Question

Find the value of \( \frac{dy}{dx} \) of the curve \( x = t^2 + 3t - 8 \), \( y = 2t^2 - 2t - 5 \) at the point \( (2, -1) \).

Hint:

For parametric equations, use \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \) to find derivatives efficiently.

Answer 1

To find \( \frac{dy}{dx} \), we use parametric 6266899d2bbfcb1799af2df0: \[ \frac{dx}{dt} = \frac{d}{dt} (t^2 + 3t - 8) = 2t + 3. \] \[ \frac{dy}{dt} = \frac{d}{dt} (2t^2 - 2t - 5) = 4t - 2. \] \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t - 2}{2t + 3}. \] At \( t = 2 \), \[ \frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}. \]