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onegirl Group Title

Find an equation of the tangent line y = f(x) at x = a. f(x) = sin 4x, a = pi/8

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    Hmm I think you've done a few of these types of problems. So I'm just curious, which result makes more sense to you. Seeing our tangent line in slope-intercept form: \(\large y=mx+b\), Or in point-slope form: \(\large y-y_o=m(x-x_o)\).

    • one year ago
  2. onegirl Group Title
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    in slope intercept from

    • one year ago
  3. zepdrix Group Title
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    Ok we will have 2 pieces we need to find. \(\large m\) is the slope of our tangent line. Take the derivative of your function, and evaluate it at \(\large x=\pi/8\). That will be your \(\large m\) value.

    • one year ago
  4. onegirl Group Title
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    okay hold on

    • one year ago
  5. onegirl Group Title
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    when i find the derivative a = pi/8 is not int he equation right?

    • one year ago
  6. onegirl Group Title
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    the*

    • one year ago
  7. onegirl Group Title
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    i got sin(4)

    • one year ago
  8. zepdrix Group Title
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    Take the derivative of sin(4x) first. Don't plug a=pi/8 in until after you've taken a derivative.

    • one year ago
  9. onegirl Group Title
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    okay

    • one year ago
  10. onegirl Group Title
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    i got 4 cos(4x)

    • one year ago
  11. zepdrix Group Title
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    ok good, now plug in x=pi/8 :D

    • one year ago
  12. onegirl Group Title
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    ok

    • one year ago
  13. onegirl Group Title
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    i got 0 after i plugged in and solved

    • one year ago
  14. zepdrix Group Title
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    Ok sounds good. So we've determined that the slope \(\large m\) of our tangent line will be 0. \[\large y=mx+b \qquad \rightarrow \qquad y=0x+b\] Now we just need to find the \(\large b\) value.

    • one year ago
  15. zepdrix Group Title
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    To find it, we'll need to plug in a coordinate pair that falls on the line. Well, since our line is tangent to the curve at x=pi/8, we can plug x=pi/8 into that function to get a corresponding y value. This will be the coordinate pair we'll plug into our tangent line.

    • one year ago
  16. onegirl Group Title
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    ok

    • one year ago
  17. onegirl Group Title
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    so i should plug in pi/8 into y = mx + b?

    • one year ago
  18. zepdrix Group Title
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    no, plug it into your original function, before you took the derivative of it.

    • one year ago
  19. onegirl Group Title
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    ohh okay

    • one year ago
  20. onegirl Group Title
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    so sin 4(pi/8) ?

    • one year ago
  21. onegirl Group Title
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    okay so i got 1

    • one year ago
  22. zepdrix Group Title
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    So that gives us a coordinate pair of \(\large \left(\dfrac{\pi}{8},\;1\right)\). Plug that into your tangent function to solve for \(\large b\).

    • one year ago
  23. onegirl Group Title
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    ok so y = 0x + b so i should plug it in place of b?

    • one year ago
  24. zepdrix Group Title
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    No, your coordinate pair is \(\large (x,\;y)\). Plug it into those.

    • one year ago
  25. onegirl Group Title
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    ohh ok got it

    • one year ago
  26. onegirl Group Title
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    okay so 1 = 0(pi/8) + b

    • one year ago
  27. zepdrix Group Title
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    So what is your b value? :3

    • one year ago
  28. onegirl Group Title
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    Lol sorry i got 1 :)

    • one year ago
  29. onegirl Group Title
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    b = 1

    • one year ago
  30. zepdrix Group Title
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    Ok sounds good. So if we plug that missing \(\large b\) into our tangent line equation, we get,\[\large y=0x+1\]Which can be simplified to,\[\large y=1\]

    • one year ago
  31. onegirl Group Title
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    okay

    • one year ago
  32. zepdrix Group Title
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    I guess it's really 3 steps. We had to find ~\(\large m\) ~\(\large b\) ~A Coordinate Pair.

    • one year ago
  33. zepdrix Group Title
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    It's not too difficult, just a lot of silly letters :) Keep practicing!

    • one year ago
  34. onegirl Group Title
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    okay so our equation is complete? and yes just lot to do and thanks

    • one year ago
  35. zepdrix Group Title
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    Yes, that's our final answer.

    • one year ago
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