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

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Aicharya · Accepted Answer

$\because x=t^5+5 t^3+20 t+7$ $\begin{aligned} & \frac{d x}{d t}=5 t^4+15 t^2+20 \\ & \text { and } y=4 t^3-3 t^2-18 t+3 \Rightarrow \frac{d y}{d t}=12 t^2-6 t-18\end{aligned}$ Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{6\left(2 t^3-t-3\right)}{5\left(t^4+3 t^2+4\right)}$ The curve is decreasing if, $\frac{d y}{d x} < 0 \Rightarrow \frac{6\left(2 t^2-t-3\right)}{5\left(t^4+3 t^2+4\right)} < 0$ $\begin{aligned} & \Rightarrow 2 t^2-t-3 < 0 \\ & \Rightarrow(t+1)(2 t-3) < 0\end{aligned}$

$\Rightarrow \mathrm{t} \in\left(-1, \frac{3}{2}\right)$

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

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