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Thread: Finding the gradient

  1. #1
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    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:
    $\displaystyle \frac{dy}{dx}=2ax+bx^-2$
    when x=2
    $\displaystyle \frac{dy}{dx}= 4a-\frac{b}{4}=-5 $(1)

    when x=2,y=-2
    $\displaystyle \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 $\displaystyle y=\frac{2x-4}{x^2}$ at the point where the curve crosses the x-axis.

    P.S
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  2. #2
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    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:
    $\displaystyle \frac{dy}{dx}=2ax+bx^-2$
    when x=2
    $\displaystyle \frac{dy}{dx}= 4a-\frac{b}{4}=-5 $(1)

    when x=2,y=-2
    $\displaystyle \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 $\displaystyle y=\frac{2x-4}{x^2}$ at the point where the curve crosses the x-axis.

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

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

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

    $\displaystyle -4 = 8a + b$

    $\displaystyle b = -8a - 4$.


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

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

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

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

    $\displaystyle -20 = 16a - b$

    $\displaystyle b = 16a - 20$.


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

    $\displaystyle -8a - 4 = 16a - 20$

    $\displaystyle 16 = 24a$

    $\displaystyle a = \frac{2}{3}$.


    And since $\displaystyle b = -8a - 4$

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

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

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

    $\displaystyle b = -\frac{28}{3}$.
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  3. #3
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    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:
    $\displaystyle \frac{dy}{dx}=2ax+bx^-2$
    when x=2
    $\displaystyle \frac{dy}{dx}= 4a-\frac{b}{4}=-5 $(1)

    when x=2,y=-2
    $\displaystyle \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 $\displaystyle y=\frac{2x-4}{x^2}$ at the point where the curve crosses the x-axis.

    P.S
    2) the curve crosses the x-axis at the point where $\displaystyle y=0$. find $\displaystyle dy/dx$ at those point.
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