Parametric Functions and Vectors: Get Stronger Answer Key

Giselle Miller

Parametric Functions and Vectors: Get Stronger Answer Key

Parametric Equations

1. A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, [latex]x=f\left(t\right)[/latex] and [latex]y=f\left(t\right)[/latex].

2. Choose one equation to solve for [latex]t[/latex], substitute into the other equation and simplify.

3. [latex]y=-2+2x[/latex]

4. [latex]y=3\sqrt{\frac{x - 1}{2}}[/latex]

5. [latex]x=2{e}^{\frac{1-y}{5}}[/latex] or [latex]y=1 - 5ln\left(\frac{x}{2}\right)[/latex]

6. [latex]x=4\mathrm{log}\left(\frac{y - 3}{2}\right)[/latex]

7. [latex]x={\left(\frac{y}{2}\right)}^{3}-\frac{y}{2}[/latex]

8. [latex]y={x}^{3}[/latex]

9. [latex]{\left(\frac{x}{4}\right)}^{2}+{\left(\frac{y}{5}\right)}^{2}=1[/latex]

10. [latex]{y}^{2}=1-\frac{1}{2}x[/latex]

11. [latex]y={x}^{2}+2x+1[/latex]

12. [latex]y={\left(\frac{x+1}{2}\right)}^{3}-2[/latex]

13. [latex]y=-3x+14[/latex]

14. [latex]y=x+3[/latex]

15. [latex]\begin{array}{l}x\left(t\right)=t\hfill \\ y\left(t\right)=2\sin t+1\hfill \end{array}[/latex]

16. [latex]\begin{array}{l}x\left(t\right)=\sqrt{t}+2t\hfill \\ y\left(t\right)=t\hfill \end{array}[/latex]

17. [latex]\begin{array}{l}x\left(t\right)=4\cos t\hfill \\ y\left(t\right)=6\sin t\hfill \end{array}[/latex]; Ellipse

18. [latex]\begin{array}{l}x\left(t\right)=\sqrt{10}\cos t\hfill \\ y\left(t\right)=\sqrt{10}\sin t\hfill \end{array}[/latex]; Circle

19. [latex]\begin{array}{l}x\left(t\right)=-1+4t\hfill \\ y\left(t\right)=-2t\hfill \end{array}[/latex]

20. [latex]\begin{array}{l}x\left(t\right)=4+2t\hfill \\ y\left(t\right)=1 - 3t\hfill \end{array}[/latex]

21.

[latex]t[/latex]	[latex]x[/latex]	[latex]y[/latex]
1	-3	1
2	0	7
3	5	17

22. answers may vary: [latex]\begin{array}{l}x\left(t\right)=t - 1\hfill \\ y\left(t\right)={t}^{2}\hfill \end{array}\text{ and }\begin{array}{l}x\left(t\right)=t+1\hfill \\ y\left(t\right)={\left(t+2\right)}^{2}\hfill \end{array}[/latex]

23. answers may vary: , [latex]\begin{array}{l}x\left(t\right)=t\hfill \\ y\left(t\right)={t}^{2}-4t+4\hfill \end{array}\text{ and }\begin{array}{l}x\left(t\right)=t+2\hfill \\ y\left(t\right)={t}^{2}\hfill \end{array}[/latex]

Parametric Equations: Graphs

1. plotting points with the orientation arrow and a graphing calculator

2. The arrows show the orientation, the direction of motion according to increasing values of [latex]t[/latex].

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

4.
Graph of the given equations - looks like an upward opening parabola.

5.
Graph of the given equations - a line, negative slope.

6.
Graph of the given equations - looks like a sideways parabola, opening to the right.

7.
Graph of the given equations - looks like the left half of an upward opening parabola.

8.
Graph of the given equations - looks like a downward opening absolute value function.

9.
Graph of the given equations - a vertical ellipse.

10.
Graph of the given equations- line from (0, -3) to (3,0). It is traversed in both directions, positive and negative slope.

11.
Graph of the given equations- looks like an upward opening parabola.

12.
Graph of the given equations- looks like a downward opening parabola.

13.
Graph of the given equations- horizontal ellipse.

14.
Graph of the given equations- looks like the lower half of a sideways parabola opening to the right

15.
Graph of the given equations- looks like an upwards opening parabola

16.
Graph of the given equations- looks like the upper half of a sideways parabola opening to the left

17. Take the opposite of the [latex]x\left(t\right)[/latex] equation.

18. [latex]y\left(x\right)=-16{\left(\frac{x}{15}\right)}^{2}+20\left(\frac{x}{15}\right)[/latex]

19. [latex]\begin{cases}x\left(t\right)=64t\cos \left(52^\circ \right)\\ y\left(t\right)=-16{t}^{2}+64t\sin \left(52^\circ \right)\end{cases}[/latex]

20. approximately 3.2 seconds

21. 1.6 seconds

Vectors

1. They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.

2. Component form is a way to write a vector using its horizontal and vertical distances.

3. The first number always represents the coefficient of the i, and the second represents the j.

4. [latex]\langle 7,−5\rangle[/latex]

5. not equal

6. equal

7. equal

8. [latex]7\boldsymbol{i}−3\boldsymbol{j}[/latex]

9. [latex]−6\boldsymbol{i}−2\boldsymbol{j}[/latex]

10. [latex]\boldsymbol{u}+\boldsymbol{v}=\langle−5,5\rangle,\boldsymbol{u}−\boldsymbol{v}=\langle−1,3\rangle,2\boldsymbol{u}−3\boldsymbol{v}=\langle 0,5\rangle[/latex]

11. [latex]−10\boldsymbol{i}–4\boldsymbol{j}[/latex]

12. [latex]−\frac{2\sqrt{29}}{29}\boldsymbol{i}+\frac{5\sqrt{29}}{29}\boldsymbol{j}[/latex]

13. [latex]–\frac{2\sqrt{229}}{229}\boldsymbol{i}+\frac{15\sqrt{229}}{229}\boldsymbol{j}[/latex]

14. [latex]–\frac{7\sqrt{2}}\boldsymbol{i}+\frac{\sqrt{2}}{10}\boldsymbol{j}[/latex]

15. [latex]|\boldsymbol{v}|=7.810,\theta=39.806^{\circ}[/latex]

16. [latex]|\boldsymbol{v}|=7.211,\theta=236.310^{\circ}[/latex]

17. −6

18. −12

19.

20.
Plot of u+v, u-v, and 2u based on the above vectors. In relation to the same origin point, u+v goes to (0,3), u-v goes to (2,-1), and 2u goes to (2,2).

21.
Plot of vectors u+v, u-v, and 2u based on the above vectors.Given that u's start point was the origin, u+v starts at the origin and goes to (2,-3); u-v starts at the origin and goes to (4,-1); 2u goes from the origin to (6,-4).