Consider the ellipse $E$ with equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), and the points \( P(a \cos \theta, b \sin \theta) \), \( Q(a \cos(\theta + \phi), b \sin(\theta + \phi)) \), and \( R(a \cos(\theta - \phi), b \sin(\theta - \phi)) \) on $E$ - HSC - SSCE Mathematics Extension 2 - Question 5 - 2001 - Paper 1

To show that the chord $QR$ is parallel to the tangent at $P$, we need to find the slopes of the line segment connecting points $Q$ and $R$ and compare it to the slope of the tangent at $P$.

The coordinates of points $Q$ and $R$ are: $Q(a \cos(\theta + \phi), b \sin(\theta + \phi))$ and $R(a \cos(\theta - \phi), b \sin(\theta - \phi))$.

Calculating the slope of $QR$: [ \text{slope}{QR} = \frac{b \sin(\theta - \phi) - b \sin(\theta + \phi)}{a \cos(\theta - \phi) - a \cos(\theta + \phi)} = \frac{b (\sin(\theta - \phi) - \sin(\theta + \phi))}{a (\cos(\theta - \phi) - \cos(\theta + \phi))} ] Using the sine subtraction and addition formulas: [ \sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b) ] This means: [ \text{slope}{QR} = \text{slope}_{tangent at P} ] Thus, $QR$ is parallel to the tangent at point $P$.

Since we have established that the chord $QR$ is parallel to the tangent at point $P$—and because the tangent line at point $P$ bisects all chords passing through that point—this implies that line segment $OP$ bisects the chord $QR$. Therefore, using the property of the ellipse that any tangent at a point divides the chord associated with that point in equal halves, we can conclude: [ OP \text{ bisects } QR. ]

According to Newton's second law, the acceleration $a$ of an object is given by the force acting on it divided by its mass. In this case, the net force on the submarine is the engine force $F$ minus the resistive force which is proportional to the square of the speed, $kv^2$. Therefore, we can express this as: [ ma = F - kv^2 ] Dividing both sides by the mass $m$ provides: [ \frac{dv}{dt} = \frac{F - kv^2}{m} ]

Starting with the equation derived in part (i), we have: [ \frac{dv}{dt} = \frac{1}{m}(F - kv^2) ] Rearranging and integrating with respect to $v$ gives: [ dt = \frac{m}{F - kv^2} dv ] When we integrate from $v_1$ to $v_2$ and from $0$ to $t$, the integral becomes: [ t = \int_{v_1}^{v_2} \frac{m}{F - kv^2} dv ] Using the substitution method or recognizing it as logarithmic integration leads us to: [ t = \frac{m}{2k} \log_e \left(\frac{F - kv_1^2}{F - kv_2^2}\right) ]

To find the number of ways to divide 22 students into groups of sizes 4, 5, 6, and 7, we can use the multinomial coefficient: [ \frac{22!}{4!5!6!7!} \text{ ways.} ]

Once the groups have been formed, each group can be arranged among themselves. The ways to arrange each group are $4!$, $5!$, $6!$, and $7!$, respectively. Moreover, since the arrangement is circular, we can arrange the groups in $3!$ ways (as one position is fixed): [ \text{Total arrangements} = \frac{22!}{4!5!6!7!} (4!)(5!)(6!)(7!) \times 3! = 22! \times 3! ]