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Math Help - Find eguation of tangent

  1. #1
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    Find eguation of tangent

    Find eguation of tangent to each curve at a given point:

    2x^2y^3 + 3xy+5=0 (1,-1)
    The only thing that throws me off is two variables

    y=2sin(x) + cos^2x (pi/6, 7/4)
    I am pretty sure I know how to do this, just want to double check.

    Thx in advance
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  2. #2
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    Quote Originally Posted by BobCalc View Post
    Find eguation of tangent to each curve at a given point:

    2x^2y^3 + 3xy+5=0 (1,-1)
    The only thing that throws me off is two variables

    Mr F says: Use implicit differentiation to get dy/dx. Then proceed in the usual way.

    y=2sin(x) + cos^2x (pi/6, 7/4)
    I am pretty sure I know how to do this, just want to double check. Mr F says: If you want your work double checked, please post that work (all of it, not just your answer).

    Thx in advance
    ..
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  3. #3
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    Mr F says: Use implicit differentiation to get dy/dx. Then proceed in the usual way.

    2x^2y^3 + 3xy+5
    Ok here is the dy/dx 12xy^3+ 3
    I just do not know what to do because there are 2 variables x and y, should I isolate for y?

    y=2sin(x) + cos^2x (pi/6, 7/4)

    dy/dx= 2cox - 2sinxcox

    After plugging in pi/6, I get root 3- root 3/2- This is the only thing I think I can be wrong at.

    And thanks for the fast response mr.F
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  4. #4
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    Quote Originally Posted by BobCalc View Post
    2x^2y^3 + 3xy+5
    Ok here is the dy/dx 12xy^3+ 3 Mr F says: Your expression for dy/dx is wrong. Have you been taught implicit differentiation? Note that y is treated as an implicit function of x. So you have to use things like chain rule and product rule here.

    Once you have the correct expression that gives dy/dx, you substitute x = 1 and y = -1 into it and solve for dy/dx.

    y=2sin(x) + cos^2x (pi/6, 7/4)

    dy/dx= 2cox - 2sinxcox

    After plugging in pi/6, I get root 3- root 3/2- This is the only thing I think I can be wrong at. Mr F says: This is correct and simplifies to {\color{blue}\frac{\sqrt{3}}{2}}. Please show all your working in getting the equation of the tangent.

    And thanks for the fast response mr.F
    ..
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  5. #5
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    Hopefully this differentiation is correct. Math is not my strong point in school, so sorry for making mistakes.

    4xy^3 + 6y^2x^2 +3y+3x=0

    I sub in In (1,-1)

    -4+6+3-3=0
    2

    So m is 2?
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  6. #6
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    I really need someone to check my differentiation, my teacher told me that similar question is going to be on the test, so I really need to know.

    Find eguation of tangent to each curve at a given point:

    2x^2y^3 + 3xy+5=0 (1,-1)

    4xy^3 + 6y^2x^2 +3y+3x=0

    I sub in In (1,-1)

    -4+6+3-3=0
    2

    So m is 2? Any help would be greatly appreciated
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  7. #7
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    Quote Originally Posted by BobCalc View Post
    I really need someone to check my differentiation, my teacher told me that similar question is going to be on the test, so I really need to know.

    Find eguation of tangent to each curve at a given point:

    2x^2y^3 + 3xy+5=0 (1,-1)

    4xy^3 + 6y^2x^2 +3y+3x=0

    I sub in In (1,-1)

    -4+6+3-3=0
    2

    So m is 2? Any help would be greatly appreciated
    I asked if you had been taught implicit differentiation. If you do not understand how to apply that technique, you are not going to be able to do these questions. Go back to your class notes and/or textbook and review all the examples you have been given. Here is one more example:

    2x^2y^3 + 3xy + 5 = 0

    Differentiate both sides with respect to x, treating y as an implicit function of x.

    The derivative of 2x^2 y^3 is 4x y^3 + 2x^2 3 y^2 \frac{dy}{dx} = 4x y^3 + 6 x^2 y^2 \frac{dy}{dx}.

    In the above I have used the product rule to differentiate (2x^2) (y^3) and I have used the chain rule to differentiate y^3. Note: \frac{d y^3}{dx} = \frac{d y^3}{dy} \cdot \frac{dy}{dx} = 3y^2 \frac{dy}{dx}.

    The derivative of 3xy is 3y + 3x \frac{dy}{dx}. I used the product rule here.


    So you end up with: 4x y^3 + 6 x^2 y^2 \frac{dy}{dx} + 3y + 3x \frac{dy}{dx} = 0.

    Substitute x = 1 and y = -1: -4 + \frac{dy}{dx} - 3 + 3 \frac{dy}{dx} = 0.

    Solve for \frac{dy}{dx}. That is your value of m.
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  8. #8
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    I am very sorry, I though you asked if I knew just differentiation. This is my first year taking calculus and I never learned implicit differentiation.Its not in my textbook either, because I am done with calc and there are no questions like that.

    Do I just go -4-1 -3-3

    Do I just sub in -1/1 for dy/dx

    Once again my bad
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  9. #9
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    Quote Originally Posted by BobCalc View Post
    I am very sorry, I though you asked if I knew just differentiation. This is my first year taking calculus and I never learned implicit differentiation.Its not in my textbook either, because I am done with calc and there are no questions like that.

    Do I just go -4-1 -3-3

    Do I just sub in -1/1 for dy/dx

    Once again my bad
    The last two lines of post #7 tell you what to do.
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