4 (15 marks) Use a SEPARATE writing booklet. (a) A curve is defined implicitly by $ oot{2}{x} + oot{2}{y} = 1$. Use implicit differentiation to find $\frac{dy}{dx}$. (ii) Sketch the curve $ oot{2}{x} + oot{2}{y} = 1$. (iii) Sketch the curve $ oot{2}{|x|} + oot{2}{|y|} = 1$. (b) A bend in a highway is part of a circle of radius $r$, centre $O$. Around the bend the highway is banked at an angle $\alpha$ to the horizontal. A car is travelling around the bend at a constant speed $v$. Assume that the car is represented by a point $P$ of mass $m$. The forces acting on the car are a lateral force $F$, the gravitational force $mg$ and a normal reaction $N$ to the road, as shown in the diagram. (i) By resolving forces, show that $F = mg\sin\alpha - \frac{mg^2}{r}\cos\alpha$. (ii) Find an expression for $v$ such that the lateral force $F$ is zero.

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4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2010 - Paper 1

Question 4

4 (15 marks) Use a SEPARATE writing booklet. (a) A curve is defined implicitly by $ oot{2}{x} + oot{2}{y} = 1$. Use implicit differentiation to find $\frac{dy}{... show full transcript

Worked Solution & Example Answer:4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2010 - Paper 1

Step 1

A curve is defined implicitly by $\root{2}{x} + \root{2}{y} = 1$. Use implicit differentiation to find $\frac{dy}{dx}$.

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Answer

To find $\frac{dy}{dx}$ , we start with the implicit equation:

$\sqrt{x} + \sqrt{y} = 1$

Differentiating both sides with respect to $x$ results in:

$\frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = 0$

Using the chain rule on $\sqrt{y}$ gives:

$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0$

Rearranging this yields:

$\frac{1}{2\sqrt{y}}\frac{dy}{dx} = -\frac{1}{2\sqrt{x}}$

Multiplying through by $2\sqrt{y}$ provides:

$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$

Step 2

Sketch the curve $\root{2}{x} + \root{2}{y} = 1$.

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Answer

When sketching the curve defined by $\sqrt{x} + \sqrt{y} = 1$ , we first identify the intercepts:

Setting $y = 0$ , we find $x = 1$ .
Setting $x = 0$ , we find $y = 1$ .

The points $(1, 0)$ and $(0, 1)$ define the axes. The curve is thus bounded by these points and is composed in the first quadrant, curving downwards towards the axes.

Step 3

Sketch the curve $\root{2}{|x|} + \root{2}{|y|} = 1$.

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Answer

For the curve $\sqrt{|x|} + \sqrt{|y|} = 1$ , the approach is similar to the previous part. It is symmetric across the $x$ and $y$ axes due to the absolute values.

The points of intersection with the axes stay the same at $(1, 0)$ , $(0, 1)$ and we also have points $(-1, 0)$ and $(0, -1)$ . The curve will appear in all four quadrants, connecting these intercepts symmetrically.

Step 4

By resolving forces, show that $F = mg\sin\alpha - \frac{mg^2}{r}\cos\alpha$.

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Answer

To resolve the forces acting on the car at point $P$ , we split them into vertical and horizontal components.

The forces include:

Vertically: the normal force $N$ and the weight $mg$ , giving: $N = mg\cos\alpha$
Horizontally: the lateral force $F$ and the radial component due to gravity, giving: $F = mg\sin\alpha - \frac{mg^2}{r}\cos\alpha$

Therefore, combining these results allows us to establish the relationship above.

Step 5

Find an expression for $v$ such that the lateral force $F$ is zero.

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Answer

To find $v$ where $F = 0$ , we set:

$0 = mg\sin\alpha - \frac{mg^2}{r}\cos\alpha$

Rearranging gives:

$mg\sin\alpha = \frac{mg^2}{r}\cos\alpha$

Dividing by $mg$ results in:

$\sin\alpha = \frac{mg}{r}\cos\alpha$

To solve for $v$ , we remember that the centripetal acceleration $a_c = \frac{v^2}{r}$ .

a substitution for $v$ gives:

$v^2 = rg(\sin\alpha + \frac{g\cos\alpha}{m})$

Finally, we solve for $v$ as:

$v = \sqrt{g\cdot\sin\alpha\cdot r}$

4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2010 - Paper 1

Worked Solution & Example Answer:4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2010 - Paper 1

A curve is defined implicitly by $\root{2}{x} + \root{2}{y} = 1$. Use implicit differentiation to find $\frac{dy}{dx}$.

Sketch the curve $\root{2}{x} + \root{2}{y} = 1$.

Sketch the curve $\root{2}{|x|} + \root{2}{|y|} = 1$.

By resolving forces, show that $F = mg\sin\alpha - \frac{mg^2}{r}\cos\alpha$.

Find an expression for $v$ such that the lateral force $F$ is zero.

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