Question

Find \(\frac{{dy}}{{dx}}\)and\(\frac{{{d^2}y}}{{d{x^2}}}\). For which values of t is the curve concave upward?

\(x = {e^t},y = t{e^{ - t}}\)

Short answer

The given curve is concave upward for \(t > \frac{3}{2}\,\).

Step-by-step solution

Find \(\frac{{dy}}{{dx}},\frac{{{d^2}y}}{{d{x^2}}}\)

For this question, we can compute from the function of x:

\(\begin{array}{l}x = {e^t}\\\frac{{dx}}{{dt}} = {e^t}\end{array}\)

For this question, we can compute from the function of y:

\(\begin{array}{l}y = t{e^{ - t}}\\\frac{{dy}}{{dt}} = {e^{ - t}}(1 - t)\end{array}\)

By using the chain rule, we can get

\(\begin{array}{l}\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\\\frac{{dy}}{{dx}} = \frac{{{e^{ - t}}(1 - t)}}{{{e^t}}}\\\frac{{dy}}{{dx}} = {e^{ - 2t}}(1 - t)\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{e^{ - 2t}}(2t - 3)}}{{{e^t}}}\\\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 3t}}(2t - 3)\end{array}\)

observe \(\frac{{{d^2}y}}{{d{x^2}}}\)

When \(\frac{{{d^2}y}}{{d{x^2}}} > 0\), the curve of the function is concave upward

So by letting \(\frac{{{d^2}y}}{{d{x^2}}} > 0\) , we will get:

\(\begin{array}{l}{e^{ - 3t}}(2t - 3) > 0\\ \Rightarrow 2t - 3 > 0\,\\ \Rightarrow 2t > 3\,\\ \Rightarrow t > \frac{3}{2}\,\end{array}\)

Hence, the given curve is concave upward for \(t > \frac{3}{2}\,\).