NCEA Calculus 2023 THREE(e - Solution & Explanation

Question

A power line hangs between two poles.

The equation of the curve

y = f(x)

that models the shape of the
power line can be found by solving the differential equation:

a\,\dfrac{d^2y}{dx^2} = \sqrt{1+\left(\dfrac{dy}{dx}\right)^2}

Use differentiation to verify that the function

y = \dfrac{a}{2}\left(e^{\frac{x}{a}} + e^{-\frac{x}{a}}\right)

satisfies the above differential equation, where

a

is a positive
constant.

Official Answer

y=\dfrac{a}{2}\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)

=\left(\dfrac{a}{2}e^{\frac{x}{a}}+\dfrac{a}{2}e^{-\frac{x}{a}}\right)

\dfrac{dy}{dx}=\dfrac{1}{2}e^{\frac{x}{a}}-\dfrac{1}{2}e^{-\frac{x}{a}}

\dfrac{d^2y}{dx^2}=\dfrac{1}{2a}e^{\frac{x}{a}}+\dfrac{1}{2a}e^{-\frac{x}{a}}

\left(\dfrac{dy}{dx}\right)^2=\dfrac{1}{4}e^{\frac{2x}{a}}+\dfrac{1}{4}e^{-\frac{2x}{a}}-\dfrac{1}{2}

\#

(1)

LHS

=a\dfrac{d^2y}{dx^2}

=\dfrac{1}{2}e^{\frac{x}{a}}+\dfrac{1}{2}e^{-\frac{x}{a}}

\#

(2)

RHS

=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}

=\sqrt{1+\dfrac{1}{4}e^{\frac{2x}{a}}+\dfrac{1}{4}e^{-\frac{2x}{a}}-\dfrac{1}{2}}

=\sqrt{\dfrac{1}{4}e^{\frac{2x}{a}}+\dfrac{1}{4}e^{-\frac{2x}{a}}+\dfrac{1}{2}}

\#

(3)

=\sqrt{\left(\dfrac{1}{2}e^{\frac{x}{a}}+\dfrac{1}{2}e^{-\frac{x}{a}}\right)^2}

=\dfrac{1}{2}\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)

=a\dfrac{d^2y}{dx^2}

=\text{LHS as required}

Grading Criteria

Video Explanation

2023 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 67:09)

2023 THREE(e)