An aerial view of a stretch of road is shown in the diagram below. The road can be described by the function $y = x^2 + 2$, $x ext{ (dotted lines) as shown in the diagram.}$ Benny sits at a vantage point $B(0; 3)$ and observes a car, $P$, travelling along the road. Calculate the distance between Benny and the car, when the car is closest to Benny.

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An aerial view of a stretch of road is shown in the diagram below - NSC Mathematics - Question 9 - 2017 - Paper 1

Question 9

An aerial view of a stretch of road is shown in the diagram below. The road can be described by the function $y = x^2 + 2$, $x ext{ (dotted lines) as shown in the ... show full transcript

Worked Solution & Example Answer:An aerial view of a stretch of road is shown in the diagram below - NSC Mathematics - Question 9 - 2017 - Paper 1

Step 1

Calculate the distance PB

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Answer

The coordinates of point $P$ on the road can be given by $P(x; x^2 + 2)$ , where the distance PB can be expressed as:

$PB = ext{sqrt}ig((x - 0)^2 + (x^2 + 2 - 3)^2\big) = ext{sqrt}ig(x^2 + (x^2 - 1)^2\big)$

This expands to:

$PB^2 = x^2 + (x^4 - 2x^2 + 1) = x^4 - x^2 + 1$

Step 2

Find the derivative of PB²

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Answer

To find the minimum distance, we will take the derivative of $PB^2$ with respect to $x$ and set it to zero:

$\frac{d(PB^2)}{dx} = 4x^3 - 2x = 0$

This gives:

$x(2x^2 - 1) = 0$ So, either $x = 0$ or $x = ±\frac{1}{\sqrt{2}}$ .

Step 3

Determine the minimum distance

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Answer

We will evaluate $PB$ at $x = \frac{1}{\sqrt{2}}$ to find the minimum distance:

$PB^2 = \left(\frac{1}{\sqrt{2}}\right)^4 - \left(\frac{1}{\sqrt{2}}\right)^2 + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$

Thus, by taking the square root:

$PB = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.87$

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