Question

Find dy/dx and d² y/dx² . For which values of t is the curve concave upward?

13. x = e^t , y = te^-t

Answer 1

$$ x=e^{t} ; y=t e^{-t} $$

Aim: To find the values of \(t\), for which \(\frac{d^{2} y}{d x^{2}}>0\)

Formulae used:

\(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)

$$ \begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{d}{d x} \cdot \frac{d t}{d t}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{d}{d t} \cdot \frac{d t}{d x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \\ &\Rightarrow \\ &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d}{d t}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\mathrm{dx} / \mathrm{dt}} \end{aligned} $$

$$ \begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{d}{d t}\left(t \mathrm{e}^{-\mathrm{t}}\right) \\ &\frac{\mathrm{dy}}{\mathrm{dt}}=(1-t) \mathrm{e}^{-\mathrm{t}} \\ &\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{e}^{\mathrm{t}} \end{aligned} $$

Hence, Put in (2)

$$ \begin{gathered} \frac{d y}{d x}=\frac{(1-t) e^{-t}}{e^{t}} \\ \frac{d y}{d x}=(1-t) e^{-2 t} \end{gathered} $$

Now use (4), (6) in (3) for evaluating \(\frac{d^{2} y}{d x^{2}}\)

$$ \begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d}{d t}\left((1-\mathrm{t}) \mathrm{e}^{-2 \mathrm{t}}\right)}{e^{t}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{\mathrm{e}^{-2 \mathrm{t}} \frac{d}{d t}(1-\mathrm{t})+(1-\mathrm{t}) \frac{d}{d t}\left(\mathrm{e}^{-2 \mathrm{t}}\right)}{e^{t}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{\mathrm{e}^{-2 \mathrm{t}}(-1)+(1-\mathrm{t})\left(-2 \mathrm{e}^{-2 \mathrm{t}}\right)}{e^{t}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{(2 \mathrm{t}-3)}{e^{3 t}} \end{aligned} $$

For \(\frac{d^{2} y}{d x^{2}}>0\)

$$ \frac{(2 t-3)}{e^{3 t}}>0 $$

Since,

$$ e^{3 t}>0 \text { Always. } $$

Hence,

$$ \begin{aligned} &(2 t-3)>0 \\ &\Rightarrow \end{aligned} $$

When,

$$ \mathrm{t}>\frac{3}{2}, \frac{d^{2} y}{d x^{2}}>0 $$

Hence,

\(t \in\left(\frac{3}{2}, \infty\right)\) for curve to be concave upward.

\(t \in\left(\frac{3}{2}, \infty\right)\) for the curve to be concave upward.