Question

The curve represented by $x = t^5 + 5t^3 + 20t + 7$ and $y = 4t^3 - 3t^2 - 18t + 3$ is decreasing in the interval

• A. $(-2, -1)$
• B. $\left(\dfrac{3}{2}, 2\right)$
• C. $(-1, \dfrac{3}{2})$
• D. $(-2, 2)$

Hint:

Monotonicity of Parametric Curves:

• If $\fracdxdt>0$, the sign of $\fracdydx$ matches $\fracdydt$.
• To find decreasing interval: solve $\fracdydt<0$.

Answer 1

The Correct Option is C

Given parametric curve: $x(t), y(t)$. \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \Rightarrow \text{Curve is decreasing when } \frac{dy}{dt}<0 \text{ (since } \frac{dx}{dt}>0) \] Compute: \[ \frac{dx}{dt} = 5t^4 + 15t^2 + 20>0 \text{ for all } t \] \[ \frac{dy}{dt} = 12t^2 - 6t - 18 = 6(2t^2 - t - 3) \] Solve $\frac{dy}{dt}<0$: \[ 2t^2 - t - 3<0 \Rightarrow (2t + 3)(t - 1)<0 \Rightarrow t \in (-1, \frac{3}{2}) \]