Question: 3. Given that $y^3 = xy - x^2$, show that $\frac{dy}{dx} = \frac{y - 2x}{3y^2 - x}$. Hence show that the curve $y^3 = xy - x^2$ has a stationary point when $x = \frac{1}{8}$. [7]
As shown in Fig. 4, boxes P and Q are descending vertically supported by a parachute. Box P has mass 75 kg. Box Q has mass 25 kg and hangs from box P by means of a light vertical wire. Air resistance on the boxes should be neglected.
At one stage the boxes are slowing in their descent with the parachute exerting an upward vertical force of 1030N on box P. The acceleration of the boxes is $a$ m s$^{-2}$ upwards and the tension in the wire is $T$ N.
(i) Draw a labelled diagram showing all the forces acting on box P and another diagram showing all the forces acting on box Q. [2]
(ii) Write down separate equations of motion for box P and for box Q. [3]
(iii) Calculate the tension in the wire. [2]
1. **State the problem:**
We are given the curve defined implicitly by $$y^3 = xy - x^2$$ and asked to find $\frac{dy}{dx}$, show the formula for the derivative, and find the stationary point when $x=\frac{1}{8}$. Then, for the physics problem, analyze forces on boxes P and Q, write equations of motion, and find the tension $T$.
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### Part 1: Calculus Problem
2. **Find $\frac{dy}{dx}$ for the curve $y^3 = xy - x^2$**
Use implicit differentiation with respect to $x$:
$$\frac{d}{dx}(y^3) = \frac{d}{dx}(xy - x^2)$$
Apply the chain and product rules:
$$3y^2 \frac{dy}{dx} = y + x \frac{dy}{dx} - 2x$$
3. **Rearrange to isolate $\frac{dy}{dx}$:**
Bring all terms involving $\frac{dy}{dx}$ to one side:
$$3y^2 \frac{dy}{dx} - x \frac{dy}{dx} = y - 2x$$
Factor out $\frac{dy}{dx}$:
$$\left(3y^2 - x\right) \frac{dy}{dx} = y - 2x$$
Divide both sides by $3y^2 - x$:
$$\frac{dy}{dx} = \frac{y - 2x}{3y^2 - x}$$
4. **Show the stationary point when $x=\frac{1}{8}$:**
At a stationary point, $\frac{dy}{dx} = 0$, so numerator must be zero:
$$y - 2x = 0 \implies y = 2x$$
Substitute $y=2x$ into original equation:
$$ (2x)^3 = x(2x) - x^2$$
$$8x^3 = 2x^2 - x^2 = x^2$$
Divide both sides by $x^2$ (assuming $x \neq 0$):
$$8x = 1 \implies x = \frac{1}{8}$$
Then $y = 2 \times \frac{1}{8} = \frac{1}{4}$.
So stationary point is at $$\left(\frac{1}{8}, \frac{1}{4}\right)$$.
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### Part 2: Physics Problem
5. **Draw labelled diagrams:**
- For box P (mass 75 kg): Forces are weight down $75g$, tension $T$ down (from box Q), and parachute force 1030 N up.
- For box Q (mass 25 kg): Forces are weight down $25g$ and tension $T$ up.
6. **Write equations of motion:**
Let acceleration $a$ be upwards.
- For box P:
$$\text{Upward forces} - \text{Downward forces} = m_P a$$
$$1030 - 75g - T = 75a$$
- For box Q:
$$\text{Upward forces} - \text{Downward forces} = m_Q a$$
$$T - 25g = 25a$$
7. **Calculate tension $T$:**
From box Q equation:
$$T = 25a + 25g$$
Substitute into box P equation:
$$1030 - 75g - (25a + 25g) = 75a$$
Simplify:
$$1030 - 75g - 25a - 25g = 75a$$
$$1030 - 100g = 75a + 25a$$
$$1030 - 100g = 100a$$
Solve for $a$:
$$a = \frac{1030 - 100g}{100}$$
Use $g = 9.8$ m/s$^2$:
$$a = \frac{1030 - 980}{100} = \frac{50}{100} = 0.5 \text{ m/s}^2$$
Now find $T$:
$$T = 25a + 25g = 25(0.5) + 25(9.8) = 12.5 + 245 = 257.5 \text{ N}$$
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**Final answers:**
- $$\frac{dy}{dx} = \frac{y - 2x}{3y^2 - x}$$
- Stationary point at $$\left(\frac{1}{8}, \frac{1}{4}\right)$$
- Acceleration $$a = 0.5 \text{ m/s}^2$$ upwards
- Tension $$T = 257.5 \text{ N}$$