NCEA Calculus 2019 THREE(e - Solution & Explanation

Question

The graph below shows the function

y = 2\sqrt{36 - x^2}

, and the tangent to that function at point P.
The tangent intersects the

x

-axis at the point

(8,0)

.

Loading diagram...

Find the

x

-coordinate of point P.
*You must use calculus and show any derivatives that you need to find when solving this
problem.*

Official Answer

y=2\sqrt{36-x^2}

\dfrac{dy}{dx}=(36-x^2)^{-\frac{1}{2}}\cdot-2x

=\dfrac{-2x}{\sqrt{36-x^2}}

Gradient of tangent

=\dfrac{-2\sqrt{36-x^2}}{(8-x)}

=\dfrac{2\sqrt{36-x^2}}{x-8}

\therefore \dfrac{2\sqrt{36-x^2}}{x-8}=\dfrac{-2x}{\sqrt{36-x^2}}

2(36-x^2)=16x-2x^2

72-2x^2=16x-2x^2

72=16x

x=4.5

Or alternatively:

y=\dfrac{-2x}{\sqrt{36-x^2}}(x-8)

y=\dfrac{-2x^2}{\sqrt{36-x^2}}+\dfrac{16x}{\sqrt{36-x^2}}

Substituting for

y

:

2\sqrt{36-x^2}=\dfrac{-2x^2}{\sqrt{36-x^2}}+\dfrac{16x}{\sqrt{36-x^2}}

2(36-x^2)=-2x^2+16x

36-x^2=-x^2+8x

36=8x

x=4.5

Grading Criteria

Correct $\dfrac{dy}{dx}$ of curve.
AND
Correct gradient of tangent.
OR
Correct equation of tangent involving expression for $\dfrac{dy}{dx}$ .

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2019 NZQA Exam - Worked Answers by infinityplusone(starts at 57:30)

2019 THREE(e)