Question

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

\(x = {t^3} + 1,y = {t^2} - t\)

Short answer

The given curve is concave upward for \(0 < t < 1\).

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:

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

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 the chain rule, we can get

\(\begin{array}{l}\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\\\frac{{dy}}{{dx}} = \frac{{2t - 1}}{{3{t^2}}}\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{2(3{t^2}) - 6t(2t - 1)}}{{{{\left( {3{t^2}} \right)}^2}}}\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{6{t^2} - 12t + 6t}}{{9{t^4}}}\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{6{t^2} - 6t}}{{9{t^4}}}\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{6{t^2} - 6t}}{{9{t^4}}}\left( {\frac{1}{{3{t^2}}}} \right)\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 6t(t - 1)}}{{9{t^4}\left( {3{t^2}} \right)}}\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 2(t - 1)}}{{9{t^5}}}\end{array}\)

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:

\(\begin{array}{l}\frac{{ - 2(t - 1)}}{{9{t^5}}} > 0\\ \Rightarrow 9{t^5} > 0\,\,\,\,\,\,\,and\,\,\,\, - 2(t - 1) > 0\,\,\\ \Rightarrow t > 0\,\,\,\,\,\,\,and\,\,\,\,(t - 1) > 2\\ \Rightarrow t > 0\,\,\,\,\,\,\,and\,\,\,\,t < 1\,\,\,\,\end{array}\)

Hence, the given curve is concave upward for \(0 < t < 1\).