Question

Составьте уравнение касательной и нормали к кривой в точке, соответствующей значению параметра t=t0

Answer 1

$21)\; \; \{\; x=asin^3t\; \; ,\; \; y=acos^3t\; \}\; ,\; \; t_0=\pi /3\\\\x'_{t}=3\, a\, sin^2t\cdot cost\\\\y'_{t}=3\, a\, cos^2t\cdot (-sint)\\\\y'_{x}=\frac{y'_{t}}{x'_{t}}=\frac{-3\, a\, cos^2t\cdot sint}{3\, a\, sin^2t\cdot cost}=-\frac{cost}{sint}=-ctgt\\\\t_0=\pi /3\; \; \to \; \; y'_{x}=-ctg\frac{\pi }{3}=-\frac{\sqrt3}{3}$

$22)\; \; \{\; x=arcsin\frac{t}{\sqrt{1+t^2}}\; \; ,\; \; y=arccos\frac{1}{\sqrt{1+t^2}} \; \}\; ,\; t_0=-1\\\\x'_{t}=\frac{1}{\sqrt{1-\frac{t^2}{1+t^2}}}\cdot \frac{\sqrt{1+t^2}-t\cdot \frac{2t}{2\sqrt{1+t^2}}}{1+t^2}=\frac{\sqrt{1+t^2}}{\sqrt{1+t^2-t^2}}\cdot \frac{1+t^2-t^2}{\sqrt{1+t^2}\cdot (1+t^2)}=\frac{1}{1+t^2}\\\\y'_{t}=-\frac{1}{\sqrt{1-\frac{1}{1+t^2}}}\cdot \frac{-\frac{2t}{2\sqrt{1+t^2}}}{1+t^2}=\frac{\sqrt{1+t^2}}{\sqrt{1+t^2-1}}\cdot \frac{t}{\sqrt{1+t^2}\cdot (1+t^2)}=\frac{t}{t(1+t^2)}=\frac{1}{1+t^2}$

$y'_{x}=\frac{1}{1+t^2}:\frac{1}{1+t^2}=1\\\\y'_{x}(-1)=1\\\\\\23)\; \; \{\; x=\sqrt3\, cost\; \; ,\; \; y=sint\; \}\; ,\; \; t_0=\frac{\pi}{3}\\\\x'_{t}=-\sqrt3\, sint\\\\y'_{t}=cost\\\\y'_{x}=\frac{cost}{-\sqrt3\, sint}=-\frac{ctgt}{\sqrt3}\\\\t_0=\frac{\pi}{3}:\; \; y'_{x}=-\frac{ctg\pi /3}{\sqrt3}=-\frac{\sqrt3/3}{\sqrt3}=-\frac{1}{3}$