\)	\(x=2\cos \ t\)	\(y=4\sin \ t\)
   0	\(x=2\cos (0)=2\)	\(y=4\sin (0)=0\)
   \(\frac{\pi }{6}\)	\(x=2\cos (\frac{\pi }{6})=\sqrt{3}\)	\(y=4\sin (\frac{\pi }{6})=2\)
   \(\frac{\pi }{3}\)	\(x=2\cos (\frac{\pi }{3})=1\)	\(y=4\sin (\frac{\pi }{3})=2\sqrt{3}\)
   \(\frac{\pi }{2}\)	\(x=2\cos (\frac{\pi }{2})=0\)	\(y=4\sin (\frac{\pi }{2})=4\)
   \(\frac{2\pi }{3}\)	\(x=2\cos (\frac{2\pi }{3})=-1\)	\(y=4\sin (\frac{2\pi }{3})=2\sqrt{3}\)
   \(\frac{5\pi }{6}\)	\(x=2\cos (\frac{5\pi }{6})=-\sqrt{3}\)	\(y=4\sin (\frac{5\pi }{6})=2\)
   \(\pi\)	\(x=2\cos (\pi )=-2\)	\(y=4\sin (\pi )=0\)
   \(\frac{7\pi }{6}\)	\(x=2\cos (\frac{7\pi }{6})=-\sqrt{3}\)	\(y=4\sin (\frac{7\pi }{6})=-2\)
   \(\frac{4\pi }{3}\)	\(x=2\cos (\frac{4\pi }{3})=-1\)	\(y=4\sin (\frac{4\pi }{3})=-2\sqrt{3}\)
   \(\frac{3\pi }{2}\)	\(x=2\cos (\frac{3\pi }{2})=0\)	\(y=4\sin (\frac{3\pi }{2})=-4\)
   \(\frac{5\pi }{3}\)	\(x=2\cos (\frac{5\pi }{3})=1\)	\(y=4\sin (\frac{5\pi }{3})=-2\sqrt{3}\)
   \(\frac{11\pi }{6}\)	\(x=2\cos (\frac{11\pi }{6})=\sqrt{3}\)	\(y=4\sin (\frac{11\pi }{6})=-2\)
   \(2\pi\)	\(x=2\cos (2\pi )=2\)	\(y=4\sin (2\pi )=0\)

   shows the graph.

   By the symmetry shown in the values of \(x\) and \(y,\) we see that the parametric equations represent an ellipse. The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing \(t\) values.

4. Graph the parametric equations: \(x=5\cos \ t,\) \(y=3\sin \ t.\)

5. Graph the parametric equations \(x=5\cos \ t\) and \(y=2\sin \ t.\) First, construct the graph using data points generated from the parametric form. Then graph the rectangular form of the equation. Compare the two graphs.

   जवाब दिखाएँ

   Construct a table of values like that in .

   \(t\)	\(x=5\cos \ t\)	\(y=2\sin \ t\)
   \(\text{0}\)	\(x=5\cos (0)=5\)	\(y=2\sin (0)=0\)
   \(\text{1}\)	\(x=5\cos (1)\approx 2.7\)	\(y=2\sin (1)\approx 1.7\)
   \(\text{2}\)	\(x=5\cos (2)\approx -2.1\)	\(y=2\sin (2)\approx 1.8\)
   \(\text{3}\)	\(x=5\cos (3)\approx -4.95\)	\(y=2\sin (3)\approx 0.28\)
   \(\text{4}\)	\(x=5\cos (4)\approx -3.3\)	\(y=2\sin (4)\approx -1.5\)
   \(\text{5}\)	\(x=5\cos (5)\approx 1.4\)	\(y=2\sin (5)\approx -1.9\)
   \(-1\)	\(x=5\cos (-1)\approx 2.7\)	\(y=2\sin (-1)\approx -1.7\)
   \(-2\)	\(x=5\cos (-2)\approx -2.1\)	\(y=2\sin (-2)\approx -1.8\)
   \(-3\)	\(x=5\cos (-3)\approx -4.95\)	\(y=2\sin (-3)\approx -0.28\)
   \(-4\)	\(x=5\cos (-4)\approx -3.3\)	\(y=2\sin (-4)\approx 1.5\)
   \(-5\)	\(x=5\cos (-5)\approx 1.4\)	\(y=2\sin (-5)\approx 1.9\)

   Plot the \((x,y)\) values from the table. See .

   Next, translate the parametric equations to rectangular form. To do this, we solve for \(t\) in either \(x(t)\) or \(y(t),\) and then substitute the expression for \(t\) in the other equation. The result will be a function \(y(x)\) if solving for \(t\) as a function of \(x,\) or \(x(y)\) if solving for \(t\) as a function of \(y.\)

   \[\begin{array}{lllll}x=5\cos \ t & \\ \frac{x}{5}=\cos \ t & \text{Solve for }\cos \ t. \\ y=2\sin \ t\begin{array}{llll} & & & \end{array}\ & \text{Solve for }\sin \ t. \\ \frac{y}{2}=\sin \ t & \end{array}\]

   Then, use the Pythagorean Theorem.

   \[\begin{array}{l}{\cos }^{2}t+{\sin }^{2}t=1 \\ {(\frac{x}{5})}^{2}+{(\frac{y}{2})}^{2}=1 \\ \frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\end{array}\]

6. Graph the parametric equations \(x=t+1\) and \(y=\sqrt{t},\) \(t\ge 0,\) and the rectangular equivalent \(y=\sqrt{x-1}\) on the same coordinate system.

   जवाब दिखाएँ

   Construct a table of values for the parametric equations, as we did in the previous example, and graph \(y=\sqrt{t},\) \(t\ge 0\) on the same grid, as in .

7. Sketch the graph of the parametric equations \(x=2\cos \ \theta \ \text{and}\ y=4\sin \ \theta ,\) along with the rectangular equation on the same grid.

   जवाब दिखाएँ

   The graph of the parametric equations is in red and the graph of the rectangular equation is drawn in blue dots on top of the parametric equations.

8. Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of \(45^{\circ}\) to the horizontal, making contact 3 feet above the ground.

   1. ⓐ Find the parametric equations to model the path of the baseball.
   2. ⓑ Where is the ball after 2 seconds?
   3. ⓒ How long is the ball in the air?
   4. ⓓ Is it a home run?
   जवाब दिखाएँ

   1. ⓐ

      Use the formulas to set up the equations. The horizontal position is found using the parametric equation for \(\ x.\) Thus,

      \[\begin{array}{l}x=({v}_{0}\cos \ \theta )t \\ x=(140\cos (45^{\circ}))t\end{array}\]

      The vertical position is found using the parametric equation for \(y.\) Thus,

      \[\begin{array}{l}y=-16{t}^{2}+({v}_{0}\sin \ \theta )t+h \\ y=-16{t}^{2}+(140\sin (45^{\circ}))t+3\end{array}\]
   2. ⓑ

      Substitute 2 into the equations to find the horizontal and vertical positions of the ball.

      \[\begin{array}{l}x=(140\cos (45^{\circ}))(2) \\ x=198\text{ feet} \\ \\ y=-16{(2)}^{2}+(140\sin (45^{\circ}))(2)+3 \\ y=137\text{ feet}\end{array}\]

      After 2 seconds, the ball is 198 feet away from the batter’s box and 137 feet above the ground.

   3. ⓒ

      To calculate how long the ball is in the air, we have to find out when it will hit ground, or when \(y=0.\) Thus,

      \[\begin{array}{ll}y=-16{t}^{2}+(140\sin ({45}^{∘}))t+3 & \\ y=0 & \text{Set }y(t)=0\text{ and solve the quadratic}. \\ t=6.2173 & \end{array}\]

      When \(t=6.2173\) seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.)

   4. ⓓ

      We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity’s sake, let’s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let’s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when x = 400 feet. So we will set x = 400, solve for \(t,\) and input \(t\) into \(y.\)

      \[\begin{array}{l}\begin{array}{l}x=(140\cos (45^{\circ}))t \\ 400=(140\cos (45^{\circ}))t \\ t=4.04\end{array} \\ \\ \begin{array}{l}y=-16{(4.04)}^{2}+(140\sin (45^{\circ}))(4.04)+3 \\ y=141.8\end{array}\end{array}\]

      The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run. See .

9. What are two methods used to graph parametric equations?

   जवाब दिखाएँ

   plotting points with the orientation arrow and a graphing calculator

10. What is one difference in point-plotting parametric equations compared to Cartesian equations?

11. Why are some graphs drawn with arrows?

    जवाब दिखाएँ

    The arrows show the orientation, the direction of motion according to increasing values of \(t.\)

12. Name a few common types of graphs of parametric equations.

13. Why are parametric graphs important in understanding projectile motion?

    जवाब दिखाएँ

    The parametric equations show the different vertical and horizontal motions over time.

14. \(\{\begin{array}{l}x(t)=t \\ y(t)={t}^{2}-1\end{array}\)

    \(t\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
    \(x\)
    \(y\)

15. \(\{\begin{array}{l}x(t)=t-1 \\ y(t)={t}^{2}\end{array}\)

    \(t\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
    \(x\)
    \(y\)

16. \(\{\begin{array}{l}x(t)=2+t \\ y(t)=3-2t\end{array}\)

    \(t\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
    \(x\)
    \(y\)

17. \(\{\begin{array}{l}x(t)=-2-2t \\ y(t)=3+t\end{array}\)

    \(t\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)
    \(x\)
    \(y\)

18. \(\{\begin{array}{l}x(t)={t}^{3} \\ y(t)=t+2\end{array}\)

    \(t\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
    \(x\)
    \(y\)

19. \(\{\begin{array}{l}x(t)={t}^{2} \\ y(t)=t+3\end{array}\)

    \(t\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
    \(x\)
    \(y\)

20. \(\{\begin{array}{l}x(t)=t \\ y(t)=\sqrt{t}\end{array}\)

21. \(\{\begin{array}{l}x(t)=-\ \sqrt{t} \\ y(t)=t\end{array}\)

22. \(\{\begin{array}{l}x(t)=5-|t| \\ y(t)=t+2\end{array}\)

23. \(\{\begin{array}{l}x(t)=-t+2 \\ y(t)=5-|t|\end{array}\)

24. \(\{\begin{array}{l}x(t)=4\text{sin}\ t \\ y(t)=2\cos \ t\end{array}\)

25. \(\{\begin{array}{l}x(t)=2\text{sin}\ t \\ y(t)=4\text{cos}\ t\end{array}\)

26. \(\{\begin{array}{l}x(t)=3{\cos }^{2}t \\ y(t)=-3\sin \ t\end{array}\)

27. \(\{\begin{array}{l}x(t)=3{\cos }^{2}t \\ y(t)=-3{\sin }^{2}t\end{array}\)

28. \(\{\begin{array}{l}x(t)=\text{sec}\ t \\ y(t)=\tan \ t\end{array}\)

29. \(\{\begin{array}{l}x(t)=\text{sec}\ t \\ y(t)={\tan }^{2}t\end{array}\)

30. \(\{\begin{array}{l}x(t)=\frac{1}{{e}^{2t}} \\ y(t)={e}^{-\ t}\end{array}\)

31. \(\{\begin{array}{l}x(t)=t-1 \\ y(t)=-{t}^{2}\end{array}\)

32. \(\{\begin{array}{l}x(t)={t}^{3} \\ y(t)=t+3\end{array}\)

33. \(\{\begin{array}{l}x(t)=2\cos \ t \\ y(t)=-\sin \ t\end{array}\)

    जवाब दिखाएँ

34. \(\{\begin{array}{l}x(t)=7\cos \ t \\ y(t)=7\sin \ t\end{array}\)

35. \(\{\begin{array}{l}x(t)={e}^{2t} \\ y(t)=-{e}^{\ t}\end{array}\)

36. \(x={t}^{2},\ y\ =\ 3t,\ 0\le t\le 5\)

37. \(x=2t,\ y=\ {t}^{2},\ -5\le t\le 5\)

38. \(x=t,\) \(y=\sqrt{25-{t}^{2}},\) \(0

39. \(x(t)=-t,y(t)=\sqrt{t},\) \(t\ge 0\)

40. \(x=-2\cos \ t,\) \(y=6\ \sin \ t,\) \(0\le t\le \pi\)

अपने आप की कोशिश करें

Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

अधिक में Trigonometry

The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions