Difference between revisions of "TU Wien:Mathematik 2 UE (diverse)/Übungen WS07/Beispiel 105"
Jump to navigation
Jump to search
(→Links) |
m |
||
| Line 1: | Line 1: | ||
<div {{Angabe}}> | <div {{Angabe}}> | ||
| − | Es sei <amsmath>g_u(u,v)=\frac{\partial}{\partial_u}g(u,v)=u^2-v</amsmath> und <amsmath>g_v(u,v)=\frac{\partial}{\partial_v}g(u,v)=-u+v^3</amsmath>. Man bestimme <amsmath>h(t)=\frac{d}{dt}g(2t,t^2+1)</amsmath>. | + | Es sei <amsmath>g_u(u,v)=\frac{\partial}{\partial_u}g(u,v)=u^2-v</amsmath> und <amsmath>g_v(u,v)=\frac{\partial}{\partial_v}g(u,v)=-u+v^3</amsmath>.<br/> |
| + | Man bestimme <amsmath>h(t)=\frac{d}{dt}g(2t,t^2+1)</amsmath>. | ||
</div> | </div> | ||
| − | == | + | ==Lösung von [[Benutzer:saufnix|saufnix]]== |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
Formel: <amsmath>F'=g_{u}u'+g_{v}v'</amsmath> für <amsmath>F(t)=g(u(t),v(t))</amsmath> | Formel: <amsmath>F'=g_{u}u'+g_{v}v'</amsmath> für <amsmath>F(t)=g(u(t),v(t))</amsmath> | ||
| Line 20: | Line 14: | ||
<amsmath>h(t) = g_{u}(2t,t^{2}+1)\cdot u(t)'+g_{v}(2t,t^{2}+1)\cdot v(t)'=ln(2t\cdot sin(2t)-(t^{2}+1))\cdot2+tan(-2t+(t^{2}+1)^{3}\cdot2t</amsmath> | <amsmath>h(t) = g_{u}(2t,t^{2}+1)\cdot u(t)'+g_{v}(2t,t^{2}+1)\cdot v(t)'=ln(2t\cdot sin(2t)-(t^{2}+1))\cdot2+tan(-2t+(t^{2}+1)^{3}\cdot2t</amsmath> | ||
| + | |||
| + | ==Links== | ||
| + | *[http://www.informatik-forum.at/showthread.php?t=53370 Diskussion im Informatik-Forum (SS07)] | ||
| + | *[http://www.informatik-forum.at/showthread.php?t=43007 Diskussion im Informatik-Forum (SS06)] | ||
Ähnliche Beispiele: | Ähnliche Beispiele: | ||
*[[WS07 Beispiel 106]] | *[[WS07 Beispiel 106]] | ||
Revision as of 02:51, 14 November 2007
Es sei <amsmath>g_u(u,v)=\frac{\partial}{\partial_u}g(u,v)=u^2-v</amsmath> und <amsmath>g_v(u,v)=\frac{\partial}{\partial_v}g(u,v)=-u+v^3</amsmath>.
Man bestimme <amsmath>h(t)=\frac{d}{dt}g(2t,t^2+1)</amsmath>.
Lösung von saufnix
Formel: <amsmath>F'=g_{u}u'+g_{v}v'</amsmath> für <amsmath>F(t)=g(u(t),v(t))</amsmath>
<amsmath>u(t)=2t\Rightarrow u(t)'=2</amsmath>
<amsmath>v(t)=t^{2}+1\Rightarrow v(t)'=2t</amsmath>
<amsmath>h(t) = g_{u}(2t,t^{2}+1)\cdot u(t)'+g_{v}(2t,t^{2}+1)\cdot v(t)'=ln(2t\cdot sin(2t)-(t^{2}+1))\cdot2+tan(-2t+(t^{2}+1)^{3}\cdot2t</amsmath>
Links
Ähnliche Beispiele: