Question

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

\(x = {t^2} + 1,y = {t^2} + t\)

Short answer

The given curve is concave upward for \(t < 0\).

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:

\(\begin{array}{l}x = {t^2} + 1\\\frac{{dx}}{{dt}} = 2t\end{array}\)

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

\(\begin{array}{l}y = {t^2} + t\\\frac{{dy}}{{dt}} = 2t + 1\end{array}\)

By using thechain rule, we can get

\(\begin{array}{l}\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\\\frac{{dy}}{{dx}} = \frac{{2t + 1}}{{2t}}\\\frac{{dy}}{{dx}} = 1 + \frac{1}{{2t}}\\\frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{4{t^3}}}\end{array}\)

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:

\(\frac{{ - 1}}{{4{t^3}}} > 0 \Rightarrow t < 0\)

Hence, the given curve is concave upward for \(t < 0\).