3. (a) Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes. (b) The diagram shows the graph of $y=f(x)$. Draw separate one-third page sketches of the graphs of the following: (i) $y = |f(x)|$ (ii) $y = (f(x))^2$ (iii) $y = \frac{1}{\sqrt{f(x)}}$ (c) Find the equation of the tangent to the curve defined by $x^2 - xy + y^3 = 5$ at the point $(2, -1)$. (d) The base of a solid is the region in the $xy$ plane enclosed by the curves $y = x^4$, $y = -x^4$ and the line $x = 2$. Each cross-section perpendicular to the $x$-axis is an equilateral triangle. (i) Show that the area of the triangular cross-section at $x=h$ is $\,\sqrt{3}h^2$. (ii) Hence find the volume of the solid.

SSCE HSC Mathematics Extension 2 Vector equation of a curve: 3. (a) Sketch the curve $y = \fr

3. (a) Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes - HSC - SSCE Mathematics Extension 2 - Question 3 - 2004 - Paper 1

Question 3

$3.-(a)-Sketch-the-curve-$y-=-\frac{4x^2}{9}$-showing-all-asymptotes-HSC-SSCE Mathematics Extension 2-Question 3-2004-Paper 1.png$

3. (a) Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes. (b) The diagram shows the graph of $y=f(x)$. Draw separate one-third page sketches of the grap... show full transcript

Worked Solution & Example Answer:3. (a) Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes - HSC - SSCE Mathematics Extension 2 - Question 3 - 2004 - Paper 1

Step 1

Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes.

96%

114 rated

Answer

To sketch the curve, we first find the asymptotes. The function $y = \frac{4x^2}{9}$ is a parabolic function that opens upwards. Since it is a rational function, we look for vertical and horizontal asymptotes. Here, there are no vertical asymptotes as the denominator only becomes zero when $x=\pm\infty$ , and there are no restrictions in the domain. The horizontal asymptote, however, can be found as $x$ approaches $\,\pm\infty$ , which yields $y \to \infty$ . Therefore, the curve will approach the x-axis as it moves towards infinity along both directions.

Step 2

Draw separate one-third page sketches of the graphs of the following: (i) $y = |f(x)|$

99%

104 rated

Answer

To sketch $y = |f(x)|$ , we take the graph of $f(x)$ and reflect any part of the graph that is below the x-axis upwards. This ensures all values of the resulting graph are non-negative.

Step 3

Draw separate one-third page sketches of the graphs of the following: (ii) $y = (f(x))^2$

96%

101 rated

Answer

To sketch $y = (f(x))^2$ , we square the values of $f(x)$ . This means that all values will be non-negative and any x-value where $f(x)$ was zero will remain zero, while all other points will be transformed upwards.

Step 4

Draw separate one-third page sketches of the graphs of the following: (iii) $y = \frac{1}{\sqrt{f(x)}}$

98%

120 rated

Answer

For the sketch of $y = \frac{1}{\sqrt{f(x)}}$ , we need to ensure that $f(x)$ is positive where the function exists. The graph will approach infinity as $f(x)$ approaches zero, indicating vertical asymptotes where $f(x)$ intersects the x-axis.

Step 5

Find the equation of the tangent to the curve defined by $x^2 - xy + y^3 = 5$ at the point $(2, -1)$.

97%

117 rated

Answer

To find the tangent, we first use implicit differentiation on the equation. Differentiating yields: $2x - (x \frac{dy}{dx} + y) + 3y^2 \frac{dy}{dx} = 0$ . Substituting the point $(2, -1)$ into this yields: $2(2) - (2 \frac{dy}{dx} - 1) + 3(-1)^2 \frac{dy}{dx} = 0$ . This simplifies to find $rac{dy}{dx}$ and we get the slope. Finally, we use the point-slope form of the equation to write the equation of the tangent line.

Step 6

Show that the area of the triangular cross-section at $x=h$ is $\,\sqrt{3}h^2$.

97%

121 rated

Answer

The area of an equilateral triangle with side length $s$ is given by $A = \frac{\sqrt{3}}{4}s^2$ . We find $s$ by determining the height of the solid area at $x=h$ . The coordinates at that point will define the side length of the equilateral triangle, hence we derive that the area is as stated.

Step 7

Hence find the volume of the solid.

96%

114 rated

Answer

The volume can be found by integrating the area of the triangular cross-section from $0$ to $2$ (the limits of integration), leading to: $V = \int_0^2 \sqrt{3}h^2 \, dx.$ This integral provides the total volume of the solid.

3. (a) Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes - HSC - SSCE Mathematics Extension 2 - Question 3 - 2004 - Paper 1

Worked Solution & Example Answer:3. (a) Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes - HSC - SSCE Mathematics Extension 2 - Question 3 - 2004 - Paper 1

Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes.

Draw separate one-third page sketches of the graphs of the following: (i) $y = |f(x)|$

Draw separate one-third page sketches of the graphs of the following: (ii) $y = (f(x))^2$

Draw separate one-third page sketches of the graphs of the following: (iii) $y = \frac{1}{\sqrt{f(x)}}$

Find the equation of the tangent to the curve defined by $x^2 - xy + y^3 = 5$ at the point $(2, -1)$.

Show that the area of the triangular cross-section at $x=h$ is $\,\sqrt{3}h^2$.

Hence find the volume of the solid.

Join the SSCE students using SimpleStudy...