Results 1 to 3 of 3

Math Help - Finding the gradient

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Finding the gradient

    Hi
    I few question i am having troubling finding the gradient:

    1) Given that the curve y=ax^2+\frac{b}{x} has a gradient of -5 at the point (2, -2) find the value of a and b.

    this is how i approached it:
    \frac{dy}{dx}=2ax+bx^-2
    when x=2
    \frac{dy}{dx}= 4a-\frac{b}{4}=-5 (1)

    when x=2,y=-2
    \frac{dy}{dx}= 4a-\frac{b}{4}=-2 (2)


    (1)-(2)
    this is where a got stuck because when i added and minus it didn't work.

    2)Find the gradient of the curve y=\frac{2x-4}{x^2} at the point where the curve crosses the x-axis.

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by Paymemoney View Post
    Hi
    I few question i am having troubling finding the gradient:

    1) Given that the curve y=ax^2+\frac{b}{x} has a gradient of -5 at the point (2, -2) find the value of a and b.

    this is how i approached it:
    \frac{dy}{dx}=2ax+bx^-2
    when x=2
    \frac{dy}{dx}= 4a-\frac{b}{4}=-5 (1)

    when x=2,y=-2
    \frac{dy}{dx}= 4a-\frac{b}{4}=-2 (2)


    (1)-(2)
    this is where a got stuck because when i added and minus it didn't work.

    2)Find the gradient of the curve y=\frac{2x-4}{x^2} at the point where the curve crosses the x-axis.

    P.S
    1) You know that (2, -2) lies on the curve y = ax^2 + \frac{b}{x}.

    So -2 = a(2)^2 + \frac{b}{2}

    -2 = 4a + \frac{b}{2}

    -4 = 8a + b

    b = -8a - 4.


    You also know that the gradient is -5 at this point.

    So \frac{dy}{dx} = 2ax - \frac{b}{x^2}

    -5 = 2a(2) - \frac{b}{2^2}

    -5 = 4a - \frac{b}{4}

    -20 = 16a - b

    b = 16a - 20.


    But since b = -8a - 4 as well, we have

    -8a - 4 = 16a - 20

    16 = 24a

    a = \frac{2}{3}.


    And since b = -8a - 4

    b = -8\left(\frac{2}{3}\right) - 4

    b = -\frac{16}{3} - 4

    b = -\frac{16}{3} - \frac{12}{3}

    b = -\frac{28}{3}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    Quote Originally Posted by Paymemoney View Post
    Hi
    I few question i am having troubling finding the gradient:

    1) Given that the curve y=ax^2+\frac{b}{x} has a gradient of -5 at the point (2, -2) find the value of a and b.

    this is how i approached it:
    \frac{dy}{dx}=2ax+bx^-2
    when x=2
    \frac{dy}{dx}= 4a-\frac{b}{4}=-5 (1)

    when x=2,y=-2
    \frac{dy}{dx}= 4a-\frac{b}{4}=-2 (2)


    (1)-(2)
    this is where a got stuck because when i added and minus it didn't work.

    2)Find the gradient of the curve y=\frac{2x-4}{x^2} at the point where the curve crosses the x-axis.

    P.S
    2) the curve crosses the x-axis at the point where y=0. find dy/dx at those point.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding The Gradient Of A Matrix Equation
    Posted in the Calculus Forum
    Replies: 12
    Last Post: May 9th 2011, 07:57 PM
  2. Replies: 1
    Last Post: March 3rd 2010, 05:02 AM
  3. Finding Gradient
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 14th 2009, 01:54 AM
  4. Finding the gradient function.... lost
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 29th 2009, 06:01 AM
  5. [SOLVED] finding the gradient vector
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 27th 2009, 08:15 PM

Search Tags


/mathhelpforum @mathhelpforum