NCEA Calculus 2024 THREE(e - Solution & Explanation

Question

The diagram below shows part of the graph of the function

f(x) = e^{-x^2}

, where

x \geq 0

.

Loading diagram...

The point P lies on the curve and the point Q lies on the

x

-axis so that OP = PQ, where O is
the origin.

Prove that the largest possible area of the triangle OPQ is

\dfrac{1}{\sqrt{2e}}

.

*You do not need to show that the area you have found is a maximum.*

*You must use calculus and show any derivatives that you need to find when solving this problem.*

Official Answer

Let

P

have coordinates

(x,y)

Area of triangle

OPQ = xe^{-x^2}

\dfrac{dA}{dx} = e^{-x^2}(1-2x^2)

For max / min,

\dfrac{dA}{dx}=0

e^{-x^2}(1-2x^2)=0

Either

e^{-x^2}=0

No solutions

Or

(1-2x^2)=0

i.e.

x=\pm\dfrac{1}{\sqrt{2}}=\pm 0.7071

But ignore

x=-\dfrac{1}{\sqrt{2}}

as

x>0

Then area

=\dfrac{1}{\sqrt{2}}\times e^{-\frac{1}{2}}=\dfrac{1}{\sqrt{2}}\times\dfrac{1}{e^{\frac{1}{2}}}

Area

=\dfrac{1}{\sqrt{2e}}

as required.

Grading Criteria

Achievement (u)

Correct expression for $\dfrac{dA}{dx}$ .

Merit (r)

Finding $x=\dfrac{1}{\sqrt{2}}$ , and evidence of ignoring $x=-\dfrac{1}{\sqrt{2}}$ with evidence of a calculus method.

Excellence T1

E7 / T1 Correct proof, but with one minor error.

Excellence T2

E8 / T2 Correct proof of exact area value, with evidence of a calculus method.

Video Explanation

2024 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 52:18)

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2024 THREE(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2