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Math Help - Trigonometry-related calculus questions

  1. #1
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    Trigonometry-related calculus questions

    I would appreciate if someone could check my solutions for the following:

    find dy/dx of y=sqrt(sec(5x))
    my answer is (5/2)(sec5xtan5x)^(-1/2)

    find dy/dx of y+sin(y)=x
    my answer is (1/(1+cos(y)))

    Thank you for looking , thank you even more if you help!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turillian@gmail.com View Post
    I would appreciate if someone could check my solutions for the following:

    find dy/dx of y=sqrt(sec(5x))
    my answer is (5/2)(sec5xtan5x)^(-1/2)

    find dy/dx of y+sin(y)=x
    my answer is (1/(1+cos(y)))

    Thank you for looking , thank you even more if you help!
    the first is incorrect i'm afraid.
    Attached Thumbnails Attached Thumbnails Trigonometry-related calculus questions-chain.gif  
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  3. #3
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    Any chance I could get a pointer on where I went wrong with the first one?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turillian@gmail.com View Post
    Any chance I could get a pointer on where I went wrong with the first one?
    better yet, i gave you the solution

    do you see where you went wrong? you forgot to put back the sec(5x) when you differentiated. and the sec(5x)tan(5x) is not the part that should be raised to the power
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  5. #5
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    The image didn't load the first time I viewed your response, I thereby withdraw my previous comment, and submit a new one in it's place:
    Thanks a lot

    I have another question, what software do you use to type in all the math formulas? It seems to work pretty well
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turillian@gmail.com View Post
    The image didn't load the first time I viewed your response, I thereby withdraw my previous comment, and submit a new one in it's place:
    Thanks a lot
    it was my bad, i forgot to post the image, when i went back and looked at the post i noticed it wasn't there, so i uploaded it then, sorry
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  7. #7
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    Related rates calculus question.

    I'm having trouble getting the following word question:

    A police officer in a patrol car is approaching an intersection at 25m/s. When he is 210 m from the intersection, a truck crosses the intersection travelling at right angles to the police car's path at a rate of 25m/s. If the oficer focusses his spotlight on the truck, how fast is the light beam turning 3 seconds later assuming that both vehicles continue at their same rates.

    I would appreciate if someone could take a quick look at this question, maybe just give me a hint of what kind of approach to take, I have a feeling I can figure it out once I have a start.

    The following two I thought I had figured out, but then I checked my answer with a graphing calculator and it didn't make any sense.

    Find dy/dx of 1-2cos^2(x) = 4sin(x)cos(x)+y^2
    I simplified this one to sin^2(x) = sin(4x)+y^2
    and then tried to differentiate from there but it didn't work out, and I tried to differentiate the original as well. Was the original simplification incorrect?

    Find dy/dx of (1/(1+cos(x)))
    I tried this out and had something that looked good, but then I checked with a calculator and the derivative didn't mirror the slope of the equation at all.

    Thanks for looking. Hopefully I can get all this stuff figured out soon and answer some questions for other people.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by turillian@gmail.com View Post
    I'm having trouble getting the following word question:

    A police officer in a patrol car is approaching an intersection at 25m/s. When he is 210 m from the intersection, a truck crosses the intersection travelling at right angles to the police car's path at a rate of 25m/s. If the oficer focusses his spotlight on the truck, how fast is the light beam turning 3 seconds later assuming that both vehicles continue at their same rates.

    I would appreciate if someone could take a quick look at this question, maybe just give me a hint of what kind of approach to take, I have a feeling I can figure it out once I have a start.
    you asked for a hint, so that's what i'll give. see the diagram below, this is a related rates problem.

    Find dy/dx of 1-2cos^2(x) = 4sin(x)cos(x)+y^2
    I simplified this one to sin^2(x) = sin(4x)+y^2
    and then tried to differentiate from there but it didn't work out, and I tried to differentiate the original as well. Was the original simplification incorrect?
    yes, the simplification was incorrect.

    1 - 2cos^2(x) = 4sin(x)cos(x) + y^2 ..........since cos(2x) = 2cos^2(x) - 1 = -(1 - 2cos^2(x)) and sin(2x) = 2sin(x)cos(x)
    => -cos(2x) = 2sin(2x) + y^2
    => y^2 = -cos(2x) - 2sin(2x)
    try differentiating now

    Find dy/dx of (1/(1+cos(x)))
    y = 1/(1 + cos(x)) = (1 + cos(x))^-1
    By the chain rule
    => dy/dx = -(1 + cos(x))^-2 * (-sin(x))
    => dy/dx = sin(x)(1 + cos(x))^-2
    Attached Thumbnails Attached Thumbnails Trigonometry-related calculus questions-rate.gif  
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  9. #9
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    I couldn't ask for better help, thank you!
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  10. #10
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    I'm working on that word problem currently. I would appreciate if someone could take a quick look at my solutions to the following problems. I have a feeling I messed something up, because I'm good at messing things up.

    dy/dx of y=[csc^2(x)]-[3x^2][cot^2(x)]
    dy/dx=2(cscx)d/dx(cscx) - 6x(2(cotx))d/dx(cotx)
    dy/dx=2(cscx)(-cscxcotx) - 12x(cotx)(-csc^2(x))

    dy/dx of y = csc((x^2-2)^3)
    dy/dx= [-csc(x^2-2)^3][cot(x^2-2)^3][d/dx(x^2-2)^3]
    dy/dx= [-csc(x^2-2)^3][cot(x^2-2)^3][3(x^2-2)^2][d/dx(x^2)]
    dy/dx= [-csc(x^2-2)^3][cot(x^2-2)^3][3(x^2-2)^2][2x]
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  11. #11
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    Re:

    Re:
    Attached Thumbnails Attached Thumbnails Trigonometry-related calculus questions-dx.gif  
    Last edited by qbkr21; May 12th 2007 at 01:05 PM.
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  12. #12
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    Re:

    Re:
    Attached Thumbnails Attached Thumbnails Trigonometry-related calculus questions-dx-2.gif  
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  13. #13
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    Quote Originally Posted by qbkr21 View Post
    Re:
    There is nothing wrong with his notation of dy/dx. He specified what the function y(x) is. Basically what you are objecting to is the difference between
    [d/dx](y)
    and
    d(y)/dx

    There is no difference.

    -Dan
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  14. #14
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    Thanks for fixing my solution. I'm still a little bit fuzzy on the specifics of leibniz notation, thanks for the clarification.
    Last edited by turillian@gmail.com; May 14th 2007 at 08:59 AM.
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