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Thread: Find value of k of a function >>> -k< y <k

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    Post Find value of k of a function >>> -k< y <k

    This question is from edexcel IAL paper c34 no.13



    13. A curve C has parametric equations:

    . . . . .[tex]x\, =\, 6\, \cos(2t),\, y\, =\, 2\, \sin(t),\, -\dfrac{\pi}{2}\, <\, t\, <\, \dfrac{\pi}{2}[/tex]

    (a) Show that [tex]\, \dfrac{dy}{dx}\, =\, \lambda\, \csc(t),\, [/tex] giving the exact value of the constant [tex]\, \lambda.[/tex]

    (b) Find an equation of the normal to C at the point where [tex]\, t\, =\, \dfrac{\pi}{3}.\, [/tex] Give you answer in the form y = mx + c, where m and c are simplified surds.


    The cartesian equation for the curve C can be written in the form

    . . . . .[tex]x\, =\, f(y),\, -k\, <\, y\, <\, k[/tex]

    where f (y) is a polynomial in y and k is a constant.

    (c) Find f (y).

    (d) State the value of k.




    My problem is part (d) where the answer is k= - 2 but I don't know how to find it. (WHY is NEGATIVE?)

    I cannot solve f(y) by letting f(y)=0. (which give k= 2^1/2

    Is that using the range of f(y)? How to find the range of f(y)

    BELOW IS THE ANSWER KEY.


    Thank you very much!!!!!!!!!!!!!!!!
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    Last edited by stapel; 01-26-2017 at 10:30 AM. Reason: Typing out the text in the graphic.

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    Elite Member stapel's Avatar
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    Quote Originally Posted by Simonchan View Post
    This question is from edexcel IAL paper c34 no.13



    13. A curve C has parametric equations:

    . . . . .[tex]x\, =\, 6\, \cos(2t),\, y\, =\, 2\, \sin(t),\, -\dfrac{\pi}{2}\, <\, t\, <\, \dfrac{\pi}{2}[/tex]

    (a) Show that [tex]\, \dfrac{dy}{dx}\, =\, \lambda\, \csc(t),\, [/tex] giving the exact value of the constant [tex]\, \lambda.[/tex]

    (b) Find an equation of the normal to C at the point where [tex]\, t\, =\, \dfrac{\pi}{3}.\, [/tex] Give you answer in the form y = mx + c, where m and c are simplified surds.


    The cartesian equation for the curve C can be written in the form

    . . . . .[tex]x\, =\, f(y),\, -k\, <\, y\, <\, k[/tex]

    where f (y) is a polynomial in y and k is a constant.

    (c) Find f (y).

    (d) State the value of k.




    My problem is part (d) where the answer is k= - 2 but I don't know how to find it. (WHY is NEGATIVE?)

    I cannot solve f(y) by letting f(y)=0.
    Why would you set f(y) equal to zero? What, exactly, have been your steps and your reasoning?

    For instance, you started with x(t), applied the appropriate trig identity, and substituted y(t) into the function. This gave you "x = 6(1 - (1/2)y^2), which is x in terms of y. You noted the original domains for x(t) and y(t). You used this to determine the range of y(t), which is then the domain of x(y). And... then what? At what point did you try to find the zeroes of the function, and why?

    Please be complete. Thank you!

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