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El_Tucan

  • 3 years ago

find all values in [0,pi) so that dy/dx = 0 for the function y = sin(x^2)

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  1. myininaya
    • 3 years ago
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    Are you having issues finding y'?

  2. myininaya
    • 3 years ago
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    Do you know the chain rule?

  3. waterineyes
    • 3 years ago
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    Ma'am will explain you, don't worry..

  4. El_Tucan
    • 3 years ago
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    yeah i get using the chain rule to find the derivative but i don't know what is meant by "finding values so that dy/dx = 0"

  5. myininaya
    • 3 years ago
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    y'=f(x) y'=0 implies you set f(x)=0 and solve for x

  6. myininaya
    • 3 years ago
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    Just set the derivative equal to zero and solve for x

  7. myininaya
    • 3 years ago
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    on the interval [0,pi)

  8. El_Tucan
    • 3 years ago
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    i have y' = 2x * cos(x^2) correct? so 2x * cos(x^2) = 0?

  9. myininaya
    • 3 years ago
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    What did you get for y'?

  10. myininaya
    • 3 years ago
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    Right! :)

  11. myininaya
    • 3 years ago
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    So we have x=0 or cos(x^2)=0

  12. myininaya
    • 3 years ago
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    You need to solve cos(x^2)=0 for x on [0,pi)

  13. El_Tucan
    • 3 years ago
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    i plugged cos(0^2) in my graphing calculator which is 1, is that right? im sorry :(

  14. waterineyes
    • 3 years ago
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    You are to find values for which it becomes 0 not 1..

  15. myininaya
    • 3 years ago
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    For what values satisfy cos(u)=0 look at unit circle.

  16. El_Tucan
    • 3 years ago
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    gotcha, thanks :)

  17. waterineyes
    • 3 years ago
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    Really??

  18. myininaya
    • 3 years ago
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    You are actually solving cos(u)=0 for u on 0<u<=pi^2 Keep in mind that pi^2 is between 3pi and 7 pi/2 So you actually want to find values for u that satisfy cos(u)=0 on 0<u<3pi

  19. El_Tucan
    • 3 years ago
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    sorry waterineyes ;) i got it, thank you myininaya, the square roots of pi/2, 8pi/2, & 5pi/2 are the zero's within that interval

  20. waterineyes
    • 3 years ago
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    Sorry for what dear?? I was just confirming that...

  21. El_Tucan
    • 3 years ago
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    i thought you were picking on me :) im working on my math esteem lol

  22. waterineyes
    • 3 years ago
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    There is nothing like that, I was just confirming that you got it or not...

  23. myininaya
    • 3 years ago
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    Hey @El_Tucan that 8pi/2 should be something else

  24. myininaya
    • 3 years ago
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    Those other two values you got were right with the little square root thingy that you said :)

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