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Thread: Calculas!

  1. #1
    Member Awsom Guy's Avatar
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    Calculas!

    I have another question to post here.
    Write the equation of the straight line passing through the point (2,1) and normal to the curve 4y=x^2.

    I know one thing x=2 and y=1 but I have no idea where to start or to do anything. How does the equation shown derivate.
    Please help me get started.

    The answer provided is x+y-3=0.
    Thanks
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  2. #2
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    Quote Originally Posted by Awsom Guy View Post
    I have another question to post here.
    Write the equation of the straight line passing through the point (2,1) and normal to the curve 4y=x^2.

    I know one thing x=2 and y=1 but I have no idea where to start or to do anything. How does the equation shown derivate.
    Please help me get started.

    The answer provided is x+y-3=0.
    Thanks
    The curve is $\displaystyle 4y = x^2$

    or $\displaystyle y = \frac{1}{4}x^2$.


    It is easy to see that $\displaystyle (2, 1)$ lies on this curve.


    $\displaystyle \frac{dy}{dx} = \frac{1}{2}x$

    and when $\displaystyle x = 2, \frac{dy}{dx} = 1$.


    So the tangent will have gradient = 1, which means the normal will have gradient $\displaystyle -\frac{1}{1} = -1$.


    So the normal has gradient -1 and passes through (2, 1)

    $\displaystyle y = mx + c$

    $\displaystyle 1 = -1(2) + c$

    $\displaystyle 1 = -2 + c$

    $\displaystyle c = 3$.


    So the equation of the normal is $\displaystyle y = -x + 3$.
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  3. #3
    Member Awsom Guy's Avatar
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    sorry one more little thing...

    one little question is the derivative of 2 equal to 1.
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  4. #4
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    Quote Originally Posted by Awsom Guy View Post
    one little question is the derivative of 2 equal to 1.

    no, $\displaystyle \frac{d}{dx}(2)= 0$
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  5. #5
    Member Awsom Guy's Avatar
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    Then why is the above derivative of 2 = 1. explain please.
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  6. #6
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    The derivative is $\displaystyle \frac{1}{2}x$.

    What does this equal when $\displaystyle x = 2$?
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  7. #7
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    I think you are confsued, Prove it has said if

    $\displaystyle y = \frac{1}{4}x^2$ then $\displaystyle \frac{dy}{dx} = \frac{1}{2}x$

    This is the only derivate that has been taken.

    The following is not a derivative, Prove it is saying at the point $\displaystyle x = 2, $ then $\displaystyle \frac{dy}{dx} = 1$

    I.e from the first statement

    $\displaystyle \frac{dy}{dx} = \frac{1}{2}x$

    Put $\displaystyle x = 2 $ here gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}\times 2 =1$
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  8. #8
    Member Awsom Guy's Avatar
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    Right thanks for explaining that. I appreciate that.
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  9. #9
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    The derivative of $\displaystyle \frac{1}{4}x^2$, $\displaystyle \frac{1}{2}x$ at x= 2 is 1. The "derivative of 2" means the derivative of the constant function "f(x)= 2" which is 0 because the derivative measures the rate of change and a constant function doesn't change.
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