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Math Help - Crazy AP style derivative problems

  1. #1
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    Crazy AP style derivative problems

    Big homework assignment this week, and luckily i was able to do most of the problems. However, there are these 3 that i just can't go right.

    1. Equation is 2(y^3) + 6(x^2)y- 12(x^2) + 6y = 1

    a. Show that the derivative of the above equation is equal to (4x -2xy)/(x^2 + y^2 + 1)

    b. Write an equation of each horizontal tangent line to the curve.

    c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates of point P.


    2. Equation is x^2 + 4(y^2) = 7 = 3xy

    a. Show that the derivative of the above equation is equal to (3y - 2x)/(8y - 3x)

    b. Show that there is a point P with x coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y coordinate of P.

    c. Find the value of the second derivative at the point P found in part b. Does the curve have a local maximum, a local minimum, or neither at the point P? Justify.


    3. Equation is x(y^2) - (x^3)y = 6

    a. Show that the derivative of the above euation is equal to (3(x^2)y - y^2)/(2xy - x^3)

    b. Find all the points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

    c. Find the x coordinate of each point on the curve where the tangent line is vertical.


    Not easy, i know! Hopefully someone can help me out with some of these. Thanks in advance!!!!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    I will give you the derivatives of each

    Quote Originally Posted by Blue Griffin View Post
    Big homework assignment this week, and luckily i was able to do most of the problems. However, there are these 3 that i just can't go right.

    1. Equation is 2(y^3) + 6(x^2)y- 12(x^2) + 6y = 1

    a. Show that the derivative of the above equation is equal to (4x -2xy)/(x^2 + y^2 + 1)

    b. Write an equation of each horizontal tangent line to the curve.

    c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates of point P.


    2. Equation is x^2 + 4(y^2) = 7 = 3xy

    a. Show that the derivative of the above equation is equal to (3y - 2x)/(8y - 3x)

    b. Show that there is a point P with x coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y coordinate of P.

    c. Find the value of the second derivative at the point P found in part b. Does the curve have a local maximum, a local minimum, or neither at the point P? Justify.


    3. Equation is x(y^2) - (x^3)y = 6

    a. Show that the derivative of the above euation is equal to (3(x^2)y - Y^2)/(2xy - x^3)

    b. Find all the points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

    c. Find the x coordinate of each point on the curve where the tangent line is vertical.


    Not easy, i know! Hopefully someone can help me out with some of these. Thanks in advance!!!!
    2y^3+6x^2y-12x^2+6y=1...then 6y^2\cdot{y'}+6x^2\cdot{y'}+12xy-24x+6y'=0...now solve for y'.... x^2+4y^2-7=3xy...then 2x+8y\cdot{y'}=3x\cdot{y'}+3y..once again solve for y'.... xy^2-x^3y=6...then 2xy\cdot{y'}+y^2-x^3\cdot{y'}+3x^2y=0....and solve for y'...I am sure you can do the rest
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Blue Griffin View Post
    Big homework assignment this week, and luckily i was able to do most of the problems. However, there are these 3 that i just can't go right.

    1. Equation is 2(y^3) + 6(x^2)y- 12(x^2) + 6y = 1

    a. Show that the derivative of the above equation is equal to (4x -2xy)/(x^2 + y^2 + 1)

    b. Write an equation of each horizontal tangent line to the curve.

    c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates of point P.


    2. Equation is x^2 + 4(y^2) = 7 = 3xy

    a. Show that the derivative of the above equation is equal to (3y - 2x)/(8y - 3x)

    b. Show that there is a point P with x coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y coordinate of P.

    c. Find the value of the second derivative at the point P found in part b. Does the curve have a local maximum, a local minimum, or neither at the point P? Justify.


    3. Equation is x(y^2) - (x^3)y = 6

    a. Show that the derivative of the above euation is equal to (3(x^2)y - y^2)/(2xy - x^3)

    b. Find all the points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

    c. Find the x coordinate of each point on the curve where the tangent line is vertical.


    Not easy, i know! Hopefully someone can help me out with some of these. Thanks in advance!!!!
    They are all simialr so lets do 3.

    x(y^2) - (x^3)y = 6

    using implicit differentation we get

    \underbrace{1 \cdot y^2+x(2y\frac{dy}{dx})}_{Product Rule}-(\underbrace{3x^2y+x^3\frac{dy}{dx}}_{Product Rule})=0

    (2xy-x^3)\frac{dy}{dx}=3x^2y-y^2 \iff \frac{dy}{dx}= \frac{3x^2y-y^2}{2xy-x^3}

    for b) eval at x=1

     1(y^2)-(1^3)y=6 \iff y^2-y-6=0 \iff (y-3)(y+2)=0

    so we get the points (1,3) and (1,-2)

    for c) set the derivative equal to zero.

    0=\frac{3x^2y-y^2}{2xy-x^3} \iff 0=3x^2y-y^2

    0=y(3x^2-y)

    so y=0 \mbox{ or } y=3x^2

    y=0 is not on the curve (why?)

    Now we can plug  y=3x^2 into the original equation.

    x(y^2) - (x^3)y = 6 \iff x(3x^2)^2-x^3(3x^2)=6

    9x^5-3x^5=6 \iff x^5=1 \iff x=1

    Try a few of the others and see what you can do.

    Good luck.
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  4. #4
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    Ok, your advice helped a lot, and i was able to figure out 2. It was easier than i thought....

    Need a bit of help with the rest of one, though. I don't really know where to start with b or c.
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by Blue Griffin View Post
    Ok, your advice helped a lot, and i was able to figure out 2. It was easier than i thought....

    Need a bit of help with the rest of one, though. I don't really know where to start with b or c.

    Horizontal tangents occur when the derivative is zero

    so set the derivative equal to zero and solve for the points

    you will need to use the similar proceedure that I used in 3 to find the points on the graph.

    set the derivative equal to -1 and see what you can do from there.

    Good luck.
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