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onegirl

  • 3 years ago

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

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  1. zepdrix
    • 3 years ago
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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: \(\large y-y_o=m(x-x_o)\).

  2. onegirl
    • 3 years ago
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    in slope intercept from

  3. zepdrix
    • 3 years ago
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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.

  4. onegirl
    • 3 years ago
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    okay hold on

  5. onegirl
    • 3 years ago
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    when i find the derivative a = pi/8 is not int he equation right?

  6. onegirl
    • 3 years ago
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    the*

  7. onegirl
    • 3 years ago
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    i got sin(4)

  8. zepdrix
    • 3 years ago
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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.

  9. onegirl
    • 3 years ago
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    okay

  10. onegirl
    • 3 years ago
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    i got 4 cos(4x)

  11. zepdrix
    • 3 years ago
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    ok good, now plug in x=pi/8 :D

  12. onegirl
    • 3 years ago
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    ok

  13. onegirl
    • 3 years ago
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    i got 0 after i plugged in and solved

  14. zepdrix
    • 3 years ago
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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.

  15. zepdrix
    • 3 years ago
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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.

  16. onegirl
    • 3 years ago
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    ok

  17. onegirl
    • 3 years ago
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    so i should plug in pi/8 into y = mx + b?

  18. zepdrix
    • 3 years ago
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    no, plug it into your original function, before you took the derivative of it.

  19. onegirl
    • 3 years ago
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    ohh okay

  20. onegirl
    • 3 years ago
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    so sin 4(pi/8) ?

  21. onegirl
    • 3 years ago
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    okay so i got 1

  22. zepdrix
    • 3 years ago
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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 \(\large b\).

  23. onegirl
    • 3 years ago
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    ok so y = 0x + b so i should plug it in place of b?

  24. zepdrix
    • 3 years ago
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    No, your coordinate pair is \(\large (x,\;y)\). Plug it into those.

  25. onegirl
    • 3 years ago
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    ohh ok got it

  26. onegirl
    • 3 years ago
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    okay so 1 = 0(pi/8) + b

  27. zepdrix
    • 3 years ago
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    So what is your b value? :3

  28. onegirl
    • 3 years ago
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    Lol sorry i got 1 :)

  29. onegirl
    • 3 years ago
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    b = 1

  30. zepdrix
    • 3 years ago
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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,\[\large y=1\]

  31. onegirl
    • 3 years ago
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    okay

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

  33. zepdrix
    • 3 years ago
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    It's not too difficult, just a lot of silly letters :) Keep practicing!

  34. onegirl
    • 3 years ago
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    okay so our equation is complete? and yes just lot to do and thanks

  35. zepdrix
    • 3 years ago
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    Yes, that's our final answer.

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