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

Simonchan

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Jan 26, 2017
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This question is from edexcel IAL paper c34 no.13

[hr][/hr]
13. A curve C has parametric equations:

. . . . .\(\displaystyle x\, =\, 6\, \cos(2t),\, y\, =\, 2\, \sin(t),\, -\dfrac{\pi}{2}\, <\, t\, <\, \dfrac{\pi}{2}\)

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

(b) Find an equation of the normal to C at the point where \(\displaystyle \, t\, =\, \dfrac{\pi}{3}.\, \) 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

. . . . .\(\displaystyle x\, =\, f(y),\, -k\, <\, y\, <\, k\)

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

(c) Find f (y).

(d) State the value of k.


[HR][/HR]
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.
Screen Shot 2017-01-26 at 21.05.05.jpg

Thank you very much!!!!!!!!!!!!!!!!
 

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stapel

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This question is from edexcel IAL paper c34 no.13

[HR][/HR]
13. A curve C has parametric equations:

. . . . .\(\displaystyle x\, =\, 6\, \cos(2t),\, y\, =\, 2\, \sin(t),\, -\dfrac{\pi}{2}\, <\, t\, <\, \dfrac{\pi}{2}\)

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

(b) Find an equation of the normal to C at the point where \(\displaystyle \, t\, =\, \dfrac{\pi}{3}.\, \) 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

. . . . .\(\displaystyle x\, =\, f(y),\, -k\, <\, y\, <\, k\)

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

(c) Find f (y).

(d) State the value of k.


[HR][/HR]
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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