2021 TWO(c)

Merit
Question
A curve has the equation y=(x2+3x+2)cos3xy = (x^2 + 3x + 2)\cos 3x.

Find the equation of the normal to the curve at the point where the curve crosses the yy-axis.

*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
y=(x2+3x+2)cos3xy=(x^2+3x+2)\cos 3x
dydx=(2x+3)cos3x(x2+3x+2)3sin3x\dfrac{dy}{dx}=(2x+3)\cos 3x-(x^2+3x+2)3\sin 3x
Crosses yy-axis x=0, y=2, dydx=3\Rightarrow x=0,\ y=2,\ \dfrac{dy}{dx}=3

Normal gradient is 13-\dfrac{1}{3}

Equation of normal:
y2=13(x0)y-2=-\dfrac{1}{3}(x-0)

y=13x+2y=-\dfrac{1}{3}x+2

3y+x6=03y+x-6=0
Grading Criteria

Achievement (u)

  • Correct derivative.

Merit (r)

  • Correct solution with correct derivative.

Excellence T1

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Excellence T2

-
Video Explanation
NCEA Level 3 Calculus Differentiation 2021 NZQA Exam - Worked Answers by infinityplusone(starts at 36:45)
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