NCEA Calculus 2018 ONE(d - Solution & Explanation

Question

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A car is being pulled along by a rope attached to the tow-bar at the back of the car.

The rope passes through a pulley, the top of which is 3 m further from the ground than the tow-bar.

The pulley is

x

m horizontally from the tow-bar, as shown in the diagram above.

The rope is being winched in at a speed of

0.6

m s

^{-1}

.

The wheels of the car remain in contact with the ground.

At what speed is the car moving when the length of the rope,

L

, between the tow-bar and the pulley is

5.4

m?

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

Official Answer

\dfrac{dL}{dt}=0.6\ \text{m s}^{-1}

L^2=x^2+3^2

x=\sqrt{L^2-9}

\dfrac{dx}{dL}=\dfrac{1}{2}(L^2-9)^{-\frac{1}{2}}\cdot 2L

=\dfrac{L}{\sqrt{L^2-9}}

\dfrac{dx}{dt}=\dfrac{dL}{dt}\times\dfrac{dx}{dL}

=0.6\times\dfrac{L}{\sqrt{L^2-9}}

When

L=5.4

\dfrac{dx}{dt}=0.6\times\dfrac{5.4}{\sqrt{5.4^2-9}}

=0.722\ \text{m s}^{-1}

Grading Criteria

Achievement (u)

Correct expression for $\dfrac{dx}{dL}$ or $\dfrac{dL}{dx}$ .

Merit (r)

Correct solution with correct derivatives.

Excellence T1

-

Excellence T2

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2018 NZQA Exam - Worked Answers by infinityplusone(starts at 9:02)

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2018 ONE(d)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2