QUESTION 3 [TOTAL MARKS: 5 MARKS] A robot's trajectory is described parametrically as \( x=4 t^{2}-3 t+2 t \) and \( y=t^{3}-t \) where \( x \) is horizontal distance (in metres), \( y \) is the vertical distance (in metres) and \( t \) is the time (in seconds). i) Find the function of \( \frac{d y}{d x} \). [3 marks] ii) Using your answer from QUESTION 3 i), find \( \frac{d y}{d x} \) at 1 seconds.

To solve the problem, we begin with the parametric equations provided: 1. \( x(t) = 4t^2 - 3t + 2t = 4t^2 - t \) 2. \( y(t) = t^3 - t \) Now, we need to find \( \frac{dy}{dx} \), which can be calculated using the chain rule: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \] Let's first compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). i) To find \( \frac{dx}{dt} \): \[ \frac{dx}{dt} = \frac{d}{dt}(4t^2 - t) = 8t - 1 \] To find \( \frac{dy}{dt} \): \[ \frac{dy}{dt} = \frac{d}{dt}(t^3 - t) = 3t^2 - 1 \] Now, we can write \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 1}{8t - 1} \] --- ii) Now we need to find \( \frac{dy}{dx} \) at \( t = 1 \). First, we substitute \( t = 1 \) into the expressions for \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \): \[ \frac{dx}{dt}\bigg|_{t=1} = 8(1) - 1 = 8 - 1 = 7 \] \[ \frac{dy}{dt}\bigg|_{t=1} = 3(1)^2 - 1 = 3 - 1 = 2 \] Now we can substitute these values into the derivative: \[ \frac{dy}{dx}\bigg|_{t=1} = \frac{2}{7} \] Thus, the answer is: i) \( \frac{dy}{dx} = \frac{3t^2 - 1}{8t - 1} \) ii) \( \frac{dy}{dx}\bigg|_{t=1} = \frac{2}{7} \)