NCEA Calculus 2021 TWO(e - Solution & Explanation

Question

The graph below shows the curve

y = \sqrt{2x - 4}

, and the tangent to the curve at point P.
The tangent passes through the point

(-2, 1)

.

Loading diagram...

Find the coordinates of point P.

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

Official Answer

y=\sqrt{2x-4}

\dfrac{dy}{dx}=\dfrac{1}{\sqrt{2x-4}}

Gradient of tangent

=\dfrac{y-1}{x+2}

\dfrac{1}{\sqrt{2x-4}}=\dfrac{\sqrt{2x-4}-1}{x+2}

x+2=2x-4-\sqrt{2x-4}

\sqrt{2x-4}=x-6

2x-4=x^2-12x+36

x^2-14x+40=0

(x-4)(x-10)=0

x=4

or

x=10

Rejecting

x=4

by checking the surd equation

x=10

\sqrt{16}=4

True

x=4

\sqrt{4}=-2

False

One solution:

x=10

Therefore, the coordinates of point P are

(10,4)

OR

Rejecting

x=4

by checking the gradient:

At

(10,4)

,

\dfrac{dy}{dx}=\dfrac{1}{\sqrt{16}}=\dfrac{1}{4}

Gradient:

\dfrac{y-1}{x+2}=\dfrac{3}{12}=\dfrac{1}{4}

At

(4,2)

,

\dfrac{dy}{dx}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}

Gradient:

\dfrac{y-1}{x+2}=\dfrac{1}{6}

One solution:

x=10

Therefore, the coordinates of point P are

(10,4)

Grading Criteria

Correct derivative: $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{2x-4}}$ and $\sqrt{2x-4}=x-6$

T1: Correct solution with correct derivative: $\mathrm{P}\ (10,4)$ without any justification for $x\ne 4$

T2: Correct solution with correct derivative: $\mathrm{P}\ (10,4)$ : $x\ne 4$ must be justified with respect to either the surd equation or the gradient of the tangent.

Video Explanation

NCEA Level 3 Calculus Differentiation 2021 NZQA Exam - Worked Answers by infinityplusone(starts at 44:24)

2021 TWO(e)