Author Topic: Methods SAC 2 practice question HELP (Read 882 times) Tweet Share

aakritisharma · « **on:** May 14, 2013, 03:09:00 pm »

okay so this is a practice question for my upcoming methods sac and im having some problems with it.

a design is based on a cubic function y=a(x-b)^3 + c. it has a stationary point at x=60. the design begins ABOVE the point (0,200) and ends at (180,110) at which the gradient is -1.8. fully define the equation.

please help asap.

thank you much appreciated

Daenerys Targaryen · « **Reply #1 on:** May 15, 2013, 05:31:15 pm »

$y=a(x-b)^3 + c$

$\frac{dy}{dx}=3ax^2-6abx+3ab^2$

$\frac{dy}{dx}=0 \text{ when x=60}$

$3a(60)^2-6ab(60)+3ab^2=0$

$3600a-120ab+ab^2=0 \text{ -[1]}$

Also

$\frac{dy}{dx}=-1.8 \text{ when x=180}$

$3a(180)^2-6ab(180)+3ab^2=-1.8 \text{ -[2]}$

Also

$\text{If (180,110) is on the line; Let x=180 and y=110}$

$110=a(180-b)^3+c \text{ -[3]}$

$\text{Solve [1], [2] and [3]}$

$a=-0.000042, b=60, c=182$

$\therefore y=-0.000042(x-60)^3+182$

$\text{If line is above y=200 to begin with then }-0.000042(x-60)^3+182\g200$

$x=-15.6$

$\therefore y=-0.000042(x-60)^3+182;-15.6\leq x \leq180$

~~EDIT: I think I am wrong, I dont think it actually think (0,200) is on the point, and i never used the -1.8. Aq!~~
EDIT: error fixed (i think)

aakritisharma · « **Reply #2 on:** May 19, 2013, 05:02:39 pm »

thanks for the help hatersgonnahate but my teacher made a mistake. the stationary point was supposed to be at x=80. thats why the y intercept was below (0,200) not above.

anyways thanks

much appreciated

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Author Topic: Methods SAC 2 practice question HELP (Read 882 times) Tweet Share

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Methods SAC 2 practice question HELP

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