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Math Help - Curves. tangent lines, etc.

  1. #1
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    Curves. tangent lines, etc.

    Consider the curve defined by -8x^2+5xy+y^3=-149

    I know that dy/dx = (16x-5y)/(5x+3y^2)
    I also know that the equation of the line tangent to the curve at point (4,-1) is y+1=3(x-4).

    However, now I'm asked the following..

    1. There's a number k so that the point (4.2,k) is on the curve. Using the tangent line, approximate the value of k.
    2. Write an equation that can be solved to find the actual value of k so that point (4.2,k) is on the curve.
    3. Solve the equation found in part (d) for the actual value of k.


    I don't even know where to BEGIN. I'm not looking for the answer, I'm looking for someone patient enough to teach me and guide me to the solution.
    Please help, thanks.
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  2. #2
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    There's a number k so that the point (4.2,k) is on the curve. Using the tangent line, approximate the value of k.
    y + 1 = 3(x - 4)

    k + 1 = 3(4.2 - 4)

    k \approx \, ?


    Write an equation that can be solved to find the actual value of k so that point (4.2,k) is on the curve
    -8(4.2)^2 + 5(4.2)k + k^3 = -149


    Solve the equation found in part (d) for the actual value of k.
    get out your calculator and solve for k ... you should get a value close to your estimate.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    y + 1 = 3(x - 4)

    k + 1 = 3(4.2 - 4)

    k \approx \, ?


    -8(4.2)^2 + 5(4.2)k + k^3 = -149


    get out your calculator and solve for k ... you should get a value close to your estimate.

    Thank you SO much. I mean, you pretty much worked out the entire problem for me, but I understand what you did perfectly.
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