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2 years ago

Let v1=(1, 6) and v2=(5, -3)

Mathematics

1 answer:

astraxan [27]2 years ago

3 0

Answer:

See Explanation;

Step-by-step explanation:

The given vectors are;

$v_1=\:\:$ and $v_2=\:\:$

a) To add or subtract two vectors, we add or subtract their corresponding components

$v_1+v_2=\:\:+\:\:$

$v_1+v_2=\:\:$

$v_1-v_2=\:\:-\:\:$

$v_1-v_2=\:\:$

$v_2-v_1=\:\:-\:\:$

$v_2-v_1=\:\:$

b) See attachment for graphs(1,6

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A. Find the amplitude.

Feliz [49]

Answers:

a) Amplitude = 2
b) Period = pi
c) Vertical shift = -2, which means it has been shifted down 2 units.
d) Horizontal shift = 3pi/8, this shifting is to the right.
e) There is  one  cycle between 0 and 2pi.
f) The equation of the graph is y = 2*sin(2(x-3pi/8))-2

========================================================

Explanations:

Part (a)

The highest point is when y = 0 and the lowest point is when y = -4. The vertical distance between the peak and valley is 4 units, which cuts in half to 2. This is the amplitude. It's the vertical distance from the center to either the peak or valley.

Note: Amplitude is always positive as it measures a distance.

---------------------

Part (b)

For x > 0, the first valley or lowest point occurs between 0 and pi/4. It appears to be the midpoint of the two values. So that would be (0+pi/4)/2 = pi/8.

The next valley occurs between pi and 5pi/4. Compute the midpoint to get (pi+5pi/4)/2 = (4pi/4+5pi/4)/2 = (9pi/4)/2 = 9pi/8

So we have the graph go from one valley x = pi/8 to the next valley over x = 9pi/8. This is a distance of pi units because 9pi/8-pi/8 = 8pi/8 = pi

The graph repeats itself every pi units, so the period is pi.

---------------------

Part (c)

The midline is normally through y = 0, aka the x axis. However, the graph shows the midline is through y = -2. This means the graph has been shifted down 2 units.

---------------------

Part (d)

This will depend on whether you use sine or cosine. This is entirely because cosine is a phase-shifted version of sine, and vice versa. I'll go with sine.

The parent sine function y = sin(x) goes through the origin (0,0)

However, as part (c) mentioned, we shifted the graph 2 units down. So we have y = sin(x)-2. But plugging x = 0 into this leads to the point (0,-2)

This doesn't match what the graph says. The graph shows the point (3pi/8, -2) on the red curve. The x coordinate 3pi/8 is the midpoint of pi/4 and pi/2

This must mean we need to shift the sine graph 3pi/8 units to the right.

---------------------

Part (e)

Start at the lowest point when x = pi/8. If you start the cycle here, then it ends when x = 9pi/8. See part (b).

So far we've completed one cycle. If we start at x = 9pi/8, then the next valley or lowest point is slightly beyond or to the right of x = 2pi. This means we run out of room and we haven't completed a full cycle.

Overall, one full cycle is between 0 and 2pi.

---------------------

Part (f)

Again I'm going to use sine instead of cosine. Refer back to part (d).

The general sine curve equation is

y = A*sin(B(x-C))+D

where

|A| = amplitude
B handles the period, specifically T = 2pi/B where T is the period. We can solve for B to get B = 2pi/T
C = horizontal phase shift
D = vertical shift, and ties together with the midline equation

In this case, we found that