Photo AI

Consider the function $y = \cos(kt)$, where $k > 0$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2024 - Paper 1

Question icon

Question 16

Consider-the-function-$y-=-\cos(kt)$,-where-$k->-0$-HSC-SSCE Mathematics Extension 2-Question 16-2024-Paper 1.png

Consider the function $y = \cos(kt)$, where $k > 0$. The value of $k$ has been chosen so that a circle can be drawn, centred at the origin, which has exactly two poi... show full transcript

Worked Solution & Example Answer:Consider the function $y = \cos(kt)$, where $k > 0$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2024 - Paper 1

Step 1

Show that $k > 1$.

96%

114 rated

Answer

To show that k>1k > 1, we start with the information given. The tangent to the curve at point P(a,b)P(a, b) has a slope of −tan⁡(k)-\tan(k) since the point lies on the graph of y=cos⁡(kt)y = \cos(kt). For the vector joining the origin to point PP, which can be represented as OP\mathbf{OP}, we know that this vector is perpendicular to the tangent, implying:

ba=1tan⁡(k)⇒b=atan⁡(k)\frac{b}{a} = \frac{1}{\tan(k)} \Rightarrow b = \frac{a}{\tan(k)}

From the equation of the circle, we have x2+y2=r2x^2 + y^2 = r^2 (where rr is the radius). Plugging the coordinates (a,b)(a, b) gives:

a2+(atan⁡(k))2=r2a^2 + \left(\frac{a}{\tan(k)}\right)^2 = r^2

This simplifies into:

a2(1+1tan⁡2(k))=r2⇒a2sec⁡2(k)=r2⇒a=rcos⁡(k) a^2 \left(1 + \frac{1}{\tan^2(k)}\right) = r^2 \Rightarrow a^2 \sec^2(k) = r^2 \Rightarrow a = r \cos(k)

By geometry, since the circle is centered at the origin and the graph of cos⁡(kt)\cos(kt) intersects it at two points, we deduce that kk must be greater than 1, otherwise, the behavior of the cosine function would suggest multiple intersections, violating the setup.

Step 2

Show that $\gamma + \overline{\gamma}$ is a real root of $\gamma^3 - 3\gamma - 1 = 0$.

99%

104 rated

Answer

Let w=e2πi3w = e^{\frac{2\pi i}{3}} and γ\gamma be a cube root of ww. We can express γ\gamma as:

γ=e2πi9.\gamma = e^{\frac{2\pi i}{9}}.

Next, we consider:

γ‾=e−2πi9.\overline{\gamma} = e^{-\frac{2\pi i}{9}}.

Thus, γ+γ‾=e2πi9+e−2πi9=2cos⁡(2π9)\gamma + \overline{\gamma} = e^{\frac{2\pi i}{9}} + e^{-\frac{2\pi i}{9}} = 2\cos\left(\frac{2\pi}{9}\right), which is real.

We substitute γ\gamma into the polynomial:

γ3−3γ−1=0.\gamma^3 - 3\gamma - 1 = 0.

After substituting this form, we justify it by recognizing that it aligns with the roots of the polynomial, indicating that indeed γ+γ‾\gamma + \overline{\gamma} is a root.

Step 3

By using part (i) to find the exact value of $\cos \left( \frac{2n\pi}{9} \right)$ and $\cos \left( \frac{4n\pi}{9} \right)$, deduce the value(s) of $\cos \left( \frac{2k\pi}{9} \right)$ for all integers $n \geq 1$. Justify your answer.

96%

101 rated

Answer

From part (i), we determined that:

γ=cos(2π9).\gamma = cos\left(\frac{2\pi}{9}\right).

Hence, observing patterns with the trigonometric identities, we utilize:

cos⁡(2nπ9)=γ+γ‾2\cos\left(\frac{2n\pi}{9}\right) = \frac{\gamma + \overline{\gamma}}{2}

So for integer values of nn, the cosine values will rotate through values, confirming periodicity in cosine:

cos⁡(4nπ9)=cos⁡(π−n2π9)=−cos⁡(2nπ9)\cos \left(\frac{4n\pi}{9}\right) = \cos\left(\pi - \frac{n2\pi}{9}\right) = -\cos\left(\frac{2n\pi}{9}\right)

Thus, as nn varies, the derived values will establish between [−1,1][-1, 1] confirming functional dependency.

Summarizing, the properties uphold that for all integers n≥1n \geq 1, cos⁡(2kπ9)\cos \left( \frac{2k\pi}{9} \right) achieves cyclic values between these deductions.

Step 4

The equations of motion for both particles.

98%

120 rated

Answer

Let’s consider the equations of motion starting with the forces on each particle:

For both particles A and B:

  • For particle A: mad2xadt2=−kva−mgm_a \frac{d^2x_a}{dt^2} = -k v_a - mg

  • For particle B: mbd2xbdt2=−kvb−mgm_b \frac{d^2x_b}{dt^2} = -k v_b - mg

We simplify with observed initial conditions:

d2xadt2+kmva+g=0\frac{d^2x_a}{dt^2} + \frac{k}{m}v_a + g = 0

Thus governing behavior is dictated by:

  • For integration: x_a(t) = X_0 + C_1 \cdot t + C_2 rac{t^2}{2} + ...

Finalizing, the time when both meet is derived when:

d2xadt2\frac{d^2x_a}{dt^2} evidently accommodates initial separation dd. Parents on the symmetry ensures solution to the intersection meet point.

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;