Results 1 to 5 of 5

Thread: Need help to understand how the slope was derived in this problem

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    12

    Need help to understand how the slope was derived in this problem

    I am kind of struggling to understand how the slope is derived in the following problem:
    "Find the equation of the tangent line to the curve y = x^2 through the point (7, 49)."

    The answer is that the slope is 2x but I don't understand how they got there since there is only one set of points given.

    Thank you for the help,

    Alex
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    $\displaystyle y'=\mbox{exponent}*x^{\mbox{exponent -1}}$

    $\displaystyle y'=2x^{2-1}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    The slope at any point $\displaystyle \displaystyle x$ on the curve $\displaystyle \displaystyle y = x^2$ is given by $\displaystyle \displaystyle \frac{dy}{dx} = 2x$.

    So at the point $\displaystyle \displaystyle x = 7$, the slope is $\displaystyle \displaystyle 2\cdot 7 = 14$.


    Equation of tangent: $\displaystyle \displaystyle y = mx + c$.

    You have a coordinate that lies on your tangent $\displaystyle \displaystyle (x, y) = (7, 49)$ and the slope $\displaystyle \displaystyle m = 14$. Substitute and solve for $\displaystyle \displaystyle c$, then you can write the equation of your tangent line.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    How well do you understand the concept behind differentiation?

    If $\displaystyle y=x^2,$

    $\displaystyle \displaystyle\frac{dy}{dx} = 2x$

    $\displaystyle
    \displaystyle\frac{dy}{dx}$ is 'the rate of change of y with respect to x'. It is used to tell us the equation of the gradient of the curve at any point.

    In the question, $\displaystyle \frac{dy}{dx} = 2x$. So take any point on your curve. Let's say $\displaystyle (3,9)$ This has x co-ordinate 3.

    Substituting this into $\displaystyle \frac{dy}{dx} = 2x$ gives $\displaystyle \frac{dy}{dx} = 6$. What this means is that at the point $\displaystyle (3,9)$ the gradient of the curve $\displaystyle y=x^2$ is 6.

    Also, I don't think that $\displaystyle 2x$ is the answer to that question.

    We have a point (7,49) and we know that $\displaystyle \frac{dy}{dx} = 2x$. So the gradient of the tangent is $\displaystyle 2\times 7 = 14$.

    So we know that when x = 7, y = 14 and m is 14. Substitute this into y = mx + c to find c and you'll find the equation of the tangent.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    from first principles ...

    $\displaystyle \displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

    $\displaystyle f(x) = x^2$

    $\displaystyle f(x+h) = (x+h)^2$

    $\displaystyle \displaystyle f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$

    $\displaystyle \displaystyle f'(x) = \lim_{h \to 0} \frac{x^2+2xh+h^2 - x^2}{h}$

    $\displaystyle \displaystyle f'(x) = \lim_{h \to 0} \frac{2xh+h^2}{h}
    $

    $\displaystyle \displaystyle f'(x) = \lim_{h \to 0} \frac{h(2x+h)}{h}
    $

    $\displaystyle \displaystyle f'(x) = \lim_{h \to 0} 2x+h = 2x$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 17th 2011, 06:59 PM
  2. How to understand this problem?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 20th 2011, 03:17 AM
  3. a problem can't understand it
    Posted in the Algebra Forum
    Replies: 10
    Last Post: Aug 7th 2010, 08:27 AM
  4. Replies: 0
    Last Post: Feb 27th 2010, 11:57 AM
  5. I don't understand this problem
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Jan 29th 2009, 02:14 PM

Search Tags


/mathhelpforum @mathhelpforum