Math

posted by .

Show that the tangent to the curve 25x^5+5x^4-45x^3-5x^2+2x+6y-24=0 at the point (-1,1) is also a normal at two points of the curve.

  • Math -

    whewww!

    125x^4 + 20x^3 - 135x^2 - 10x + 2 + 6dy/dx = 0
    at (-1,1)
    125 - 20 - 135 + 10 - 2 + 6dy/dx = 0

    6dy/dx = 22, so the slope of the tangent at (-1,1) is 22/6 = 11/3
    equation of tangent:
    1= (11/3)(-1) + b
    b = 1 + 11/3 = 14/3

    y = (11/3)x + 14/3 = (11x+14)/3

    sub that into the original equation
    25x^5+5x^4-45x^3-5x^2+2x+6y-24=0
    25x^5+5x^4-45x^3-5x^2+2x+6(11x+14)/3-24=0
    25x^5+5x^4-45x^3-5x^2 + 24x + 4=0

    after dividing out the factor (x+1) we are left with
    25x^4 - 20x^3 - 25x^2 +20x+4=0

    tried to solve this, even had Wolfram page try it
    http://www.wolframalpha.com/input/?i=25x%5E4+-+20x%5E3+-+25x%5E2+%2B20x%2B4%3D0

    it gave me x = appr -.9507 and appr -.169872

    If we sub those values into the derivative equation we should get -3/11 or -.2727..
    I did not get that.

    Can't find my error
    OR,
    those two points don't exist
    (you might print out my solution and go through it to see if find any errors)

  • Math -

    -6y = 25x^5+5x^4-45x^3-5x^2+2x+24
    y' = -1/6 (125x^4 + 20x^3 - 135x^2 - 10x + 2)
    y'(-1) = 3
    normal has slope -1/3

    Go to http://rechneronline.de/function-graphs/

    and plot

    -1/6 * (25x^5+5x^4-45x^3-5x^2+2x-24)
    and its derivative
    as well as
    3x+4

    The line is not normal to the curve anywhere. Typo somewhere here?

  • Math - PS -

    When plotting the graphs, set
    x: -2 to 2
    y: -10 to 10

    that will show things quite clearly.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. 12th Grade Calculus

    1. a.) Find an equation for the line perpendicular to the tangent curve y=x^3 - 9x + 5 at the point (3,5) [* for a. the answer that I obtained was y-5 = -1/18 (x-3) ] b.) What is the smallest slope on the curve?
  2. Calculus

    Consider the curve y^2+xy+x^2=15. What is dy/dx?
  3. calculus!URGENT

    Show that the tangent line to the curve y=x^3 at the point x=a also hits the curve at the point x=-2a. Any help?
  4. calculus

    1. Given the curve a. Find an expression for the slope of the curve at any point (x, y) on the curve. b. Write an equation for the line tangent to the curve at the point (2, 1) c. Find the coordinates of all other points on this curve …
  5. calculus

    Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show that dy/dx= (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 …
  6. Calculus

    Please help this is due tomorrow and I don't know how to Ive missed a lot of school sick Consider the curve given by the equation x^3+3xy^2+y^3=1 a.Find dy/dx b. Write an equation for the tangent line to the curve when x = 0. c. Write …
  7. Business Statistics

    In each case, sketch the two specified normal curves on the same set of axes: a A normal curve with m 20 and s 3, and a normal curve with m 20 and s 6. b A normal curve with m 20 and s 3, and a normal curve with m 30 and s 3. …
  8. Calculus

    Consider the curve given by y^2 = 2+xy (a) show that dy/dx= y/(2y-x) (b) Find all points (x,y) on the curve where the line tangent to the curve has slope 1/2. (c) Show that there are now points (x,y) on the curve where the line tangent …
  9. Calculus AB

    Could someone please help me with these tangent line problems?
  10. Math

    . Given that x²cos y_sin y=0,(0,π). A. Verify that the given points on the curve. B.use implicit differention to find the slope of the above curve at the given point. C.find the equation of tangent and normal to the curve at that.

More Similar Questions