Any Math Wizards Out There?

General Forums Off-topic Chat
1

this concerns high school level math, and graphing trig functions

I’m stuck :confused::slight_smile:

you are given a graph of a cosine function, minimum point at (8 pi /3, -2), maximums at (2 pi/3, 4) and (14 pi /3, 4)

(basically looks like a “v” just right of the y axis)

the function is of the form

y = a cos(bx + c) + d

you are asked to find a, b, c, and d, and the smallest positive zero of the function

so I’ve found

a = 3

b = 1/2

c = -2 pi /3

d = 1

so the curve crosses the x axis somewhere between 4 pi/3 and 8 pi /3

how can I find that exact point?

thanks for reading:cool:

2

To find the zero?

Setup an equation:
3cos(.5x + -2pi/3) + 1 = 0

and solve for x.

ORRRR
Since it’s beginning trig, you can probably get away with entering another line on your graphing calc that says y = 0, then find the intersection with your sinusoid and that’s where it crosses zero.

3

And now that I check it, it looks like you messed up your a,b,c,d values. Lemme try it…

1 Like
4

Ok, what happened is that b has to be applied to all of the input i.e.
instead of “a cos(bx + c) + d”, it’s “a cos(b[x + c]) + d” or as we learned it:
“a cos b(x + c) + d”

Also I’m too lazy to edit my post so I’m double and triple posting yes I know.

5

K, and your zero is at 5.91566. I don’t know if that’s a pi approximation, but there ya go.

6

thanks vk :smiley:

gimme a few minutes to digest your posts

7

And to get the zero, analytically…

  1. 3cos(.5[x -2pi/3]) + 1 = 0
  2. 3cos(.5[x -2pi/3]) = -1
  3. cos(.5[x -2pi/3]) = -1/3
  4. .5(x - 2pi/3) = arccos-1/3
  5. x - 2pi/3 = 2arccos-1/3
  6. x = 2arccos-1/3 + 2pi/3
  7. x = approx 5.91566

Keep in mind that by arccos, I mean the cos^-1 button on your calculator, the inverse cosine function. I’m giving you the “first” value of it right now because I don’t think you guys have gotten to inverse trig functions judging by the ppl I’ve tutored at school.

8

your answer looks correct since the graph crosses just left of 6 pi/3:D

looking at y= a cos(bx + c) + d

and if y = 0

is there a “clean way” to set this thing to x=

the algebra has me :confused:

many thanks for the help:cool::D:cool:

9

you already answered my question

many thanks:D:D:D

10

As for the algebra, here’s an annotated rehash of what I did:


1.  3cos(.5[x -2pi/3]) + 1 = 0  - Given equation
2.  3cos(.5[x -2pi/3]) = -1     - Subtract 1 from both sides
3. cos(.5[x -2pi/3]) = -1/3     -  Divide both sides by 3
4. .5(x - 2pi/3) = arccos-1/3   - Take inverse cosine of both sides to nullify cos
5. x - 2pi/3 = 2arccos-1/3      - Multiply both sides by 2
6. x = 2arccos-1/3 + 2pi/3      - Add 2pi/3 to both sides
7. x = approx 5.91566           - Use calculator (cos^-1 button for arccos) to approximate values and add
11

And REMEMBER:
it’s “y= a cos b(x + c) + d”

NOT “y= a cos(bx + c) + d”

12

thanks again, its much clearer now:D

13

sending you a pm:eek:

14

woah:eek:, it’s been awhile for me. valarking, your the man! I hope your putting those skills into some new blender functionality:) I repeat what I said on another thread. “What do you know, the Valarking guy is a nice helpful guy after all.”