Výsledky cvičení

1.

1. [a)] $x=t,\; y=t^2,\; z=e^t,\; t \in (-\infty, \infty)$,
2. [b)] $x=a \cos t,\; y=a \sin t,\; z=a^2 \cos 2t,\; t \in (-\infty, \infty)$,
3. [c)] $x=t,\; y=t-t^2,\; z=1-t^2,\; t \in (-\infty, \infty)$.

2. a) $10$, $\;$ b) $e^2$, $\;$ c) $\sqrt 3 (e^\pi -1)$, $\;$ d) $3t^2+2t-28$, $\;$ e) $(5 \sin^2 t)/2$. 3. a) $t=ln \frac{s}{\sqrt 3}$ 4.

1. [a)] ${\bf d}(1)=[1+2\lambda , 1+3\lambda ,2], \; {\bf n}(1)=[1-8\lambda , 1+9\lambd... ...hcal K}(1) =\frac{1}{14} \sqrt{\frac{38}{7}}, \; {\mathcal T}(1) =\frac{3}{19}$
2. [b)] ${\bf d}(0)=[1+\lambda , -1,0], \; {\bf n}(0)=[1, \lambda , 0], \; {\bf b}(0)=[... ...) \equiv z=0, \; {\mathcal K}(0) =\frac{1}{2}, \; {\mathcal T}(0) =\frac{3}{2}$
3. [c)] ${\bf d}(0)=[1+\lambda , \lambda ,1+\lambda ], \; {\bf n}(0)=[1-\lambda , \lam... ...2z+1=0, \; {\mathcal K}(0) =\frac{\sqrt 2}{3}, \; {\mathcal T}(0) =\frac{1}{3}$

5. ${\mathcal K}= {\mathcal T}=\frac{4}{(t^2+2)^2}$ 6.

1. [a)] ${\bf d}(0) \equiv y=0, \; {\bf n}(0) \equiv x=0$; $\;$ ${\bf d}(1) \equiv 3x-y-2=0, \; {\bf n}(1) \equiv x+3y-4=0$;
2. [b)] ${\bf d}(0) \equiv y=x, \; {\bf n}(0) \equiv x=-y$; $\;$ ${\bf d}(\pi ) \equiv x+y-\pi =0, \; {\bf n}(\pi ) \equiv x-y-\pi =0$;
3. [c)] ${\bf d}(1) \equiv 2x-y+4=0, \; {\bf n}(1) \equiv x+2y-3=0$;
4. [d)] ${\bf d}(t) \equiv 2x \sin t+2y \cos t-\sin 2t=0, \; {\bf n}(t) \equiv x \cos t-y \sin t-\cos 2t=0$;
5. [e)] ${\bf d}(1/2,1/2) \equiv 4x-2y-1=0, \; {\bf n}(1/2,1/2) \equiv 2x+4y-3=0$;
6. [f)] ${\bf d}(x_0,y_0) \equiv {\displaystyle \frac {x \; x_0}{a^2}-\frac {y \; y_0}{... ...quiv {\displaystyle \frac {(x-x_0) \; a^2}{x_0}+\frac {(y-y_0) \; b^2}{y_0}=0}$;
7. [g)] ${\bf d}(\pi /4) \equiv y-1/2=0, \; {\bf n}(\pi /4) \equiv x-1/2=0$

7. a) $[(4+9x_2)^{3/2}-(4+9x_1)^{3/2}]/27$, $\;$ b) $(t_2^2-t_1^2)/2$, $\;$ c) $15/4+\ln 2$, $\;$ d) $24a$, $\;$ e) $4$. 8. a) $\vert \sin x \vert /(1+\cos ^2 x)^{3/2}$, $\;$ b) $6/[t(4+9t_2)^{3/2}]$, $\;$ c) $ab/(a^2 \sin^2 t+b^2 \cos^2 t)^{3/2}$, $\;$ d) $(2+\varphi ^2)/a(1+\varphi ^2)^{3/2}$, $\;$ e) $0$. 9. a) $x=-4t^3, \; y=3t^2+1/2$, $\;$ b) $x=2t+1/t, \; y=\ln t-t^2-1$.