Answer to Question #271920 in Calculus for Angel Nodado

Question

Find dy/dx and d²y/dx²;

x=√t, y=2t+4 ; t=1

Answer 1

$x = \sqrt t \Rightarrow \frac{{dx}}{{dt}} = \frac{1}{{2\sqrt t }}$

$y = 2t + 4 \Rightarrow \frac{{dy}}{{dt}} = 2$

Then

$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{2}{{\frac{1}{{2\sqrt t }}}} = 4\sqrt t \Rightarrow {\left. {\frac{{dy}}{{dx}}} \right|_{t = 1}} = 4$

$\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{4}{{2\sqrt t }} = \frac{2}{{\sqrt t }}$

Then

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}} = \frac{{\frac{2}{{\sqrt t }}}}{{\frac{1}{{2\sqrt t }}}} = \frac{{4\sqrt t }}{{\sqrt t }} = 4 \Rightarrow {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 1}} = 4$

Answer: ${\left. {\frac{{dy}}{{dx}}} \right|_{t = 1}} = 4$ , ${\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 1}} = 4$