NCEA Calculus 2017 TWO(d - Solution & Explanation

Question

Find the coordinates of the point P

(x,y)

on the curve

y = \sqrt{x}

that is closest
to the point

(4,0)

.

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*You do not need to prove that your solution is the minimum value.*

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

Official Answer

d^2=(4-x)^2+(\sqrt{x})^2

=16-8x+x^2+x

=16-7x+x^2

Minimum distance

\Rightarrow \frac{d(d^2)}{dx}=0

\frac{d(d^2)}{dx}=-7+2x=0

x=3.5

y=\sqrt{x}=\sqrt{3.5}

P=(3.5,\sqrt{3.5})

Alternative:

d=(x^2-7x+16)^{\frac{1}{2}}

\frac{dd}{dx}=\frac{1}{2}(x^2-7x+16)^{-\frac{1}{2}}\cdot(2x-7)

=\frac{2x-7}{2\sqrt{x^2-7x+16}}

Minimum when

\frac{dd}{dx}=0

2x-7=0

etc

Grading Criteria

Achievement (u)

Correct expression for $\dfrac{dd}{dx}$ or $\dfrac{d(d^2)}{dx}$

Merit (r)

Correct solution with correct derivative.

Excellence T1

-

Excellence T2

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2017 NZQA Exam - Worked Answers by infinityplusone(starts at 32:54)

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2017 TWO(d)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2